Talk:Monty Hall problem

Jerrywickey says:

++ A far simpler and more intuitive explanation ++

for the solution to the Monty Hall Problem might be needed. Readers struggling with understanding probabilities need an explanation that they can "feel" in their gut. The text below is such and if no one objects or if someone encourages me to do so, I will post it to the page. After all helping readers who consult Wikipedia is and should be contributor's goal.

simple clear intuitive explanation

The chance that the prize door will be chosen from three doors at random is 33% (1/3) But more importantly the choice has a 67% (2/3) chance of being the wrong door. Removing one door does not change that 67% chance that the chosen door is wrong. Even if both other doors were not opened but instead removed, there still remains the same 67% chance that opening the chosen door will reveal it to be empty.

When one other door is opened to reveal it to be empty, new information is added to the system that can be exploited to recalculate the odds that the remaining door is hiding the prize. The chosen door still has a 67% chance of being wrong. That can't change. This implies that the remaining door has only a 33% chance of being wrong while it has a 67% chance not 50% of being the prize door.

The intuitive argument against this is that "removal of one door reassigns the odds for both doors to 50% 50%; not just reassigns the chance for the door not chosen without effecting the odds of the chosen door." The error of this false assumption is easily demonstrated if the number of doors is increased.

The chance that the prize door will be chosen from ten doors at random is 10% (1/10) But more importantly the choice has a 90% (9/10) chance of being the wrong door. Removing eight doors does not change the 90% chance that the chosen door is wrong, but since the removal was selective, removing only empty doors, but not removing the chosen door nor the prize door, it becomes easy to see intuitively that which ever door remains after eight empty doors are removed has a far greater than 50% chance of being the prize door. It doesn't make sense that the prize is just as likely to be behind the chosen door as the last remaining door, because the removal was selective. The original choice was not selective. The remover knew which door held the prize, causing the remover to not remove the prize door while the chooser did not have this knowledge and made his or her choice .

This is because when making the original choice with the information available before any doors were removed, the chosen door was very unlikely to be the correct choice, 90% chance of being wrong. That chance isn't changed by the removal of eight other doors. However, if it is known that all eight removed doors were empty, then the chance that the remaining, unchosen door, is the prize door is very high. Much higher than 50%. It has a 90% chance of holding the prize because the chance of the chosen being wrong is 90%.

Jerrywickey (talk) 18:54, 25 May 2012 (UTC)

Jerrywickey, I was a 50% 50% believer, read the whole article and still was. The wording of this comment convinced me. If it has made it to the main page since may its lost in the mix. I would suggest adding it, or raising it to a more prominent location. Zath42 (talk) 14:27, 23 July 2012 (UTC)

Jerrywickey, please start a new sections after earlier discussions. It is a fact that some editors follow Morgan et al. in claiming that the chance of the door first selected by the guest could be changed by the special behavior of the host in opening a losing door. If he should be extremely biased e.g. to open his preferred door if ever possible, then he can do that in 2/3, but if in 1/3 his preferred door hides the car he then would be forced to open his strictly avoided door, showing that the chance by switching to his preferred but still closed door is max. 1 and the chance of the door first selected by the guest could converge to zero. So there is some desire to first of all show by Bayes' formula that "which one" of his two doors the host has actually opened could be of influence on the probability to win by switching. Please read also the archive of this talk page. --Gerhardvalentin (talk) 19:49, 25 May 2012 (UTC)

Jerrywickey, your explanation still does nothing for me. I can't '"feel" it in my gut' at all. You remove the wrong doors... and leave 2, it's therefore down to 50/50. But I'm a linguist, not a mathematician. What do I know? Oh yeah... I'm meant to "feel it in my gut". Malick78 (talk) 22:57, 26 May 2012 (UTC)

Perhaps I can offer some insight into this. For some time I have been incorporating this into some training (on an unrelated topic) that I have been giving to scientists and engineers. Here is what I do:

First I ask anyone who has heard of this problem before to silently watch what I am about to do.

Then I hand out the (fully unambiguous, mathematically explicit version of the standard problem) Krauss and Wang description from this page, in writing, and ask everyone to read it and put their answers on paper (unsigned) and hand it back. I count the answers and write that on a whiteboard. Usually, "no advantage to switching" is way ahead.

Then I open it up for discussion. There is always a spirited debate with much certainty on both sides. I have never seen anyone, ever, change their position based upon hearing arguments from the other side. Ever.

Then I prove who is right using the cups simulation described on the page, moving to ten cups if needed. In my experience, this is the best way to convince engineers. (I use toy cars and toy goats - had to buy ten packages of toy barnyard animals at the 99 cent store to get the goats.) I have also found that playing with me as Monty and someone from the audience as the contestant and having the contestant never switch if he thinks there is no advantage to switching works best. I have never had anyone remain unconvinced that they should switch after choosing cups and keeping score.

My point in the training is that actual data trumps logical argument, no matter how sure you are that you are right. but here is an interesting thing I have observed: a significant number of those who got it wrong and argued vigorously that they were right blame the problem description. and it doesn't matter whether I presented the Vos Savant version or the Krauss and Wang version! Just as something about the human mind makes engineers get the wrong answer and defend it to the end, something about the human mind makes engineers reject the notion that they were wrong and blame the problem description. We tend to "feel" things that are not true. --Guy Macon (talk) 02:48, 27 May 2012 (UTC)

Excellent work Guy, perhaps you should try to get it published somewhere so that we can report it here. It is also a sad fact that not one person has changed their mind on the disputed issues here. Martin Hogbin (talk) 13:24, 27 May 2012 (UTC)

Points well taken.

However, removing only empty doors is an assumption of the riddle. Any examination which explores other possibilities is not an exploration of the Monty Hall Problem. If the prize could be removed, the Mony Hall game makes no sense. It is this selectivity on which the solution must be based. If you didn't "feel" the one before then try this.

When making the original choice one has only a one third chance of choosing the winning door. What is more important to understanding the problem, however, is that also means that the choice has a two thirds chance of being wrong. No later event, removing one door included, changes those odds. After the removal of one door, or any other event aside from exposing the winning door, the chance the chosen door is wrong remains two thirds. Nothing can change that.

With the removal of one door only one other door remains. Since the winning door must be one of the two as an assumption of the riddle, then the chance that the chosen door is wrong is still two thirds, which implies that the chance that the single remaining door could be wrong is only one third. Meaning that the remaining, unchosen door, has a two thirds chance, not fifty fifty, of being the winning door.

Some might argue that "removing one door can not change the odds for one of the remaining doors, but not the other. Just designating one door as chosen doesn't give it preferential treatment." This is an erroneous assumption. The error becomes intuitively obvious if more than three doors are used.

If one were choosing from a hundred doors instead of just three, then the chance that the chosen door is wrong is 99 out of 100. Removing 98 doors does not change that probability. It does not give the chooser any more confidence that his chosen door is correct, but he does intuitively realize that the door he chose still has a ninety nine percent chance of being wrong. He also intuitively realizes that since only wrong doors were removed the one single door that remains has a much greater chance, much greater than 50% 50%, of being the winning door. How much? 99/100 Why? Because the one he chose had and still has a 99/100 chance of being wrong. After all, it was selected from a hundred choices. The selectivity of the removal of 98 wrong doors changes the odds for the one remaining door, but not for the chosen door. — Preceding unsigned comment added by Jerrywickey (talk • contribs) 13:59, 27 May 2012 (UTC)

Curiously, you are both wrong. Jerywickey, you say, 'When making the original choice one has only a one third chance of choosing the winning door. ... No later event, removing one door included, changes those odds'. That is not necessarily true, although under the standard assumptions made about the problem it is true that the odds do not change.

To take a really obvious example first, suppose that Monty tells you that the car is behind door number 2. The odds change then for sure.

Now consider a more interesting and instructive case. Suppose that Monty does not know where the car is and opens one of the two doors that you have not chosen at random and it happens to reveal goat. What are the odds then that the car is behind the door that you originally chose?

The important point to consider is whether any event that occurs after you have chosen your door but before you decide whether to swap or not gives you any information about the whereabouts of the car. Under the standard assumptions you know the Monty will reveal a goat, because he must do under the rules, you also gain no information from his choice of door when you happen to have originally chosen the door hiding the car because the host must choose randomly between the two doors available to him under the standard assumptions in that case. So, the host opening, say door 3 to reveal a goat tells you nothing you do not already know, thus your original odds of having chosen the car cannot change. In the standard version of the problem you have a 2/3 chance of winning if you swap. Martin Hogbin (talk) 16:02, 27 May 2012 (UTC)

Thank you Martin, you are absolutely correct, and you clearly articulate the dilemma of the article: -- Not sure who left this comment

So, what difference does it make if you get a new choice each time a door is opened? Sure, the odds are 1-in-3 when you have three doors. But when you get a new choice from 2 doors, the odds are even. So why do people even talk about the door that isn't part of the problem anymore? I am faced with a new choice that is only about two doors. Why so much ink spilled for a simple choice? -- Avanu (talk) 14:15, 9 September 2012 (UTC)

Well Jerrywickey, I posted a "simpler and more intuitive" solution to this problem which (IMO) was elegant in its simplicity but is now gone. I agree with your basic sentiment that a simpler and more intuitive explanation is indicated, but if contributions stemming from original thinking on ways to explain this problem are deleted outright, then clearly Wikipedia is not the place for them. The fundamental question is whether original solutions or explanations are welcome in this article, or if the article is simply a compilation of solutions proposed by others. I have posted my explanation elsewhere on the web where you're not likely to find it. In the meantime, folks can spin their wheels with overly-complex mathematical explanations which aren't clear to the lay person. Chris319 (talk) 01:53, 16 October 2012 (UTC)

Conditional or Simple solutions for the Monty Hall problem?

Should the Monty Hall problem page be edited according to the 'Simple' or the 'Conditional' solutions? --Guy Macon (talk) 21:18, 6 September 2012 (UTC)

The aim of this RfC is to resolve a longstanding and ongoing conflict involving multiple editors concerning the relative importance and prominence within the Monty Hall problem article of the 'simple' and the more complex 'conditional' solutions to the problem.

The 'simple' solutions do not consider which specific door the host opens to reveal a goat (see examples here and here). The 'conditional' solutions use conditional probability to solve the problem in the case that the host has opened a specific door to reveal a goat (see example here).

One group of editors considers that the 'simple' solutions are perfectly correct and easier to understand and that the more complex, 'conditional' solutions are an unimportant academic extension to the problem.

The other group believes that the 'simple' solutions are essentially incomplete or do not answer the question as posed and that the 'conditional' solutions are necessary to solve the problem. Both sides claim sources support their views.

That argument is unlikely to ever be resolved but two proposals have been made to resolve the dispute. Both proposals aim to give equal prominence and weight to the two types of solution.

One of the points of contention is whether either of the proposals below violates any Wikipedia policies and guidelines (in particular WP:NPOV, WP:NOR, WP:V, WP:WEIGHT, WP:EP, MOS:JARGON, WP:MOSINTRO, WP:MTAA and WP:OPINION). See the individual editor's comments below for arguments on both sides of this issue.

Proposal 1 is for the initial sections including 'Solution' and 'Aids to understanding' to be based exclusively on 'simple' solutions (with no disclaimers that they do not solve the right problem or are incomplete) then to follow that, for those interested, with a section at the same heading level giving a full and scholarly exposition of the 'conditional' solutions.

Proposal 2 is for the article to include in the initial 'Solution' section both one or more 'simple' solutions and an approachable 'conditional' solution (showing the conditional probability the car is behind Door 2 given the player picks Door 1 and the host opens Door 3 is 2/3) with neither presented as "more correct" than the other, and to include in some later section of the article a discussion of the criticism of the 'simple' solutions.

Proposal 1, Proposal 2, or Neither? (or Abstain.)

Considering all Wikipedia policies and guidelines, should the Monty Hall problem page be edited according to Proposal 1, Proposal 2, or neither? Abstain is also an acceptable answer.

Note: Because prior attempts to resolve this conflict have resulted in long discussions with many endless back and forth comments, please place any responses to other editor's comments in your own "Comments from user X" section and limit your comments to no more than 500 words. If you wish to have a threaded discussion, feel free to start a new section on this talk page but outside of this RfC or on your own talk page.

---

Comments from Elen of the Roads

(327 of 500 words)

Note: I am assigning Neither to the comments below. If this is incorrect, please indicate "Proposal #1", "Proposal #2", or "Abstain". --Guy Macon (talk) 20:16, 20 September 2012 (UTC)

Neither is my starting position on this - I believe the dichotomy is artificial and the structure of the article needs looking at in a different way. Being an Arb (lol) I would say Proposal #1 is a second choice, with Proposal #3 being an oppose.--Elen of the Roads (talk) 14:04, 21 September 2012 (UTC)

Commenting so there's no danger of ever being asked to close this, or PD a rerun Arbcom !! Seriously, the Monty Hall problem was originally presented as a parlour teaser for folks with a bit of mathematical knowledge who were familiar with game shows like Opportunity Knocks. The question was - should the contestant stick with his door or switch, and the utterly non-intuitive answer is that he should always switch, regardless of whether Monty knows where the goat is, whether Monty prefers the red door, or what Monty had for breakfast. The article has to present the puzzle and the answer in this light first. The conditional mathematicians contentions that the answer has the potential to be different if Monty has had his Weetabix must come separated, later in the article, for those who have an interest in such things. --Elen of the Roads (talk) 23:02, 6 September 2012 (UTC)

@Guy Macon - I've always liked the ten goats explanation. Surely the actual, unspoken condition is that Monty always knows where the goats are. Everyone takes this as read, because they extrapolate from how real game shows work, and if he opened the door on the car, the show will end ten minutes early. All of the other potential conditions - whether Monty opens the door on the left first, or prefers the blue door, are not part of the original problem, and do not affect the solution to the original problem. --Elen of the Roads (talk) 15:29, 7 September 2012 (UTC)

See section below Talk:Monty_Hall_problem#Does_this_version_work. Uses the approach taken in Zebra Puzzle, ie that it is the statement of the puzzle that is faulty, not the solution. Does this remove the divide? It is not true (for any given value of true) that "all the simple solutions are wrong" but it is true that Vos Savant's statement of the puzzle is incomplete, and anything claiming to be a solution to that version of the puzzle is incomplete if it does not highlight the missing assumption. --Elen of the Roads (talk) 14:18, 15 September 2012 (UTC)

Comments from Glrx

(63 of 500 words)

Neither. I'll second Elen's comments. Unfortunately, this article has taken a mathematician's viewpoint much too seriously. For WP, it is an interesting puzzle and not a graduate level exercise in probability/statistics. The simple stuff must be first and the academic views much later. There should not be an overly precise description of the problem at the beginning. I'll explain the neither vote later. Glrx (talk) 23:40, 6 September 2012 (UTC)

Comments from Ebe123

(6 of 500 words)

Neither. I echo Elen on this. ~~Ebe123~~ → report 00:50, 7 September 2012 (UTC)

Comments from czarkoff

(197 of 500 words)

Proposal 1. Though I share the view by Elen of the Roads that the first things to present readers are the problem and the counter-intuitive solution, this is exactly the purpose of the lead section. I reject the notion that the math is taken too seriously in this article – it is the subject, not a side discussion, so it should be presented and explained in the article's body. By analogy with WP:WEIGHT I would present the material to the reader in order of complexity increase, so that the solutions involving less logical operations or iterations are placed on top of more complex material.

That said, I would suggest to move the current lead section (I generally prefer the word "lede" over the phrase "lead section", but in this article it is indeed a magazine's lede, not a summary one would expect) into a first named section, which I would call "Background"; in the lead section I would overview the problem and state the solution in the first paragraph. In the second paragraph I would provide the brief descriptions of the solutions, as general as possible. The proposal 1 should be implemented in sections 2 and onward. — Dmitrij D. Czarkoff (talk•track) 01:48, 7 September 2012 (UTC)

Comments from Noleander

(187 of 500 words)

Neither. Splitting the solutions into Simple and non-Simple seems like it would just perpetuate the underlying conflict for several more years. Is it not possible to simply enumerate the several different approaches to the problem? I suggest:

• Just list the approaches, one after another. The approaches could be grouped based on similarity, but no need to name them as "Simple" or not.
• There is no reason to treat some approaches (e.g. the so-called simple ones) as "better" than others, more deserving of being in the lead. The lead should summarize all the approaches (or at least all classes of approches).
• Consider using the word "approach" instead of "solution", to indicate that some solutions may not be considered to be full or accurate.
• Emphasize (earlier) that the analysis depends heavily on the host's behavior (e.g. whether the host deliberately avoids the car when opening doors, vs. opening random doors) ... I know that is in the article body already, but it should be emphasized in the lead.

Bottom line: don't divide the approaches into two categories that could perpetuate the civil war. (Maybe this is Option 2?). --Noleander (talk) 02:28, 7 September 2012 (UTC)

Comments from ChrisGualtieri

(152 of 500 words)

Neither - Clearly the problem and show's ORIGINAL problem diverge into a thought experiment along the way. First, go with the simple answer in no uncertain terms, with the most prevalent rules without behavior such as 'evil or angelic' Monty. The problem as classically postulated. Then get into the counters and thought process behind that. Once that is done, the 'complex' matter can be dealt with separately assuming X or X case, but by all means, each of these have to be explicit in their terms about the ideology (and not math) which goes into it. I'll end up doing it myself if this can't get resolved. Either way... this conflict is going to end shortly. ChrisGualtieri (talk) 02:31, 7 September 2012 (UTC)

Additional note: Crunching the data seems to support that. If things get really long winded we can probably collapse complicated sections. Seems to be quite a lot of material after all. ChrisGualtieri (talk) 03:05, 7 September 2012 (UTC)

Comments from Rick Block

(468 of 500 words)

Proposal 2 This is a POV dispute, plain and simple.

POV 1) "Simple" solutions are the "right" way to solve the problem.

POV 2) The "right" solution is to compute the conditional probabilities the car is behind Door 1 and Door 2, given the player has selected Door 1 and the host has opened Door 3.

Proposal 1's "compromise" is to give "simple" solutions far greater WP:WEIGHT, and structurally endorse POV 1. Proposal 2 gives equal WEIGHT and endorses neither POV, remaining strictly NPOV.

Regarding WEIGHT: many, many sources present "simple" solutions. But the vast majority of these are popular, not academic, sources and many of them uncritically parrot vos Savant's ("simple") solution. Within the field of probability, the textbook solution (literally and figuratively, meaning both appears in numerous textbooks and is completely standard) is to compute the conditional probabilities. This solution, presented by the preponderance of sources in the most relevant academic field, should have at least equal WEIGHT to any other.

Regarding STRUCTURE: there is a controversial, but by no means fringe, POV expressed by numerous sources, e.g. [1] [2] [3] [4] [5] [6], criticizing "simple" solutions. To be NPOV the article must not endorse this POV (as it arguably did at one point, which is perhaps the actual source of much of the conflict). But it equally must not endorse the opposite view that "simple" solutions are universally accepted as "perfectly correct". Proposal 1 does exactly this, presenting "simple" solutions as "the solution" ("with no disclaimers that they do not solve the right problem or are incomplete") and relegating "conditional" solutions to a later section "for those interested". This creates a strong structural POV suggesting the "simple" solutions are true and undisputed, which (hardly coincidentally) exactly matches the POV of certain editors involved in this conflict.

Should a player who picks Door 1 and then sees the host open Door 3 switch to Door 2?

Furthermore, the claim "simple" solutions are easier to understand is at best dubious. Krauss and Wang say 97% of their test subjects drew an image of the (conditional!) situation where the player picked Door 1 and the host opened Door 3 (like the image to the right), and that once formed this image "prevents the problem solver from gaining access to the intuitive [simple] solution". We know vos Savant's solution was not convincing (she received thousands of letters after publishing it). As Eisenhauer says "what could and should have been a correct and enlightening answer to the problem was made unconvincing and misleading."

The resolution here is simple. Follow Wikipedia's core content policy of NPOV. Include BOTH "simple" and approachable "conditional" solutions in an initial "Solution" section, presenting both as equally valid. Discuss the differences between these types of solutions in a later section "for those interested". I.e. Proposal 2. -- Rick Block (talk) 04:30, 7 September 2012 (UTC)

Comments from Trovatore

(209 of 500 words)

Note: I am assigning Abstain to the comments below. If this is incorrect, please indicate "Proposal #1", "Proposal #2", or "Neither". --Guy Macon (talk) 20:16, 20 September 2012 (UTC)

The problem with Elen's take is that there is no such thing as an answer independent of Monty. There are ("simple") answers that are based on certain assumptions of neutrality regarding Monty's strategy, assumptions that are arguably not unnatural, but are certainly not the way Monty in fact behaved.

This is not "the mathematician's view"; this is simply the fact.

That said, I think a mixture of Proposal 1 and Proposal 2 is probably appropriate. The bulk of the coverage has indeed been on the "simple" answers, with all their unexamined assumptions, and it should probably come first. However, to leave the qualification out of the lead paragraph is not reasonable. To fail to point out the hidden (and certainly inaccurate, compared to the real-life model) assumptions, is to be just plain incorrect. --Trovatore (talk) 09:32, 7 September 2012 (UTC)

Hope this is OK since I have space left. I'd like to respond to Guy Macon's point below about simulation. There is no mystery about what a given simulation will show. The question is, how do you make your simulated Monty behave? If for example you make him always show an unchosen empty door, then yes, you will get the "simple" answer. But the statement of the problem gives no warrant at all for that choice. --Trovatore (talk) 17:37, 7 September 2012 (UTC)

Comments from Guy Macon

(420 of 500 words)

Neither. The article should start with a description of what the MHP is (using the Krauss and Wang version) and the "better to switch" question without (in the lead) saying what the answer is or why. Just present the problem in the lead and let the reader decide what the answer is.

Next should be a history section, still not saying what the answer is but instead documenting what various letter writers to the American Statistician wrote in 1975, Marilyn vos Savant's answer, the answer of the 1,000 PhDs who disagreed with her, etc. The key here is not to say what the answer is, but rather to document (with citations) what various notable people say it is.

Next should be a section on simulation starting with clear instructions on how to do the simulation yourself using plastic cups (better than playing cards; goats are indistinguishable). This has a huge advantage over the other "solutions" and "aids to understanding", which often end up with two people talking past each other, each convinced that they understand the problem and have the solution and that the other person isn't thinking clearly. Again, we shouldn't give the answer but rather should report (with citations) on the result of people running simulations. One big advantage of a physical simulation (or a computer simulation that you wrote yourself) is that Monty's behavior must be defined for it to work, and alternative Monty behaviors can be easily tested.

Finally, after the Lead, History and Simulation sections, the other approaches should have a section or sections, with simple first followed by conditional. Here, for the first time, we should reveal what Wikipedia says the answer is, along with answers to different but related questions.

I have gone through physical simulations many times (mostly with engineers and scientists, and with ten cups if they assert that there is no advantage to switching even if there are ten cups), and have found it to be a reliable way to convince even the most stubborn that switching is better, and to do so in a way that no amount of verbal argument can accomplish. Furthermore, watching "Monty" select which eight cups to turn over leads to understanding -- clearly "Monty" is choosing cups with goats and avoiding the car. This also makes it easy to understand variations like "random choice Monty", where sometimes he opens the door on the car and the player loses immediately without being given a choice to switch. --Guy Macon (talk) 19:04, 8 September 2012 (UTC)

Comments from Martin Hogbin

(381 of 500 words)

I am the proposer of Proposal 1.

Firstly let me say that I completely agree with Elen of the roads that, 'Monty Hall problem was originally presented as a parlour teaser for folks with a bit of mathematical knowledge who were familiar with game shows...', and I and many other editors have long argued that the article should concentrate on this aspect. However there are some who believe that the more esoteric mathematical aspects are important and who have pressed for what they see as the only 'correct' solutions' to dominate this article. That battle has raged for years with no sign of resolution and my proposal is a compromise which should be acceptable to all.

My proposal makes no claim that any one solution is the 'correct' one and I am suggesting that we give both types of solution equal weight but:

First, we treat the problem as a simple mathematical puzzle and give the 'simple' solutions to this puzzle. This is what 90% of our readers want. We need to concentrate on making the solutions convincing and the explanations understandable. There are two things that fool most people: that the answer is 2/3 and not 1/2, and that it matters that the host knows where the car is. Once we have fully resolved these two issues most readers will be satisfied.

After we have done that, for the remaining 10% of readers and for mathematical completeness we should mention that some sources say that the simple solutions are defective in some way. We should then have a full and scholarly discussion of that claim, based on reliable sources.

• Glrx says 'The simple stuff must be first and the academic views much later'. Yes! That is exactly what I am proposing.

• Final attempt to explain my compromise proposal.

Like most others, I think the 'simple' solutions are the best and that they should be given more weight in the article, with the 'conditional' solutions being treated as an academic sideline. However, in the interests of peace and cooperation, I am suggesting that we give both equal prominence in the article but that we start the simple (even if there is some loss of rigour) and then explain the details later, just as most good text books and encyclopedia articles do. — Preceding unsigned comment added by Martin Hogbin (talk • contribs) 14:30, 12 September 2012 (UTC)

Comments from Richard Gill

(491 of 500 words.)

Neither. The important distinction is between informal, heuristic, verbal solutions which anyone can understand, and formal solutions appropriate within some academic context. If there are simple verbal equivalents of solutions based on conditional probability then they can appear early in the article alongside other simple solutions.

Marylin vos Savant's question was "should you switch?", not "what is the (this, that or another) probability?" What most readers need is to gain the intuition through being shown how to see the problem from another angle, that switching offers a huge advantage. MHP is carefully formulated so as to throw the newcomer off track. It's a trick question. It's fun. A classic brainteaser. Conditional probability is justs one of many ways to solve the problem, after just one of many possible formalizations of it has been chosen.

Another important distiction for me is that between an argument being correct and incorrect. I see no need at all that the article should reproduce arguments which are logically wrong. With careful wording, we can stick to the Truth. Richard Gill (talk) 13:44, 27 September 2012 (UTC)

Comments from Gerhard Valentin

(486 of 500 words)

Proposal 1. For the reader, it's easy to follow that  "any group"  of two doors has double chance to hide the prize, compared to one single door. The tricky story tells a brilliant honest puzzle. It is not so much important what is *the* original question and its "possible" ambiguity, but it is on the certainly *intended* paradox of a *confidential* host who observes *secrecy* regarding the car-hiding door in that one-time problem. — It is on helping people, *by modern reliable sources*, to decode and understand the puzzle.

The charm of the world famous paradox: Two still closed doors, one door hiding the prize for sure, the other door hiding a nullity for sure,
but – in one special given perspicuous scenario as per Krauss and Wang, Henze and many others – the chances are not 1/2:1/2,
but the chance of the door first selected by the guest is and remains exactly 1/3 in each and every game,
while the chance of the still closed door offered by the host is exactly 2/3 in that plausible and honest scenario.

The paradox can be decoded by careful evaluation, by game theory, principle of dominance and other approaches as per the sources, that all say that “staying” forever will diminish your probability to win the prize.

The article first should show the common assumptions: All doors are initially equally likely to hide the car in this one-time problem.
The door selected by the player has 1/3 chance, and the group of unselected doors, although there is at least one goat behind, has 2/3 chance to hide the car. And Monty, observing secrecy regarding the car-hiding door, is equally likely to open either of his two doors if he should have a choice,  in intentionally having to reveal a goat.  By this *symmetry* it is unimportant which one of his doors the host has opened, specific door numbers is irrelevant. The player's initial choice still hides the car with probability 1/3, and the other door left closed by the host therefore has probability 2/3 to hide the car in this one-time problem.

• This common and well known honest scenario (*secrecy*) should be shown first, together with a Bayes' theorem in the "posterior odds = prior odds times likelihood" version.

And then in contrast, as an  eye-opener,  immediately followed by the inconsistent *strange scenario* of a forgetful host who is showing the car in 1/3 by just randomly opening one of his two doors, deleting the probability to win by switching in that 1/3, while in the remaining 2/3  ("goat:car" and "car:goat"),  the chance of both still closed doors is 1/2:1/2 then, as per the most intuitive common appraisal.

Conditional probability theory is not "needed" to decode the paradox. Show in later sections that the MHP is often used in textbooks and in teaching conditional probability theory, with adventuresome presumptions. What matters most is a clear structure of the article. Gerhardvalentin (talk) 14:26, 21 September 2012 (UTC)

Comments from Tom Hulse

(484 of 500 words)

Proposal 2
I'm sorry Guy Macon, but I really think you should scrap this whole Rfc and start again. You've overestimated our group ability to understand what a "conditional solution" is. :) You have two groups arguing past each other on two very different questions. Comments from Elen of the Roads highlight the misunderstanding. I don't mean to pick on you Ellen, but several are following you when you believe that a "conditional solution" involves one of the tortured conditions to make the traditional "yes-you-should-switch" answer wrong. No. Instead, a conditional solution still arrives at the same yes-switch answer, but it is reached differently, with conditional probability. It says Marilyn's answer was right but her math explanation to get there was wrong. So right now all comments from Elen, Grix, Ebe123, czarkoff, ChrisGualtieri, Trovatore, Richard Gill, & Gerhard Valentin should all be disregarded as irrelvant since they don't even understand the basic premise of this Rfc. A conditional solution is using conditional probability to get the same final answer, not changing the conditions to get a different answer.
Once you understand that, read Rick Block's comments for an overview of Proposal 2, and Martin Hogbin's for a review of proposal 1. I switched my view to Proposal 2 actually when reading Martin's idea that we could admit later in the article that the simple solutions may be wrong in their math (though not in their final answer). Wrong is wrong, and the more reliable academic sources have enough weight to at least give them an equal footing, even with the massive numbers of internet parrots that just blindly repeat Marilyn's simple solution.
I do agree with Guy and the others above that the article could also benefit from a little more emphasis on simulation--Tom Hulse (talk) 07:44, 8 September 2012 (UTC)

I'll add Lambian to the long list of users who's comments are 100% irrelevant here and should not be considered, since he, like the others I mentioned, doesn't understand what this Rfc is about. He also thinks Position 2 is about about variations in the conditions of the problem (it's not at all). Boris Tsirelson misunderstands too, commenting about 'starting with the ideal case'. Both simple & conditional solutions start with the IDENTICAL PROBLEM. There is no variations in the conditions. Look more carefully. Ya'll are alsleep, lol! Condtional probability, not conditions in the problem. --Tom Hulse (talk) 07:55, 9 September 2012 (UTC)
Lambian, my opinion on Marilyn's correctness is irrelvant, as is yours. Only sources matter here, and many of the highest quality ones disagree with you, see Rick Block's post with many quality examples. He only asks for parity in discussion because of their quality weight. Regardless, you still don't understand the very premise of this whole Rfc! You refer twice to "variations" in the problem. No, that is not what this Rfc (or Position 2) is about. Please try harder to look closer at what I've said. --Tom Hulse (talk) 09:48, 9 September 2012 (UTC)

Comments from Salix alba

(218 of 500)

Proposal 1 My inclination is to stick with the status quo, this topic has been discussed to death. However I do fid the condition probability section has been poorly written and is confusing. Effort should go into fixing that. Condition probability can be explain better than it is.--Salix (talk): 17:59, 8 September 2012 (UTC)

For all its flaws the Vos Savant explanation gives the gist of the problem and a simple means when the probabilities turn out to be. Before trying for a major overhall work need to be done on expanding the Decision tree section so that readers with only basic statistical background can undersand it. Its not made clear that the tree diagram really has conditional probabilites in it with each branch of the tree representing P(car at location 1 AND host opens door 2) = P(car at location) * P(host opens door 2 | car at location). The formal solution section is just a mass of symbols making it look more complicated than it really is. In has the effect of making the whole section incomprehensible to anyone who is not familier with Bayes theorem. Fix this before going on about the order of presentation.--Salix (talk): 16:19, 15 September 2012 (UTC)

Oh the latest xkcd seems to capture some of the effect of these discussions.--Salix (talk): 16:24, 15 September 2012 (UTC)

Comments from 203.171.197.35

(48 of 500 words)

Proposal 2
Some readers may want just an overview of the problem, but most readers will want a rigorous solution. The so-called "simple solutions" are unconvincing and, at present, the only mathematically rigorous solution in the article is the one based on conditional probability; no way should we downplay that. -- 203.171.197.35 (talk) 01:50, 9 September 2012 (UTC)

Comments from Boris Tsirelson

(73 of 500 words)

I replace my original comments; hope this does not violate the rules of RfC.

Abstain.

The first solution in the article should be the most simple (that is, accessible for the widest possible audience) among all available correct solutions. (Fortunately simple and correct solutions exist!)

And I do not care, whether or not (and to what extent, and in which sense) this solution may be called "conditional", or "unconditional", or both, or neither.

Boris Tsirelson (talk) 15:48, 13 September 2012 (UTC)

Comments from Lambiam

(302 of 500 words)

Proposal 1. In the terminology of User:Martin Hogbin/Monty Hall History, I'm a card-carrying simplist. Let me repeat what I proposed on this talk page on 2 March 2008:

What about this two-part approach:

1. Present the simple solution to the "main version" of the problem, with the least amount of fussing over the problem statement we can get away with to make it into a solution. Also present some other analysis methods of the problem, leading to the same solution. Following these presentations of the solution, briefly discuss unintended interpretations of the problem statement, and the difficulty in ruling them out unambiguously, while referring to a later section on variations.
2. Have (like now) a section with variations on the problem, such as those in which the player may know more than in the main version, depending on the host's behaviour.

The terminology "unconditional solution" is unfortunate. The notion of a "conditional" solution has no meaning for the main problem version. It is only in variations of the main version that conditional issues arise.  --Lambiam 08:28, 2 March 2008 (UTC)

I'd like to add that I gave an "unconditional" mathematically solid simple proof on Richard Gill's talk page of a proposition he has dubbed "the Holy Grail of MHP studies", namely that the 2/3 chance of success on switching can't be beaten by any strategy, even if the player's initial choice and the host's choice are not necessarily random. This shows conclusively that the conditionalism is an unnecessary complication.  --Lambiam 07:38, 9 September 2012 (UTC)

Tom Hulse wrote: "Marilyn's answer was right but her math explanation to get there was wrong". Well, no, her answer was right and so was her explanation. The simple explanation is perfectly adequate, and we should not torture and confuse the reader with unnecessarily complications, or at least postpone their introduction as long as possible.  --Lambiam 09:08, 9 September 2012 (UTC)

Comments from Coffee2theorems

(500 of 500 words)

Proposal 2. I think an approachable conditional solution is doable, if care is taken not to dot every i and cross every t in sight, and maybe starting with a very short introduction to conditional probability and explanation of why it's relevant here (in the vein: if you want to guess a man's height, use statistics for men, not for all people). More sophisticated and rigorous discussion coming in a later section would be in line with WP:UPFRONT.

The problem with the unconditional solutions is that they take considerable liberties with the problem. There are people who are just sophisticated enough to sense the holes in the argument and so fail to be convinced, yet are not sophisticated enough to see how the holes can be mended and the solutions salvaged by the symmetry proof. (concrete example: me when I was in high school)

Those readers need the conditional solutions, and they need to be mentioned before some "aids to understanding section", or the readers will likely go away thinking that the article has no real solutions to offer. Incidentally, for all its simplicity, the symmetry proof is actually more sophisticated than the conditional one. It's the kind of shortcut you can take in a recipe once you know how to cook, not when you're struggling with the basic recipe ("what does it even mean that probabilities can change?!"). -- Coffee2theorems (talk) 10:57, 9 September 2012 (UTC)

Boris: The "conditional proof" is a very straightforward application of the definition of conditional probability, so it can be understood with absolutely minimal prerequisites of probability theory, and the correspondence of the decision tree with the original problem is clear and explicit. It solves the "standard MHP" by directly computing the conditional probability which decision theory says you should compute, brooking no argument and being economical in the amount theory the reader needs to understand. It's also a very standard approach to such problems.

The symmetry argument gives you another way of obtaining the conditional probability, by showing that it is equal to the unconditional probability here. It requires all the same concepts as the proof by direct computation, and more besides. The symmetry proof sections in the article use statistical independence at least. They are also even less clear than the conditional proof sections of the article, and I think the reasoning is fundamentally more sophisticated, and thereby fundamentally less accessible.

The symmetry argument certainly should remain in the article. All I'm saying is that the conditional solution should be presented along with the unconditional solution, not much later in the article. The other proposal where no mention at all of conditional probabilities is made in the main "Solution" section would amount to presenting a lie to children as a full solution, which is clearly undesirable. Most of the "unconditionalists" don't want an "unconditional solution + symmetry argument", they want just the unconditional solution, with any mention of conditional probabilities ("unimportant academic extension") omitted, or as a "compromise" at least hidden as much out of sight as at all possible. -- Coffee2theorems (talk) 03:33, 10 September 2012 (UTC)

Comments from User:Dicklyon

(79 of 500 words)

Proposal 1 – Rick Block's description of the 2 POVs is accurate enough, but his long-time insistence to putting so much weight on the minority POV 2 is what has kept this problem from being resolved. Readers would be better served by presenting the simple POV 1 solution first, without complications. Then, a later section can pay homage to those mathematicians who like to formalize the solution more carefully by appeal to conditional probabilities using conditions that don't matter. Dicklyon (talk) 18:50, 9 September 2012 (UTC)

And of course, as others have noted, every time you invite comment on this article you'll be swamped by the noise by people who don't understand the basics of it and so can't even get close to the question at hand. Dicklyon (talk) 04:44, 11 September 2012 (UTC)

Comments from Nijdam

(331 of 500 words)

Proposal 2, being, as Rick Block explained, in compliance with Wikipedia policy, as both the simple "solutions" and the conditional solution are widely sourced. I find it however very difficult, to accept the sources of the simple "solutions", being evidently mistaken, as reliable sources. I furthermore agree with Rick Block and Tom Hulse in their analysis of the situation. The latter also clearly shows where the problem lies. I would like to add to Proposal 2, at the point where the "conditional solution" is introduced, also the (well sourced) reason (need) for this solution. At this point the simple (vos Savant) solution, may be called an acceptable intuitive explanation.

To Richard Gill I would say: Don't be naive, Vos Savant indeed asked: "Would you switch?" and the direct answer to this question would be: a yes or a no. In that way neither the simple nor the conditional solution is the answer. However, immediately following the answer "Yes, I'll switch" , will be the question "Why?". And it is this 'why' that is the core of the problem. And ... it will not surprise you, the answer is: because the conditional probability ...., etc. Or equivalently, because the odds are ...Nijdam (talk) 19:25, 9 September 2012 (UTC)

Well Richard I thought you weren't, but you seem to show different here. I did study the sources and strangely enough you yourself immediately admit the interesting question, is the "why". From scientific points of view there may be many ways to approach this question. Our discussion here however is which approach is appropriate for the common Wikipedia reader. I hope you agree with me that the game theoretic approach is not especially suited for this purpose. I, and I hope all the participants here, concentrate on a probabilistic analysis. All that's said here has to be place in that context. I hope you also restrict yourself to this. Even then a lot of missionary work has to be done. And, Richard, the only relevant probabilistic solution is the conditional one. Nijdam (talk) 20:52, 12 September 2012 (UTC)

Comments from Linas

(177 of 500 words)

Note: I am assigning Proposal #1 to the comments below. If this is incorrect, please indicate "Proposal #2", "Neither", or "Abstain". --Guy Macon (talk) 20:16, 20 September 2012 (UTC)

Both :-) Leaning to proposal 1. There is nothing 'mathematically' wrong with the 'simple' explanations, and they should come first. The Bayes Thm variant is a worthy exercise, but wow ... the current article presents it very badly, maximizing the total possible confusion. On first read, my knee-jerk reaction was that it was just flat out wrong; I fought off an urge to edit. I had to re-read the 'conditional' explanation a few times before I got it -- and I do math (& probability) for a living. After this stumbling block, I currently find it dubious that the Bayes variant is somehow 'more correct' -- mostly it just seems 'more difficult', and little else. First and foremost with math topics, explanations must be simple, easy, approachable. Don't confuse dense formulas with mathematical rigor. linas (talk) 16:01, 10 September 2012 (UTC)

BTW: Appeal to authority: I edit little here on WP except for higher math articles. I recognize Trovatore, salix alba, Boris Tsirelson and Lambiam as other regulars in my edit circles, and I find it damning that they all lean to proposal 1. linas (talk) 16:10, 10 September 2012 (UTC)

Joe Decker: Don't confuse counting arguments with frequentism. The 'simple' explanations here are counting arguments, not frequentist arguments. This has nothing to do with Bayesianism; there's no need to invoke 'priors' or 'ill-explicated pre-conditions': if something is 'ill-explicated', then turning it into a Bayesian prior doesn't magically make it 'well-explicated', nor any less 'hidden'. There is nothing wrong with using this problem to illustrate Bayesian-style reasoning, and indeed, the article already has an (opaque) section on this; none of the proposals are suggesting its removal. linas (talk) 18:25, 11 September 2012 (UTC)

Comments from Joe Decker

(113 of 500 words)

Abstain (Previous material struck as I'd misunderstood the question the RfC was asking.) --j⚛e decker^talk 16:25, 15 September 2012 (UTC)

Not including a Bayesian explanation somewhere would make me, Duda and Hart, and Baby Bayes sad. It is a particularly clean way of looking at the problem that is certainly not the simplest explanation, but which tends to get people who use it to the right answer without the sort of ill-explicated preconditions hidden within the simpler and more accessible frequentist explanations. --j⚛e decker^talk 02:05, 11 September 2012 (UTC)

linas: The RfC was a bit unclear, but I gather you're correct that nothing is intending to remove the formal Bayesian argument, which was the primary point I intended to make. --j⚛e decker^talk 19:47, 14 September 2012 (UTC)

Comments from Ningauble

(432 of 500 words)

Neither proposal directly solves the problem which has plagued this article for years: attempting to give due weight to differing POVs has resulted in giving UNDUE weight to the dispute itself – a tempest in a teapot. Proposal 2 is "more worse" in this respect.

Even the most strident and widely cited critic of vos Savant's "simple" solution, (Morgan et al. (1991), cited in the article) acknowledges in a rejoinder (printed in the same issue but not cited in the article) that if it is stipulated that Monty chooses goats uniformly at random (as is stipulated in most subsequent academic literature, and in this very article) then that is a legitimate basis for her "unconditional" solution, as he terms it. Taking their criticism out of this context is POV cherry picking.

More fundamentally, there are different "frequentist" (statistical) and "Bayesian" (information theoretic) interpretations of probability. See §3 of Richard Gill's paper for Statistica Neerlandica,[7] which contrasts these perspectives as applied to MHP. Some who deny the validity of the "simple" solutions appear to be either unaware of this difference or to be dogmatic about only one being the "true" meaning of "probability". This epistemological distinction is probably too abstruse for the general readership of this article, and is seldom directly addressed in the literature on MHP. Presenting them as conflicting views rather than complementary approaches unduly makes a mountain out of a molehill. ~ Ningauble (talk) 00:41, 12 September 2012 (UTC)

• Re. Rick Block's six examples of sources criticizing "simple" solutions: [1] as discussed above, Morgan et al. (1991) is cherry picked; [2] Gillman (1992) says it depends on Monty's selection strategy, but the current "standard" problem specifies uniform random selection; [3] Eisenhauer (2000) bases his criticism on ambiguity about Monty's behavior, long since resolved by the "standard" problem; [4] Rosenthal (2005a) explicitly states that the "simple" solution is correct, and only criticizes it for not being general enough to evaluate variant host behaviors; [5] Grinstead & Snell (2006) claim to prove that the "simple" solution does not answer the equal-goat-frequency problem by showing that it fails for the unequal-goat-frequency problem, which is simply non sequitur; and [6] Lucas et al. (2006) say that the conditional probability method handles a broad class of variant host behaviors, and do not at any point say the "simple" solution to the "standard" problem is incorrect.

  Taking these sources as expressing the POV that the "simple" solutions are incorrect is a grossly distorted reading of what the sources actually say. If any of these "criticisms" of solutions merit inclusion at all, they must be contextualized by reference to the questions to which they apply. ~ Ningauble (talk) 01:08, 12 September 2012 (UTC)

Comments from User:Jouster

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Note: I am assigning Abstain to the comments below. If this is incorrect, please indicate "Proposal #1", "Proposal #2", or "Neither". --Guy Macon (talk) 20:16, 20 September 2012 (UTC)

Object Strongly to this Being an RfC—Wikipedia is not a democracy, principles of mathematics are not subject to being voted upon, and articles should be written in the way that best communicates the most correct information to the widest possible audience. If anyone does not support the implementation of these guidelines, they should discontinue editing the article. If there's a genuine desire to reach this, on the other hand, then a scientific approach is easiest adopted: find a trusted, neutral third party, have them run A/B comparisons between the two suggested complete articles, and then test for comprehension afterwards.

Is this a terrible lot of effort? Of course! But so is me typing this. The reason I am typing it is because the readers' time, in aggregate, monstrously outweighs my own. Consequently, we owe it to the readers to spend more of our time making a better article from the readers' perspective(s), not from the perspective of a plurality of the editors.

Jouster  (whisper) 02:46, 12 September 2012 (UTC)

Comments from User Albtal

(413 of 500 words)

Proposal 1 But surely start with the MvS version which caused the "furor" around the world, and state that the crucial rule there was missing, and clearly formulate this rule (K&W not needed), and then explain the solution based on:

A strong argument for the simple solution - "strong" because it even holds for weaker conditions than those in the "standard problem":

Forget The car and the goats were placed randomly, and forget he chooses one [uniformly] at random.

The following holds: If the contestant has chosen door 1, he will win the car by switching in two of three cases: If the host opens door 2, he will choose door 3, and if the host opens door 3, he will choose door 2. So he will win the car, if it is behind door 2 or door 3. Therefore switching has a 2/3-chance of winning the car.

We all know that we don't know the strategy of the host, neither in placing the car nor in choosing his door if he has a choice. So we have to decide without this knowledge. If now somebody says But it may be that the host does not choose symmetric strategies, we say Yes, it may be, but if so, we can't know; and therefore we have to decide without this knowledge.

But if we don't know the strategies of the host, we cannot compute the conditional probabilities. May be, but we don't have to compute conditional probabilities to know that the contestant has a 2/3-chance with switching. Imagine you are the coach of the contestant, and (may be just before his final choice or earlier) he asks you: What shall I do? Do you say now: Sorry, I can't compute my conditional probabilities; I cannot help you? Or would you say, as another coach would do: Take the switch. Then you have a 2/3-chance of winning.

(I think that these considerations belong to elementary game theory, and are as simple as the "game" Have I a coin in my hand?, which is even too simple to occur in elementary books there.)

...and in a separate section: Three doors, one car, two goats; two possible "Jokers": 1. The host must open a door with a goat before the contestant picks his door. 2. The contestant determines two doors of which the host has to open one with a goat. Then picking. Which Joker should he choose? --Albtal (talk) 07:39, 13 September 2012 (UTC)

@JohnSRoberts99: The show never happened in the "MHP" mode. Monty Hall was free in his actions, and never offered a switch. And a show with the crucial MHP rule that the contestant has to determine two doors of which Monty has to open one with a goat is too boring not only for Monty Hall. And if taken place frequently results would be the same as if you throw a die counting "1 or 2" / "3, 4, 5, or 6".--Albtal (talk) 20:41, 14 September 2012 (UTC)

Comments from JohnSRoberts99

(33 of 500 words)

Note: I am assigning Abstain to the comments below. If this is incorrect, please indicate "Proposal #1", "Proposal #2", or "Neither". --Guy Macon (talk) 20:16, 20 September 2012 (UTC) (Oops. cut and pasted to the wrong section. Sorry about that.) --Guy Macon (talk) 07:30, 21 September 2012 (UTC)

Does anyone have statistics from the show itself? That might be a bit more useful than trying to look at all the way people have calculated potential outcomes. I would back Proposal 2. JSR (talk) 19:33, 14 September 2012 (UTC)

Looks like I was in favor of Proposal 2.JSR (talk) 20:46, 20 September 2012 (UTC)

Comments from Wehwalt

(50 of 500 words)

Proposal 1 Agree with Elen. The game show host knows everything. That's part of the background assumption, along with the sun rising that morning, the end of the universe not yet having occurred, and the stagehands aren't leading the goat from door to door depending on what the contestant says.--Wehwalt (talk) 22:22, 20 September 2012 (UTC)

Comments from CBM

(106 of 500 words)

I don't think I have commented yet. I prefer proposal 1, because I think it captures the spirit of the actual problem better. In general, Richard Gill's published analysis of the problem is very apt and I would be glad to see us follow it. It it matters, just to respond to something I saw on this page: I am an academic and I think the conditional probability is "a somewhat unimportant academic extension to the problem" - where "somewhat unimportant" means that it should be treated in later sections of the article and explicitly as an extension rather than something that addresses the original question. — Carl (CBM · talk) 23:03, 20 September 2012 (UTC)

Comments from Mike409

(28 of 500 words)

I hate to do this, but I have to go with Proposal 1.

Wikipedia is not meant to be first Mathematically rigorous (MTAA), unfortunate as that may be.

Mike409 (talk) 04:40, 21 September 2012 (UTC)

Comments from JonRichfield

(439 of 500 words)

Responding to this RFC is about as tricky as the MHP. If I am not doing it correctly, too bad. I certainly agree with Tom Hulse about re-thinking the RFC and probably with his reasons.

Neither. The article as it stands is better than either of the two alternatives. It may not be elegant, but it is about as near to comprehensive as most reasonable readers could demand. It also starts with simple concepts, so that impatient or naive readers can do the TLDR trick without failing to read a compelling answer. Such an impatient reader might miss some of the more interesting material further down, but that is his choice; we can't force him to stay in after class and clean the dusters. To omit any of the material further down (such as unspecified details of the host from hell/heaven) would leave a mutilated article and discussion, both betraying naive readers who try to apply the principle (well, we can't force them to read and understand -- that is their problem, but if we fail to supply salient considerations, then selling the reader short is our responsibility) and letting down readers who would like some depth of treatment. In short, we should take seriously the duty of thoroughly treating material that demands thorough treatment. The question of whether WP is supposed to be mathematically rigorous is totally different; the discussion as it stands is not primarily mathematical, and inserting a few elementary mathematical remarks and equal signs in an article on a basically mathematical topic is not to sin in terms of mathematical rigour. As it stands the article is nothing like formally demanding; the discussion is light and any symbolic expressions are highly visible and easily skimmed or skipped.

A totally different question is whether the article needs re-writing (always assuming that there is no sacrifice of depth of treatment). I don't feel strongly about that; I don't think that the current version sins seriously on any of the usual niggles about OR, POV etc, but if anyone can demonstrate a better layout or improved didactics etc, then good luck. All I say is that pruning of either the historical or theoretical content would be totally improper. If someone really feels that the article covers too wide a field, it could be split, to avoid stampeding the livestock. For example, it might calm the waters a bit to treat material such as discussed below in the Rules for Monty section. I don't specially want that, but I suspect that some people might like it. At all events however, first demonstrate that everything is in place, then split. JonRichfield (talk) 09:53, 25 September 2012 (UTC)

End of RfC

Post-dated check to avoid archiving Glrx (talk) 23:35, 31 October 2012 (UTC)

The Request for Comment has run for 30 days, and RFC Bot has cleared the rfc template. Responses to date are as follows:

Proposal 1	11
Proposal 2	6
Neither	9
Total votes	26
Abstain	4
Total comments	30

– Tabulated by Ningauble (talk) 22:52, 6 October 2012 (UTC)

We have passed the default 30 day duration for an RfC and it has been 2 weeks since anyone posted a comment. I am posting a closure request at Wikipedia:Administrators' noticeboard/Requests for closure --Guy Macon (talk) 15:02, 9 October 2012 (UTC)

Update: as I had hoped, we are going to get three sets of uninvolved eyes looking at this RfC and evaluating consensus. So far we have two. See Wikipedia:Administrators' noticeboard/Requests for closure#Talk:Monty Hall problem#Conditional or Simple solutions for the Monty Hall problem? for details. --Guy Macon (talk) 21:47, 10 October 2012 (UTC)

Update: three closers have signed up and are now discussing the closure. All three have extensive experience with evaluating consensus and closing RfDs. They seem to be very well qualified and understand the issues we are trying to resolve here. One is working on his PhD in neuroscience has a strong math and psychology background, one is an advanced mathematician, and one is a graduate student in psychology -- I can't think of a better set of skills to apply to this issue. --Guy Macon (talk) 16:27, 11 October 2012 (UTC)

A quick non-RFC note: I updated my user-page, putting in more detail than I earlier wanted to, since Guy Macon misread it. I am not "a Neuroscience PhD student," but, yes, I have a strong math and psychology background, both. Churn and change (talk) 18:16, 11 October 2012 (UTC)

Also a brief disclosure of my own background-- I have a Masters degree in psychology and my training has been in cognitive psychology. A considerable part of my training has been focused on statistics, though mostly as it relates to experimental design, data analysis, and general theoretical principles, but not a level of probability theory that would typically be covered in graduate-level mathematics departments. More importantly, other than this recent and minor edit, I have not ever been involved with this article (though have certainly been aware of its content disputes). I thank Guy Macon for his approval of our group and hope we can deliver a fair and representative close based on consensus and Wikipedia policy. I, Jethrobot ^{drop me a line} (note: not a bot!) 08:38, 13 October 2012 (UTC)

Update: still under discussion by the closers. This is a Good Thing -- clearly they want to do it right rather than just posting a quick count. --Guy Macon (talk) 09:04, 24 October 2012 (UTC)

Proposed text for Solution section, take 1

Proposed text for Solution section, take 1

Solution The player should switch. Switching wins twice as often as staying. Simple explanation Switching loses if and only if the player initially picks the car door, which happens 1/3 of the time, so switching must win 2/3 of the time (Carlton 2005). Case 1: Car behind door 1 Case 2: Car behind door 2 Case 3: Car behind door 3 Player picks door 1 Player picks door 1 Player picks door 1 No matter what the host does, switching loses Host must open door 3; switching wins Host must open door 2; switching wins One case where switching loses Two cases where switching wins The situation can also be analyzed by cases, as illustrated above. Switching loses in only one of the three equally likely cases, and wins in the other two (vos Savant 1990b; Krauss and Wang, 2003). Many people find the situation intuitively easier to understand by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990; Krauss and Wang, 2003). In this case there are 999,999 doors with goats behind them, and the player's chances of picking the car door are 1 out of 1,000,000. The game host goes down the line of doors, opening each one to show 999,998 goats in total, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times (i.e. the times the player initially picked a goat) the other door will contain the prize. Explanation with door 3 open Most people take it for granted that the host opens door 3, and base their reasoning on that fact. That assumption is incompatible with the simple explanation illustrated above, because in case 3 the host does not open door 3. Once case 3 is excluded as impossible, simple case counting results in the common incorrect answer that only two cases (case 1 with door 3 open and case 2) are possible and therefore the chances are 50:50 either way (Krauss and Wang, 2003). It is true that when door 3 is open, there are only two possible cases: either the car is behind door 1 or it is behind door 2. These two cases, however, are not equally likely! This is illustrated in the alternative explanation below, with the two relevant cases on the left side of the thick line. Car behind door 2 Car behind door 1 Car behind door 3 Player picks door 1 This happens to two in six players Player picks door 1 This happens to two in six players Player picks door 1 This happens to two in six players Host must open door 3 (two in six) Switching wins Host randomly opens door 3 (one in six) Switching loses Host randomly opens door 2 (one in six) Host must open door 2 (two in six) Three in six players see door 3 opened. Two of them win and one loses by switching. Three in six players see door 2 opened. What happens to them is irrelevant to those who see door 3 opened. The above illustration shows everything that can happen to players who pick door 1. On average, three in six end up in the exact situation described in the problem statement (door 3 is opened; left of the thick line). Two of the three win and one loses by switching; therefore the probability that switching wins in that situation is 2/3 (Chun 1991; Grinstead and Snell 2006; Morgan et al. 1991).

Comments on the proposed text, take 1

Since I criticized the other proposals, I felt it is only fair if I put one up for criticism, too. This contains material from both, the article, and some my own. The referencing could probably be improved. How do you like it?

It's intentionally somewhat minimalistic, and adding a "Discussion" subsection or something to it where things like the ignorant host variant could be covered is certainly possible. This is only meant to get the "answer to the standard problem is 2/3; switch!" part done with before the article proceeds on to other stuff. -- Coffee2theorems (talk) 17:07, 26 September 2012 (UTC)

I do not like it, for reasons discussed at length above. Martin Hogbin (talk) 18:39, 26 September 2012 (UTC)

The explanation about the most common mental model is pretty unintelligible - I guess I just don't think that way. Can we have a do over on that. And I don't think the million goats belongs in there. Perhaps the next section should be Still having trouble? which could include the million goats.--Elen of the Roads (talk) 18:45, 26 September 2012 (UTC)

I'm not too attached to the million goats thing, but I put it in there because while it isn't a solution it's certainly an explanation, and it seems to be an effective one. It's often mentioned that it helps people "get" the MHP, and K&W had this to say about it:

"Hell and Heinrichs (2000) obtained 65% switch decisions by investigating a variant of the problem with 30 doors, where 28 doors were opened after the first choice.", which is big compared to "Most studies on the Monty Hall problem only report the percentage of switch decisions, usually around 10-15%." It's actually more effective than their carefully constructed "guided intuition" version with three doors (59%).

I'm loath to include any lengthy "aids to understanding" material before the conditional solution (I'm for proposal 2, after all :), but this one was fairly short and the bang for the buck high, so if there was going to be any then that was the obvious candidate. Do you have any specific reasons why it doesn't belong?

Which part of the mental model explanation is unintelligible? The paragraph before the figure? The figure? The paragraph after it? Is there anything there that is understandable? -- Coffee2theorems (talk) 19:37, 26 September 2012 (UTC)

I think it's the way it's written above. You've run a number of things together and my thought chain goes something like

• "After reading the standard version of the problem, most people form a mental model I've never formed a mental model that looks anything like that
• The solution illustrated above is incompatible with this mental model Why the hell are you showing it to me then
• Instead of switching to a compatible mental model, it is also possible to solve the problem in a way consistent with the common one by considering only cases where door 3 is opened (left of the thick line below) OK, I'm giving up at this point

I think that the idea of mental models may go some way to explain why people get confused, and certainly belongs in the article somewhere. Interestingly, it's not the actual reason everybody I ever show this puzzle gives for thinking it makes no difference to switch. They all mentally look at the two remaining doors (their pick and the one Monty hasn't opened) and say "Well, it's fifty fifty isn't it. The car's either behind my door or that door. What's the point of switching." Every time (admittedly small sample). That's why I put the version where Monty picks at random into my layout above - it can indeed go below vos Savant's solution, but it's that which seems to convince everyone - showing that the outcome is differend because Monty won't show the car. Elen of the Roads (talk) 22:15, 26 September 2012 (UTC)

I rewrote that part now. At the same time I ditched the "mental model" wording; it wasn't even used in the same sense as in the K&W paper. Their meaning is one definite arrangement (e.g. "Car / Goat / Open door" is a mental model in their sense), whereas I used it in the sense of a model that you can mentally simulate. Better? -- Coffee2theorems (talk) 00:10, 27 September 2012 (UTC)

I like it. Except for the first paragraph of the "with door 3 open" bit. It can be deleted. Start straight away with "it is true that with door 3 opened there are two cases".

References: Selvin's (1975b) second letter to the editor has this solution. It's the second oldest solution on record. His first letter to the editor (1975a) has the other solution. Since he invented the problem I think these references are good reasons to put both these solutions right at the start of the article. Both give useful insight, probably both will appeal to many readers. Some readers will have no use for one or the other. So what?

I would like to see the Economist solution here as well. That too appeals to many people. Editors have repeatedly put it forward spontaneously. Richard Gill (talk) 07:14, 27 September 2012 (UTC)

Isn't starting with "It is true that when door 3 is open" a bit sudden? The first paragraph ties that section into the bigger picture. I think that the first sentence is fairly good at least, even if the rest of the paragraph could be improved.

What I'm trying to get across to the reader is essentially these points (without necessarily explicitly spending ink on all of them):

1. Most people (such as you the reader, in all probability) initially reason based on the assumption that door 3 is open, and that leads to this common fallacious argument and this wrong conclusion (two cases, 50:50).
2. The common fallacious argument is actually salvageable, and here's how to fix it (two cases yes, 50:50 no).
3. Now that the argument is fixed, it should be obvious what went wrong. (host is sometimes constrained, sometimes not, and that matters)

I think that showing what went wrong, not just this is how to do it right, is an important part of the solution in a paradoxical problem. Certainly there are other reasons to put the conditional solution in, but this is the one that I think has the most immediate relevance to the reader.

Maybe it would be better if the paragraph didn't refer to the previous solution? Maybe something like the following (replacing both paragraphs)?

Most people take it for granted that the host opens door 3, and base their reasoning on that assumption, imagining door 3 as already open. That leads to considering only two cases as possible - the car is either behind door 1 or door 2 - and incorrectly concluding that the chances are 50:50 either way (Krauss and Wang, 2003). The two cases are indeed the only ones possible when door 3 is open, but they are not equally likely! This is illustrated in the alternative explanation below, with the two relevant cases on the left side of the thick line.

Any better? One might also throw in an extra sentence right after the vos Savant solution, saying that

Therefore, the strategy of picking door 1 and then switching no matter what happens wins 2/3 of the time.

so that if the reader goes back and is confused about what the explanation actually does show, then they are gently nudged in the right direction (there exists an interpretation of that sentence which is true, and it's the most natural interpretation). -- Coffee2theorems (talk) 09:21, 27 September 2012 (UTC)

As to the received impression "Two still closed doors, and one of them hides the car", Ningauble said above:
partitioning the doors between "your door" and "Monty's doors" is the most persuasive explanation, and just suggesting the partition is often enough of a hint.

Being able to see two "groups of doors" (1 door + 2 doors), to distinguish and to help keep them apart: Why not saying just in the beginning:

Three doors, and one of them will hide the car.    (Instead of "1/3   2/3" we should write "Player   Host") and instead of "1  2  3" we should say (or add) "1/3  1/3  1/3"

Gerhardvalentin (talk) 10:27, 27 September 2012 (UTC)

Three doors, and just only one of them will hide the car. Only one.
Note: After the guest first has selected "his" door, the entity of three doors has irrevocably been divided into two opposing groups.
The reader should be "aware" of this fact, because it is useful to examine those two groups. Each group can hide one car at most.

The guest's group can hide one goat at most, and the host's group can hide two goats at most, but the host's group "must" hide one goat at least. (At this stage, the "biased" host who in the million doors variant leaves only  #777'777 of his doors closed, because he NEVER uses to open  #777'777 if any possible, is still not "necessary" in this step.)

What the article should do, but never does: Helping to "decode the picture" that the tricky "story" and its course / sequence / devolution tells / leaves behind, resulting in a given obvious paradox.

Help to correctly distinguish, to single out, to decompose the elements and to put them together again in an insightful manner.

Help the reader to understand and to "decode" the paradox, imo he should be able to *vary* his first impression from different perspectives. Why rejecting to offer him this support? The article could be much better than it has been for years now. Gerhardvalentin (talk) 11:31, 29 September 2012 (UTC)

Coffee2theorems, you wrote "Most people take it for granted that the host opens door 3, and base their reasoning on that assumption, imagining door 3 as already open". I find this a very odd statement. We were told that the host opened door 3! (Not opens door 3). Of course one imagines door 3 as being open. Yes, because it was opened. That's why I find this intro superfluous but also confusing since there are maybe some subtleties implied by present versus past tense. But if so, they are too subtle to come across (to me, at least). How about: Because door 3 has been opened to reveal a goat and we see two doors left closed, most people automatically imagine that the two remaining possibilities for the location of the car must be equally likely. Richard Gill (talk) 12:11, 27 September 2012 (UTC)

That seems to be what people do. And if Monty had opened a door at random, they would be right. The trick is to show them that by avoiding showing the car, Monty has made it more likely that the other door has the car behind it. Elen of the Roads (talk) 12:52, 27 September 2012 (UTC)

Exactly! And the reason why it is more likely, twice as likely in fact, is because Monty is twice as likely to open door 3 when he has no choice, as when he does have a choice. Richard Gill (talk) 13:36, 27 September 2012 (UTC)

Incidentally, the ignorant host version is easy enough to explain using the "explanation with door 3 open" figure, too. When the host chooses randomly even when the car is behind door 2, one of the two players who would win in the standard setting ends up being shown the car behind door 2. Only two in six are shown a goat behind door 3, one of who wins and another one who loses. -- Coffee2theorems (talk) 14:31, 27 September 2012 (UTC)

The first sentence is that way because it was based on this one in K&W:

"Although, semantically, door 3 in the standard version is merely named as an example ('Monty Hall opens another door, say number 3"), most participants take the opening of door 3 for granted and base their reasoning on this fact."

I just took that as the basis, nothing deep implied (FWIW, the word is "opens" in the standard problem descriptions, not "opened", and the vos Savant version doesn't use past tense anywhere; you can imagine yourself doing the analysis in front of the closed doors if you want). They also say:

"34 out of the remaining 35 participants (97%) indeed drew an open door 3, and only a single participant (3%) indicated that other constellations also remain possible according to the wording of the standard version."

So there are a few who do see it otherwise, too. It might be fine to say "Because door 3 has been opened" despite that, if we started our article with the conditional solution, but we don't! The simple solutions describe things from another point of view (you analyzing cases in front of closed doors), where it is still possible that door 2 is opened. I think the reader should be told when we are changing the point of view back to door 3 being already open, and why. Trying again:

Most people consider only the situation where door 3 is open, leaving only two possibilities: the car is either behind door 1 or door 2. They then incorrectly conclude that the chances are 50:50 either way (Krauss and Wang, 2003). These are indeed the only possibilities when door 3 is open, but they are not equally likely! This is illustrated in the alternative explanation below, with these two possibilities on the left side of the thick line.

I'm not dead set against stating that door 3 has been opened as a matter of fact instead of saying merely that most people analyze the problem from that POV, but that seems slightly presumptuous as it is not the only way to analyze it. -- Coffee2theorems (talk) 13:59, 27 September 2012 (UTC)

This is all interesting stuff an worthy of a place in the article but first we have to convince our readers of the right answer. It seems we are going round in circles and the 'combing doors' explanation is coming back into fashion. I agree with all those above, that it has great explanatory and convincing power. We need it near the start. Martin Hogbin (talk) 17:59, 27 September 2012 (UTC)

Well, I've almost lost track of where we stand, but the 'combining doors' explanation is a very bad one and as I wrote above: highly misleading. For your interest again: There is nothing to combine. Each of the not chosen doors has 1/3 chance on the car. Combining them means adding 1/3+1/3 =2/3. I have not the slightest clue how this may be helpful in any solution. Why not read what I wrote on combining purses? Easy to understand for anyone! Nijdam (talk) 22:18, 27 September 2012 (UTC)

This is about convincing readers of the right answer. I think many have this incorrect argument ("two cases, 50:50") lodged in their brain and some won't listen to anything that doesn't point out the flaw in their reasoning. The conditional explanation addresses this directly - you have the two cases, and it shows why they aren't 50:50. I hope that what looks like going in circles is actually going in spirals - steadily, if slowly, upward. -- Coffee2theorems (talk) 23:02, 27 September 2012 (UTC)

The combining doors argument is a correct argument for what it delivers, namely that switching gives the car 2/3 of the time. For many newcomers to MHP, it's *the* eye-opener. It's used by many sources, both popular and academic. Some academic mathematical sources even use it as a step towards the kind of solution some editors want to see (namely, a conditional probability statement). Because if we add symmetry to the argument, the door numbers are irrelevant, and 2/3 is a conditional as well as an unconditional probability. Equivalently, door numbers are independent of door roles and we can ignore the numbers; the only thing that is relevant is the roles of the doors. Richard Gill (talk) 07:31, 28 September 2012 (UTC)

Nijdam, we discussed this earlier and, I thought, came to some form of understanding. The crucial, sometimes unstated, fact in the 'combining doors' solution is that the host's actions of revealing a goat do not affect the probability that the car is behind the originally chosen door.

My suggestion, which I thought you had agreed to, was to state, initially without proof or justification, that the probability the car is behind door 1 is not changed by the host showing a goat.

Having done this, we can then compare this with the case where the host reveals a goat by chance. It is easy to explain how that action gives us us information which can be used to revise our probability that the car is behind door 1, resulting in no advantage in switching. We can the proceed to justify our original statement that in the standard case no information is revealed and the probability that the car is behind door 1 is unchanged.

If we make this explanation the last of the simple ones we achieve three worthy objectives: we keep the explanations simple at the start, we explain why it matters that the host knows where the car is, and we show how, in some circumstances, the probability the car is behind the originally chosen door can be changed by the host's actions, thus providing a lead into conditional probability and the Morgan solutions. Martin Hogbin (talk) 14:20, 28 September 2012 (UTC)

It seems to me that the others wanted to use the combined doors solution to illustrate the same thing as the other simple solutions do: that 2/3 of all switchers (who choose door 1) win. That is a very different usage from the one where we claim "no information is revealed". What you are talking about is something like this going into the "Explanation with door 3 open" section:

One intuitive argument groups the doors into the player's doors (door 1) and the host's doors (doors 2 and 3), as illustrated below. Initially, the chances of the car being behind the player's door are 1/3, and the chances of it being behind one of the host's doors is 2/3. This remains true after the host opens door 3, because doing so provides the player no new information about what is behind the player's door — regardless of what is behind the player's door, the host's choice appears completely random to the player.

What the others seem to be talking about is something that would go in the beginning of the "Simple explanation" section. For example, one could replace the Carlton blurb with this:

If the player is determined to always stick to their initial choice, they might as well not be offered the chance to switch at all. Sticking wins exactly when the player initially picks the car, which happens 1/3 of the time. Switching wins the other 2/3 of the time. This may be easier to grasp if the doors are grouped into player's doors (door 1) and host's doors (doors 2 and 3), as below:

The player is effectively given the choice of keeping what's behind their door, or taking everything behind the host's two doors. The host opens one of his doors to show a goat, but that doesn't change the fact that switchers always get all cars behind the host's two doors, and stickers get all cars behind their one door.

I'm sure both wordings could be improved. I think the latter approach is somewhat better, as the "information" thinking is notorious for causing various kinds of confusion, and that concept is not necessary for solving the problem. The notion of no new information being revealed about what is behind the player's door is also somewhat complicated compared to the simple case counting used in most approaches. -- Coffee2theorems (talk) 16:05, 28 September 2012 (UTC)

Let me make this crystal clear, especially for the defenders of this explanation, i.e. the combining doors. It goes like this:

1. the chosen door has 1/3 chance on the car
2. hence the other doors have together 2/3 chance on the car
3. one of these other doors is opened by the host showing a goat
4. this door clearly has 0 chance on the car
5. as both doors together has 2/3 chance, the remaining door must have 2/3 chance on the car

Many will find it very "logically", and may think they now understand the paradox. This explanation is indeed found in several (reliable??) sources. It is impossible to repair this kind of reasoning, as the basic idea, the combining of two chances of 1/3 into 2/3, gives the wrong suggestion, and does not contribute to any understanding. It especially lacks the explanation why the chance of 1/3 for the remaining door magically changes into 2/3.

It is *easy* to complete/repair the reasoning. And the repair job is in the published literature.

@Richard: There is nothing to complete/repair, as the basic idea is wrong. Of course you may see the equivalence with the simple solution, and complete it, but this does not repair the wrong idea of combining the doors. Combining may seem to bring understanding, but what is understood is wrong. The step from 4. to 5. is not logical, and to prove 5. the combining is superfluous and of no help.Tell me what insight the combining brings you. Does it enlighten you that 1/3+1/3 = 2/3 ?! Nijdam (talk) 22:47, 28 September 2012 (UTC)

The competitor chooses door 1. With probability 2/3 there is a goat behind it. The host opens a different door revealing a goat but does not inform you which door he has opened. Probability 2/3 your initial door 1 hides a goat since you have gained no new information. Finally the host informs you that it was door 3 he opened. Still probability 2/3 that your initial door 1 hides a goat because *which* door he opens is independent of whether or not your door hides a goat, by symmetry.

And indeed what effectively is going on is that the host is offering you the possiblity of a car behind doors 2 and 3 in exchange for the possibility of a car behind door 1. Richard Gill (talk) 17:53, 28 September 2012 (UTC)

While the symmetry arguments are neat and all, I do not think they help the general reader. It's just so much mumbo jumbo they could parrot to a mathematician to get them to shut up, and that's about all the mileage they will ever get out of it. You can fill in the details. I can fill in the details. Nijdam can fill in the details. But the general reader can't!

A simple test to check whether the reader understood the argument: "Prove that 2/3 of switchers who pick door 1 and see door 3 opened to show a goat end up getting the car. Show your work." (and maybe same for the ignorant host case) If you have hand-waved something about symmetry, do you seriously believe the reader can turn that into an actual proof showing true understanding? Even assuming a high-school student fairly good at math?

The second illustration in my proposal should impart that level of understanding, at least to that high school student. Much more difficult problems are usually tackled in high school, and that explanation does not even require any understanding of probability theory (much less something like non-trivial statistical independence or conditional probability; are those even covered in high school?). The more complicated alternative mathematical material can well go in a later section (which could cover all non-K&W cases and "reverse mathematics" of MHP investigating necessity of assumptions etc. — all very interesting — to your satisfaction).

Even if you don't require standards of proof that would satisfy a high school math teacher, the "no information" thing is tricky enough. I did try to argue it in the combined doors example above, but it's probably pretty opaque, even though it is the most intuitive approach I can think of (i.e. some way of expressing the line of thought "case 1: car behind your door, host behavior seems random", "case 2: goat behind your door, host behavior seems random"; host behavior indistinguishable in the two cases, he has "perfect poker face", you are none the wiser about what's behind your door). -- Coffee2theorems (talk) 19:23, 28 September 2012 (UTC)

NB. By using random doors, a kind of combining doors solution is possible, but that is not what generally is understood as the combining doors solution.Nijdam (talk) 16:29, 28 September 2012 (UTC)

I agree that in this type of combined doors solution (the first one of the two I illustrated above), the combining of doors is superfluous. Once you show door 3 open, the argument of "taking everything the other guy has" loses its force, because both of you now equally 'own' one closed door! Then we go down the rabbit hole of arguing why it's not 50:50 even though there are only two places the car could be, and I don't see how combining doors helps there at all. It's a "no information revealed" argument in disguise.

If all doors are shown closed, then the host has two closed doors, and that's why taking what he has looks like such a sweet deal. I think that is where the intuitive force of the second type of this argument comes from. Showing an open door ruins the argument; one must instead show them closed and say that "no matter what the host does, you get all cars behind his two doors" or something like that. -- Coffee2theorems (talk) 17:14, 28 September 2012 (UTC)

I think the reason combining doors is *such* an eye-opener is that nobody is taken in by a cardsharp who, accused of hiding the ace of spades in his sleeves, rolls up one sleeve (say, the left one) and declares "see, nothing up my sleeve." Everybody knows this is not evidence about whether he is hiding the purloined card. None whatsoever.

Some professor (or sophomore) might note that if the accused has a tendency to roll up his right sleeve whenever possible then choosing the left one would be highly suspicious, so therefore an objective probability for sleeve preference must be taken into account. To a subjectivist this putative unknown probability is immaterial, and the sleeve chosen does not contribute any evidence about whether he is hiding the card. None. What. So. Ever. However, as a sop to those who think it makes a difference, one could stipulate in advance that the accused is completely ambidextrous and utterly dyslexic. If the professor now wants to plug a 50:50 distribution into some formula then that is his prerogative; however, if he says that one must do so, and that it is illogical to simply use the stipulation as indicating the choice of sleeve makes no difference, then I think it just illustrates what Turing laureate Richard Hamming said in The Art of Probability for Scientists and Engineers (1991), p. 4:

"Probability is too important to be left to the experts. [...] The experts, by their very expert training and practice, often miss the obvious and distort reality seriously. [...] The desire of the experts to publish and gain credit in the eyes of their peers has distorted the development of probability theory from the needs of the average user. The comparatively late rise of the theory of probability shows how hard it is to grasp, and the many paradoxes show clearly that we, as humans, lack a well grounded intuition in the matter. Neither the intuition of the man in the street, nor the sophisticated results of the experts provides a safe basis for important actions in the world we live in." [emphasis in original]

Some mathematicians might prefer adding the words "by symmetry" to the explanation. I doubt many laypersons would find the addition of a five dollar word particularly illuminating. Personally, I think it is a *virtue* of the combined-doors approach that lumping them together points out the symmetry to laypersons without reference to abstract ideas. It is a real eye-opener, and ought not be omitted from the article. ~ Ningauble (talk) 17:43, 28 September 2012 (UTC)

Laypersons omit the door numbers from their explanations because they instinctively and correctly appreciate symmetry, and because they are intuitively using probability in its classical sense. They should be applauded for their sound intuition. I'm all behind Richard Hamming here.

If the word "symmetry" is superfluous for laypersons, no hard done in using it. At least it will keep the pedants among the professionals quiet, and it will delight the others. Alternatively, how about the word "independent"? Whether the host opens door 2 or door 3 is independent of whether there is a car or a goat behind door 1. That is both intuitive and true (and easily proven, using symmetry, if you feel a proof is needed). And it's in the literature, in the discussion of Morgan et al. Richard Gill (talk) 17:58, 28 September 2012 (UTC)

Ninguable, I am with you on this one, 'The experts, by their very expert training and practice, often miss the obvious and distort reality seriously'. Morgan is a classic example of where experts use what is essentially a conjuring trick to ruin a simple puzzle. Martin Hogbin (talk) 22:13, 28 September 2012 (UTC)

I agree that name-dropping a cryptic word like "symmetry" doesn't help. The conditional solution needs to be explained in a way that is actually understood; that is done in the "Explanation with door 3 open" section of the proposal.

The "nothing up my sleeve" analogy is interesting, I didn't think of it like that. Yet if we show one of the doors open or talk about the probability of one door changing, I think we are going to confuse many people. I've often enough seen people say that probabilities don't change and get all confused about that, and explicitly showing two closed doors is just begging for the "two doors, 50:50!" knee-jerk response. Would showing the three closed doors figure and closely following the Adams explanation like this be suggestive enough?

The player is basically given the choice of keeping what's behind their door, or taking everything behind the host's two doors. The host is saying in effect: "You can keep your one door or you can have the other two doors, one of which (a goat door) I'll open for you."

This explicitly mentions opening the goat door, but leaves making the connection to sleeve-rolling (or to whatever analogy they think of) to the reader. -- Coffee2theorems (talk) 23:01, 28 September 2012 (UTC)

Once upon a time I went looking for a FA-class reference for the "combining doors" solution. The best I could come up with was Devlin (who, as has been mentioned repeatedly here, botched it and in a follow-up column resorted to a Bayes' Theorem solution). If anyone can come up with a FA-class reference (no blogs, not some random guy with a web page) presenting this solution in a way that is better, please let us know what it is. Perhaps obvious, but I'm fine with the overall approach Coffee2theorems is suggesting here. It sounds like Elen is OK with it, Richard is OK with it, C2T (obviously) is, I am - are Martin and possibly Gerhard the only naysayers? -- Rick Block (talk) 05:06, 29 September 2012 (UTC)

A certain RD Gill fixed K Devlin's solution in print. He did not use any scary words like symmetry. [8], middle of page 5; this is an article in the online statistics and probability encyclopedia of the big national and international statistics societies, and peer review is part of the procedure of publication. The contestant initially chooses door number 1. Initially, his odds that the car is behind this door are 2 to 1 against: it is two times as likely for him that his choice is wrong as that it is right.The host opens one of the other two doors, revealing a goat. Let’s suppose that for the moment, the contestant doesn’t take any notice of which door was opened. Since the host is certain to open a door revealing a goat whether or not the car is behind door 1, the information that an unspecified door is opened revealing a goat cannot change the contestant’s odds that the car is indeed behind door 1; they are still 2 to 1 against. Now here comes the further detail which we will take account of in this solution: the contestant also gets informed which specific door was opened by the host – let’s say it was door 3. Does this piece of information influence his odds that the car is behind door 1? No: from the contestant’s point of view, the chance that the car is behind door 1 obviously can’t depend on whether the host opens door 2 or door 3 – the door numbers are arbitrary, exchangeable. Therefore, also knowing that the host opened specifically door 3 to reveal a goat, the contestant’s odds on the car being behind his initially chosen door 1 still remain 2 to 1 against. He had better switch to door 2. Richard Gill (talk) 09:42, 29 September 2012 (UTC)

I could see a section called "Explanation with a door open" going in there, explaining (without an illustration showing door 3 open) how nothing changes for a contestant who does not notice which door was opened - maybe they are blind and are told that a door was opened but not which - and then saying something like:

Many people find it intuitively clear that further being told the door number reveals no new information about what is behind the player's door. However, this is only true when the host chooses randomly among his allowed choices. In a variant where the host even slightly favors one of the doors when possible, new information is revealed by his choice and the probability does change.

Trying to actually show that no new information is revealed to those whom it is not intuitive would get way too complicated, IMO, so leaving it at "many people find it intuitive" seems best. I also feel that pointing out the dependence on the unbiased host assumption is necessary if we go down this route, because readers otherwise probably won't notice and are thus misled to think that this applies to the biased host variant as well. That is entirely analogous to people not noticing the dependence on host's restricted choice assumption and assuming the simple solutions apply to the ignorant host variant as well.

This would be the only part of the "Solution" section that actually explicitly uses belief update and has "changing probabilities". I'm worried that this is unnecessarily confusing with little gain. Many peple are not familiar with the idea that probabilities can change and some resist it. The probability interpretations also rear their thorny head. That said, it's not an impossible idea. -- Coffee2theorems (talk) 10:02, 29 September 2012 (UTC)

Coffee2theorems, I notice that you are not appreciating Devlin's solution because you apparently have a frequentist mind-set. You seem to think that we can only use symmetry if we know the host is unbiased. No. We weren't told anything at all. We don't know whether the host is biased or not. If he's biased we don't know which direction he's biased in. All in all, for us it is equally likely that he'll open Door 2 or Door 3 when he has that choice because we have been given no information to the contrary. And Vos Savant underlined the fact that we are supposed to take the door numbers as mere labels, of no significance at all, in her wording say Door 1 etc.

I notice that many mathematicians have a frequentist mind-set and I suspect this is because they have been exposed to academic probability which is usually taught from a frequentist perspective. The rest of the world is still just as Bayesian as ever.

Long ago Martin said a very wise thing: no one can think seriously about MHP without pondering on the meaning of probability. Probability interpretations raise their head, thorny or not, the moment you start to solve a popular brain-teaser by using probability notions. You had better be aware that there are different interpretations, all of them legitimate, and that there is a big mismatch between the more common (subconscious) popular interpretations (which are the same as those of the founders of probability: Huygens, Pascal, Laplace, Boole ...) and the more common present day academic interpretations.

Like it or not, probability does not belong exclusively to the experts. We need meta-experts here, not experts. Richard Gill (talk) 13:35, 29 September 2012 (UTC)

I'm actually Bayesian; I'm just trying to avoid unnecessarily imposing that on the reader at the outset. If the reader understood common notation, we could bash P(gobbledy) = gook at them with impunity and be interpretation-agnostic, confusing nobody regardless of their personal interpretation of probability. We can't do that, but we can use counting of "equally possible" (sic!) cases and easily understood frequencies ("2 people out of 3") while still remaining basically interpretation-agnostic (it's not like we are using confidence intervals or p-values here). In fact, with K&W assumptions, (almost) everyone agrees regardless of interpretation! That's as close to confusing nobody as we can reasonably expect to get.

You seem to think that laypersons are Bayesians. That is indeed the Bayesian party line, but to be honest, I don't think laypersons actually have a consistent interpretation of probability that a statistician or a philosopher would recognize. It probably also varies quite a bit from person to person. People have different backgrounds, and the use of Bayesian language confuses some people (even if they perhaps think in a quasi-Bayesian way at some level). What little exposure to probability theory people have is also usually explained in frequentist terms, even if nothing inherently frequentist is actually covered. That's one reason I wrote the "However, ..." part in those terms (another is that the K&W assumptions are in those terms as well); it is easily enough re-expressed in Bayesian terms if need be. Having to make a choice of interpretation this way is IMO a major drawback of this explanation. -- Coffee2theorems (talk) 17:27, 29 September 2012 (UTC)

Hear, hear, a certain Nijdam does not consider this a fix of the combining doors "explanation". It's a solution to the MHP, but it does not fix the fundamentally wrong idea behind the combining doors. There simply is nothing to combine in making the paradox better understandable. Nijdam (talk) 08:08, 29 September 2012 (UTC)

Let me explain to dear Nijdam why this is a fix of the combining doors explanation. It focusses on what is behind Door 1. A car or a goat. That means that it lumps together what is behind Doors 2 and 3. We focus on the question: is the car behind Door 1, or is it behind Doors 2 and 3 combined? Secondly it splits up the information which we get from the host's actions into pieces. What information do we get from the host about the question whether or not the car is behind Door 1? The answer: no information. Devlin was a bit careless about this. For those who might still be worried that he did not tell the hold story, my solution discusses what information, about the question whether or not the car is behind Door 1, is given it by being revealed *which* door, Door 2 or Door 3, is opened by the host. Answer: none, by symmetry. Or if you find that word to scary: none, because we are given to understand that the door numbers are just arbitrary labels. There is nothing special about any of the doors. Richard Gill (talk) 09:42, 29 September 2012 (UTC)

So, maybe some editors don't understand some solutions: that doesn't mean those solutions are wrong. It might mean that some editors' powers of imagination, or understanding of common English language, or knowledge of the sources, are too limited. The first important issue is whether or not those solutions are published in reliable sources. The second important issue is whether there is a consensus among editors that they might be useful to our readers. The third important issue is whether or not there is a concensus among editors that thy are notable enough that we have to inform the readers about them. If an editor disagrees with a solution then he or she should point to a reliable source which supports his point of view. Richard Gill (talk) 09:36, 29 September 2012 (UTC)

Well dearest Richard, do you, with all your knowledge of the English language, understand what the fundamentally wrong idea behind the combining doors argumentation is? BTW: what do you think of the combining purses example? Nijdam (talk) 12:16, 29 September 2012 (UTC)

In this early stage of receiving what (picture) the tricky one-time-story of devolution has left in the end, it is not necessary to imply that the "paradox" that it has left can only be viewed under one single slight aspect, and in this early stage it is never necessary to imply the unimportant illusion that the overall chance to win by switching could be "enhanced by staying" in one special "repetition" that was never implied to ever follow. Never necessary in that stage. Gerhardvalentin (talk) 13:23, 29 September 2012 (UTC)

Nijdam: I can see a lot wrong with your version of the combining doors argument, but I see no point whatsoever in writing out an argument which is obviously wrong. This is just like Morgan et al. actually changing Vos Savant's words in order to be able to claim a point and get a publication in a prestigious journal by saying she was wrong. No: Morgan et al showed their stupidity by not having the imagination to see that she was right, if interpreted in a constructive way. As some of the discussants of their paper pointed out.

You say that your version is the combining doors argument, but we don't have to buy your version of it. Especially not, since it is wrong. It's a straw man, a characature.

On the other hand I see nothing wrong with the combining doors solution in the form in which I presented it to you here, quoting moreover from a reliable source, who, I happen to know, wrote it for the express purpose of showing that Devlin was close to home with his argument - he simply made one step without motivation, and when people pressed him on it, he apparently got confused and decided it was safer to go a long way round instead of finishing his short, elegant, route to the summit. But he's not a probabilist or a statistician so I can imagine he lost his self-confidence at that point.

As to the purses, I don't understand your combining purses remarks at all. If you think it has relevance you had better find a reliable source which explains it, and hopefully explains it better. If such a source doesn't exist you'ld better write one. I suspect that you have a somewhat narrow mathematician's notion of probability which doesn't always match well to what ordinary people mean when they use the language of probability in ordinary conversation and non-technical writing. Mathematicians don't have a monopoly on the use of English. Richard Gill (talk) 13:19, 29 September 2012 (UTC)

Happily you admit there is a lot wrong with "my" version of the combining doors solution. Here is a quote from Devlin's article:

Imagine you are the contestant. Suppose the doors are labeled A, B, and C. Let's assume you (the contestant) initially pick door A. The probability that the prize is behind door A is 1/3. That means that the probability it is behind one of the other two doors (B or C) is 2/3. Monty now opens one of the doors B and C to reveal that there is no prize there. Let's suppose he opens door C. (Notice that he can always do this because he knows where the prize is located.) You (the contestant) now have two relevant pieces of information:

[1] The probability that the prize is behind door B or C (i.e., not behind door A) is 2/3.

[2] The prize is not behind door C.

Combining these two pieces of information, you conclude that the probability that the prize is behind door B is 2/3.

Apparently you admit there is a lot wrong with Devlin's way of reasoning. Nijdam (talk) 14:08, 29 September 2012 (UTC)

There is not a lot wrong in Devlin's way of reasoning. There is one missing step in Devlin's argument. Devlin admitted that there was a step missing. The missing step is easy to fill in. It has been done by yours truly, and no doubt by many others (it's easy to do). What's the point of talking about a wrong argument which is nearly right? Fix it first and then discuss whether or not it is useful in the popular early parts of the article. Let's talk about the article, not about everyone's personal favourite and personal most hated solutions. Richard Gill (talk) 14:25, 29 September 2012 (UTC)

@Richard, please show me where "my" version of the combining doors diffesr, as you stated, from Devlin's. I would not brag much about fixing Devlin's argumentation. It is in fact no more than the complete solution to the problem, which is well known and of a very elementary level. Calling it fixing is something like showing a front tyre and the rear light of a bike and saying it's a bike, but only needs some fixing. Devlin does no more than stating two more or less given facts, coming down to: every door has 1/3 chance on the car and (surprisingly, but apparently not to Devlin), door 3, the opened one, has probability 0 on the car. To me it's a contradiction, instead of a brilliant idea, and I surely hope you do not use this way of reasoning yourself when explaining the MHP. Nijdam (talk) 15:31, 3 October 2012 (UTC)

Nijdam: If I get you right your position is this: even if the door numbers are elided from the problem and we have a "you choose a door, host opens another door, wanna switch to remaining door?" formulation, one should still condition on doors, because they are distinguishable to the player (e.g. left, middle, right door; or top, middle, bottom door). You do not condition on specific goats (e.g. label them A and B yourself and condition on B, just like you label the doors 1, 2, 3 and condition on 3), because the goats are indistinguishable. I am curious: how do you justify that view?

Surely the goats are visible to the player. The goats have a size (say, p times the average goat). If the host chooses randomly, then no matter what continuous size distribution you use for the goats, the revealed goat's size almost surely won't hit the median of the hidden goat's distribution, so "which goat is bigger?" will almost surely not be 50:50. You got information distinguishing the goats; you have the "probably smaller goat" and the "probably bigger goat". Might not goat size therefore matter in general (if you took Freud too seriously, you might believe the host wants to show the bigger goat..?), and need to be conditioned on? -- Coffee2theorems (talk) 07:40, 4 October 2012 (UTC)

I think Devlin's explanation has great convincing power and should be included early on in the article but I agree that it has some problems. To avoid clogging up this page with more discussion I have started a page on the subject where anyone with an opinion is welcome to comment.Martin Hogbin (talk) 11:50, 14 October 2012 (UTC)

Devlin is not a reliable source, and was plainly mistaken in his so called explanation. It may looks convincing, but in an utterly misleading way. It is especially this kind of nonsense that should be banned out once and for all. Combining doors does not in any way contribute to understanding, other than in a false way.Nijdam (talk) 13:57, 14 October 2012 (UTC)

(editing)

@Coffee2theorems: I had the impression you know all the ins and outs of the MHP, so you will understand and support my objection to Devlin's combining doors, as being highly misleading as an explanation. What your question about my position concerns, I do not see the immediate connection with the above discussion about combining doors. If you mean, the audience, being informed about the rules of the game, and unaware of the door the player will choose, is asked in advance whether the player should switch or not after the host will show them a door with a goat, yes, even then the answer "yes, you should witch" will be motivated by considering all the conditional probabilities,m like: if the player chooses door 1, and the host opens door 3, then .... As for the goats, firstly we may just consider them as undistinguishable, or formulate the problem with two empty doors. I think most people accept the goats as no more than "bad luck". But you will know this, so your question must beef a more sophisticated nature. When the goats are distinguishable, we are stuck with two goats, from which we do not know how they are chosen. Clearly if a player wins a goat, she has to be replaced. However I see no difficulties, as long as the host, when applicable, chooses randomly between the door numbers. Is this what you're aiming at? (connection was broken)Nijdam (talk) 12:34, 4 October 2012 (UTC)

I see Devlin's combined doors explanations as variants of the "condition on the event 'door 2 or door 3 was opened'" approach, and I don't see any actually wrong statements there. The question is whether he has given a convincing argument or not.

That leads to the question of whether it is permissible to condition on the event "a door (= door 2 or 3) was opened, showing a goat (= goat A or B)", or whether you should condition also on one of "door 3", "goat B", or both. Essentially, Martin's Billy/Nanny argument; I just translated it into something observable (e.g. goat size), as otherwise it makes no sense to me.

I agree that the problem may well be expressed without goats (I explained my views on that here (01:58, 2 October comment)). What I'm aiming at is that, while the idea of conditioning on goats is surprising (why should that matter! a goat is a goat!), conditioning on doors is just as surprising to others (why should that matter! a door is a door!). I can't help but notice the similarity! Consider that the person you are arguing with (Gill) is also quite competent in this field; to me that is evidence that perhaps the issue is not so cut-and-dried after all.

The MHP can also be formulated without distinguishable doors. One way is to use a bag with a red marble and two blue marbles in it, instead of e.g. shells and peas like vos Savant's (goatless!) simulation. IIRC some years ago Gill opined that the former is still a simulation of "the MHP". I feel that it is not quite satisfactory; but less strongly than I used to. Is that purely a matter of opinion, or a matter with a clear-cut correct/incorrect answer, or something in between? That, I feel, is the core issue. Once you have a formal model instead of a fuzzy word problem, there's not going to be much disagreement among informed people; the issue is in how to get there.

It's true that the "condition on a door" arguments don't work in the case of hosts who favor e.g. the right-hand door, and that that takes people by surprise (oh, door number can matter!). But the "condition on door 3" version also doesn't work for hosts who favor bigger goats, and I bet that would take people even more by surprise (oh, goat size can matter!). Either way, you've taken a real(?) situation, built a simplistic model, and are surprised when reality breaks out of it. If reality doesn't do that (lucky!), was your reasoning nevertheless wrong, because reality could have done so and you would have been surprised if it did - you just got lucky? Can you see how reasonable people might end up having a disagreement here, depending on the details? You might argue that including a door number is a second-order correction and something fuzzy like goat size is a third-order correction (effect limited by observation uncertainty), but by then you have admitted that both are approximations instead of Absolutely Correct One True Solution(s). The first-order model, with indistinguishable doors, already captures the main part of the phenomenon (2/3 vs. 1/2), so it's not entirely unreasonable to argue that it's a legitimate option to stop there ("good enough for physicist work!"). -- Coffee2theorems (talk) 18:59, 4 October 2012 (UTC)

Well, look closer at Devlins (and others') way of reasoning (see above). Try to understand what he means when he says: combining these two pieces of Information. There is nothing to combine.there! And hence nothing is explained. It really is easy to see. Nijdam (talk) 19:44, 4 October 2012 (UTC)

It's about a brain teaser that has been told in a story about a "one-time problem". In this one-time problem, the host either got the car and only one goat in 2 out of 3, being restricted to show his only goat then. – Or, in this one-time problem, the host got no car but both goats in the rest of 1 out of 3. Then – in this one-time problem – he cannot "show" any (for ever unknown) "preference". It is of no avail and therefore unnecessary to "assume" what you never will know, instead of trusting in that what you already know for sure, just from the outset.

And, just not to fall into the dangerous 50:50 trap, it is helping to "see" that in only 1 out of 3 his two doors hide just goats, but in 2 out of 3 he has got the only prize with one of those two goats. Not to be trapped, it is helping to distinguish the door first selected on the one side, and the group of those two host's doors on the other side. There is nothing wrong. And, although it is a one-time problem ("you are in a" game show, and not "you will be in ten game shows"), in only about 33 millions out of 100 millions staying will give you the prize, but in more than 66 millions out of 100 millions switching will give the prize, because of symmetry: "no matter at all, which one" of his two doors the host did open in that 100 millions. Gerhardvalentin (talk) 21:17, 4 October 2012 (UTC)

To Nijdam: Falk says that conditioning on "which door" has been opened makes only sense "if the host IS biased" and that you KNOW about that bias. And I am adding: You of course CAN utilize a mathematically correct phantasmal cure-all, but then you must be aware of it's inexpedience to give any better advice than to "always switch". Thus: no necessity at all to give the correct answer for the MHP, but very suited and helpful to teach and to train conditional probability theory, see the plenitude of available textbooks on teaching conditional probability theory. Gerhardvalentin (talk) 21:51, 4 October 2012 (UTC)

Imagine you are the contestant. You are given the choice of three miniature doors in a bag (like three marbles in a bag, really). Suppose they are labeled A, B, C; the labels are written in a secret compartment in each miniature. You blindly pick one from the bag (secretly, door A). The probability that the car is in the bag (in both senses!) is 2/3. The host looks in the bag and secretly peeks behind the doors, picking a goat door (secretly, door C), and opens it for you. As Devlin says, "he can always do this", where "this" must mean picking a goat door (he can't very well always pick door C, it might be the car door!). In other words, this is a sure event, and so the car is behind one of the host's doors (secretly, door B or C) still with probability 2/3. The car is not behind the opened door (secretly, door C). Combining these two pieces of information, you conclude that the probability the car is behind the door in the bag (secretly, door B) is 2/3. The car is probably in the bag! Switch!

I think this is basically the same thought process, put slightly more clearly, perhaps with some unnecessary wrinkles smoothed out. ("he can always do this" is a dead giveaway, IMO) You can label the doors and talk about the labels without assuming that the contestant knows them or the distinctions they stand for. You can of course argue about clarity of phrasing things and the sensibility of a model with indistinguishable doors. (another possibility is a blind contestant told appropriate things, and that probably fits the story slightly better, but I like bags of marbles as a model better, being a classic and all; also, the A, B, C labeling here is actually entirely superfluous) -- Coffee2theorems (talk) 23:39, 4 October 2012 (UTC)

Sorry, Coffee2theorems, often people come with alternative descriptions as if they would make the problem more understandable. As if the original setting is not simple enough as it is. Anyway, if you like to use an alternative setting, like yours, with tiny doors, etc, you firstly have to show it's equivalent to the MHP, and guess ... it isn't. That's often the point, someone gives an alternative description, from which they believe it is equivalent to the MHP, and then proves their different way of solving with this alternative description. Why are people that reluctant to admit Devlin's combining is nonsense? Nothing particularly difficult there. I will spell it out again. Devlin reasons: 1) the not chosen doors 2 and 3 have together probability 2/3 on the car (what he omits to mention, or simply doesn't realize, is that each has probability 1/3); 2) the opened door 3 has probability 0 on the car. To me this is a contradiction. And to you? Nijdam (talk) 06:53, 5 October 2012 (UTC)

---

I'm not sure that Gerhard is against this. The use of the closed doors figure alone (with an explanation, to suggest the partition) at the beginning was actually his idea (10:27, 27 September 2012 comment above), I just copied it because it seemed good to me. It may of course be that I somehow misunderstood it, but I don't see how. -- Coffee2theorems (talk) 08:17, 29 September 2012 (UTC)

Proposed text for Solution section, take 2

Proposed text for Solution section, take 2

Solution The player should switch. Switching wins twice as often as staying. [many citations] The answer is easily verified experimentally. It has been verified countless times using computer simulations as well as simulations with human participants. [many citations] Simple explanation If the player is resolved to always stick to their initial choice, they might as well not be offered the option to switch at all. Sticking wins exactly when the player initially picks the car door, which happens 1/3 of the time. Switching wins the other 2/3 of the time. This may be easier to grasp if the doors are grouped into player's doors (door 1) and host's doors (doors 2 and 3), as illustrated below. The player is basically given the choice of keeping what's behind their door, or taking everything behind the host's two doors. The host is saying in effect: "You can keep your one door or you can have the other two doors, one of which (a goat door) I'll open for you." No matter what the host does, a player who always switches gets all cars behind the host's two doors. (Adams 1990) The situation can also be analyzed by cases, as illustrated below. Switching loses in only one of the three equally likely cases (case 1), and wins in the other two. A player who always switches wins 2/3 of the time. (vos Savant 1990b; Krauss and Wang, 2003) Case 1: Car behind door 1 Case 2: Car behind door 2 Case 3: Car behind door 3 Player picks door 1 Player picks door 1 Player picks door 1 No matter what the host does, switching loses Host must open door 3; switching wins Host must open door 2; switching wins One case where switching loses Two cases where switching wins Many people find the situation intuitively easier to understand by considering the same problem with a million doors instead of just three. (vos Savant 1990; Krauss and Wang, 2003) In this case there are 999,999 doors with goats behind them, and the player's chances of picking the car door are one in a million. The game host goes down the line of doors, opening each one to show 999,998 goats in total, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times (i.e. the times the player initially picked a goat) the other door will contain the prize. Explanation with door 3 open When first encountering the problem, most people consider only the situation where door 3 is open, leaving only two possibilities: the car is either behind door 1 or door 2. They then incorrectly conclude that the chances are 50:50 either way (Krauss and Wang, 2003). These are indeed the only two possibilities when door 3 is open, but they are not equally likely! This is illustrated in the alternative explanation below, with these two possibilities on the left side of the thick line. Car behind door 2 Car behind door 1 Car behind door 3 Player picks door 1 This happens to two in six players Player picks door 1 This happens to two in six players Player picks door 1 This happens to two in six players Host must open door 3 (two in six) Switching wins Host randomly opens door 3 (one in six) Switching loses Host randomly opens door 2 (one in six) Host must open door 2 (two in six) Three in six players see door 3 opened. Two of them win and one loses by switching. Three in six players see door 2 opened. What happens to them is irrelevant to those who see door 3 opened. The above illustration shows everything that can happen to players who pick door 1. On average, three in six end up in the exact situation described in the problem statement (door 3 is opened; left of the thick line). Two of the three win and one loses by switching; therefore the probability that switching wins in that situation is 2/3 (Chun 1991; Grinstead and Snell 2006; Morgan et al. 1991).

Comments on the proposed text, take 2

OK, I modified the proposal again to take some of the above comments in account, as well as other modifications. What do you think? -- Coffee2theorems (talk) 20:15, 29 September 2012 (UTC)

Very well illustrated and professed, really a good assistance. Gerhardvalentin (talk) 21:33, 4 October 2012 (UTC)

I like this choice of three solutions (plus one analogy) for exemplifying basic approaches to the problem, and they seem pretty clear. But I do have some reservations about a few particulars:

1. The present Solutions section includes a lot of other material, in multiple subsections. You have not indicated how this would be reorganized (or removed?).
2. I would prefer not to reference simulations in a section on solutions because, as I have mentioned previously, they demonstrate and persuade but they do not solve or explain. I think the immediately following section should be about simulation. That section should begin by noting that many people find the above explanations unconvincing, and give the anecdote about Paul Erdős not being sure of the result until he ran a simulation. (Moving this from the lede, where it is more detail than needed.)
3. I still don't like the "with door 3 open" gambit. Characterizing it this way and referring to it as "the situation where door 3 is open" [emphasis added] is susceptible to the polemical interpretation that the previous explanations are inapt for this situation. I would prefer to find a less prejudicial heading and to reword the pertinent clause as "... most people consider only that door 3 is open, leaving two possibilities: ..." However, the same observation about people's common error applies equally well to the preceding two explanations, so the sentence might be moved to the main section introduction (in lieu of remarks about simulations).
4. In the million doors analogy, language about "on average" and "times the player initially picked" uses an expressly frequentist interpretation, but the analogy is equally applicable to a Bayesian interpretation. The same sort of slant occurs in other paragraphs that refer to, e.g., "1/3 of the time" and "always stick" [emphasis added]. I would prefer language that gets the point across while remaining agnostic about interpretations of probability.
5. In the "six cases" illustration, it might not be immediately obvious that this refers to 3 possible car locations times 2 doors Monty might open = 6 cases. This could be mentioned in the sentence introducing the illustration. I also have some reservations about expressing this in terms of six players because, as I have mentioned previously, replacing "which door do you want?" with "which player do you want to be?" strikes me as an unnecessary multiplication of entities.

All that said, this draft is a pretty good framework to build upon, and the prose is reasonably clear. The general idea I am using in points 3 & 4, and to a certain extent in point 5, is that if we are going to cut to the chase by giving straightforward, accessible solutions without going into different interpretations of the problem and of the meaning of probability, then we need to be agnostic about the differences we are not explaining (at least until it is taken up in a later section, if at all). My recommendations in points 3 and 5 are also influenced by my opinion that the root cause of the naïve 50:50 error lies in missing the distinguished status of the two unchosen doors. ~ Ningauble (talk) 18:27, 5 October 2012 (UTC)

I have indeed not explained what to do about the rest of the material. That's because this proposal is based on the RfC, where the idea of a compact and simple initial Solution section is present in both proposals. I figured that if that part is resolved, it's a big step forward, and trying to propose two things at the same time multiplies already small probabilities of success.

One idea would be to replace everything in the Solutions section before the "Formal solution" subsection with this, and then figure out what to do with the rest. Most of the rest could, for example, be moved into a later section ("Mathematical details"?), and insert other important material (e.g. "Simulation", "Sources of confusion", "History") in between. A lot of that also requires cleanup (look at the list of trivia in the "History" section, ouch!). I think this part ought to be a separate discussion, instead of a proposal from me.

The naming of the subsections was not so easy. There are strong feelings here about the two POVs, and it felt like picking labels for the "pro-choice" and "pro-life" factions, with worries that these would be taken to imply "anti-choice" and "anti-life" by the other factions. The first subsection tells you about the general winning strategy and the second analyzes the confusing "[selected] [remaining] [open]" situation in particular. Certainly it's not intended to imply that a general strategy would somehow fail in this particular case. I suppose "Why switching wins" and "Why the 50:50 answer is wrong" would be possible names for the sections. Would that be better? That could tie into writing a short outline at the beginning, explaining the confusion and that we'll first explain the correct answer using strategic thinking ("always switch" > "always stick", overall probability is 2/3) before tackling the confusion directly.

I don't think the use of frequentist language needs to imply frequentist interpretation. The frequentists don't own the law of large numbers, any more than Bayesians own the Bayes' theorem. AFAIK frequentist language is still commonly used in elementary probability and statistics courses for high schoolers and undergraduates (even those who will be Bayesians!), as well as in newspaper articles and other popular media, because it's easy to understand. The use of Bayesian language also tends to stir the controversy here. I don't really care if this uses frequentist terminology throughout, Bayesian terminology throughout, or mixes and matches willy-nilly, as long as it's written understandably with the general reader in mind and is not downright incorrect.

The use of six players instead of six cases is because otherwise you'd need to explain why you've counted an identical case twice (but moving people around is easy to explain). The explanation that Monty might open a door he's not allowed to open is a bit confusing, especially as it's an important point of confusion in this problem that he may not reveal the car. I thought of using a coin which Monty always tosses and then just enumerating all cases, but that had sourcing problems. -- Coffee2theorems (talk) 16:09, 6 October 2012 (UTC)

Naming the explanations without expressing a point of view might be accomplished by stepping back from characterizing what they mean/answer/entail/imply (or whether they are simplistic/general/particular/abstruse), and just describe the features of the problem that they examine or focus on. The following three subsection headings come to mind:

• Two choices: your door or Monty's?
• Three possibilities: where's the car?
• Six cases: Monty's choice

This is about as neutral as I can make it while giving each a usefully descriptive handle. ~ Ningauble (talk) 17:21, 6 October 2012 (UTC)

Incidentally, regarding the six cases, I am really not sure which is less confusing to people: equal-weighted cases where some are virtual pairs representing a non-choice, or unequal-weighted cases forking only where there are real choices. I.e.:

• A forked case: Monty's choice

Virtual cases and unequal-weighted cases are both potential stumbling blocks that take a little thought or explanation. That Monty's choice only arises if the player initially guessed right certainly seems to be part of what makes the problem so baffling. ~ Ningauble (talk) 19:14, 6 October 2012 (UTC)

Having unequal-weighted cases is one of the important points in the conditional explanation. It shows that, even though you have two cases, it's still not 50:50! (that's the paradox!) Now what is the reason for that? Is counting people confusing? I don't think so: if you understand simulations, you should understand this. It is basically the description of a simulation, counting simulated people. But surely having one explanation for the "two cases but not 50:50" fact does not exclude having another! Maybe we should somehow explicitly point out that the reason "you picked the car" case is less likely is that in that case, Monty's (figurative/virtual?) coin had to land in a particular way for you to see door 3 opened (the split in the illustration), but if you picked a goat Monty didn't toss a coin (no split), so "door 3 open" is evidence against "you picked the car"? What would be clearest?

Purely proof-structurally descriptive labels feel like an overkill for this potential POV problem. I much prefer splitting things by theorems, not proofs; most people are much more interested in theorems. Would subsection names "Switching is the best strategy" and "The two cases are not equally likely" do? Maybe with some material at the beginning explaining the common misconception that the two cases are equally likely, and that we will first explain why switching is best and then why the two cases are not equally likely? I can't see any possible way to construe these as somehow POV. -- Coffee2theorems (talk) 19:25, 7 October 2012 (UTC)

Oh dear

It looks like we are back to square one. Coffee2theorems asserts above that, 'Unconditional solutions do not, however, give you the probabilities for the two doors in question', and seems to think that the simple solutions therefore do not actually the question as asked. This is what the years long dispute has been about. I suggest that if C2t wants to continue with this assertion he does so on the /Arguments page. Many (most) editors here believe that the 'simple' solutions are fine (or at least that any deficiencies in them can be initially glossed over).

This page is about improving the article, not showing our mathematical prowess. C2t insists that we must show our readers, early on in the article that the simple solutions are in some way deficient. I thought, and hoped, that we had gone beyond that point.

If we have two section entitled, or implying:

• Idiot's solutions/Real solutions
• Solutions that do not answer the question/Solutions that answer the question
• Incomplete solutions/Complete solutions

we project a lack of confidence in our initial solutions and readers are going to be less convinced by them.

Note that I am not insisting the reverse, that we say the simple solutions are perfectly complete and correct, just that we do not distract our readers with esoteric mathematical issues before we have then on-side. I have absolutely no objection to discussing these issues later on in the article. Martin Hogbin (talk) 08:39, 11 October 2012 (UTC)

You say that Coffe2theorems says "Unconditional solutions do not, however, give you the probabilities for the two doors in question". He's right. But so what? Some people aren't interested in the probabilities for the two doors in question, separately. Others realize that it is a simple step to get there, if you are so inclined. Why can't we just have an initial section with decent *arguments* why it is wise to switch doors? We need to help the reader see the answer isn't 50-50. I don't understand, Martin, why you take Coffee2theorems' remark as criticism of some solutions. It's just a fact. If you find it interesting, great. If you don't, ignore it.

Secondly, why talk about *solutions*? Let's talk about *arguments*. Arguments for switching. The word *solution* implies that this is a maths problem to which there is right solution and many wrong solutions. We are talking about a popular brain-teaser, we want arguments which help people understand what is going on. Richard Gill (talk) 12:18, 11 October 2012 (UTC)

I have been avoiding much interaction on this talk page, but I want to say that I agree with Martin Hogbin's comment completely, and it expresses what I was thinking when I commented in the RFC. Much of the "conditional" solutions appear to me to represent (interesting) academic extensions of the original problem that should be covered after the original problem is discussed. That's the genera pattern all over Wikipedia: begin with the basic case, and cover academic extensions later on. — Carl (CBM · talk) 12:51, 11 October 2012 (UTC)

I didn't say anything about the unconditional solutions not answering the question. They do answer the question: Switch! They even go the extra mile to show that always switching wins 2/3 of the time, and always staying wins 1/3 of the time, so the first strategy is twice as good as the other. You don't even need the K&W assumptions if you randomize the door numbers; randomization guarantees that the unconditional probabilities are objectively correct without any assumptions. If you really doubt that any of the other more bizarre possible strategies could possibly be better, then Gnedin will reassure you.

The problem statement says nothing about epistemic probabilities of the car location in the one door open situation, and they are fundamentally unnecessary for solving the problem (but necessary for resolving the paradox). Of course the unconditional solutions don't give these probabilities; they simply do not condition on anything.

You claim that I somehow insist that we must tell the reader that "simple" solutions (whatever that means) are deficient. I didn't say that anywhere. Your characterizations of the suggested section names are also not in any way apt. For example, how would "Game theory" translate to "Idiot's solutions"? Or "Strategic thinking", "The correct answer", "The general solution", "The correct strategy", "How to solve the problem", or..?? -- Coffee2theorems (talk) 12:58, 11 October 2012 (UTC)

Just to clear some possible points of confusion: I was thinking/suggesting of organization like this for the initial (re)solution section (choosing the exact names is difficult; these are just possibilities):

• Section: "Resolving the paradox"

• Subsection: "The optimal strategy"

• Show that always switching wins 2/3 of the time and always sticking 1/3 of the time.
• The solutions would be the same as in the "simple explanation" subsection of my current proposal, or possibly other unconditional solutions. Not e.g. Devlin.
• Hopefully we can insert a claim here that by randomization the 2/3 winning rate can be guaranteed no matter what the producer/host does; that way he can't possibly guess your initial pick (K&W is actually unnecessary for the solutions of this subsection!).

• Subsection: "A further look at the confusing case"

• If we can agree on a Devlin-type solution going in the article, I suggest it go here.
• The current illustration of a conditional solution, or equivalent.

There was no intent that it should be implied in any way that the unconditional solutions don't solve the problem (as Martin fears), or that more complicated material would be placed before the simple solutions (as Carl fears).

I'm not also suggesting that the article say anything like "the paradox cannot be resolved without conditional probability". I'm just saying that the essential reason the MHP is paradoxical is that people apply the principle of indifference twice, that's inherently a subjective/epistemic probability belief update issue, and that it's helpful if the subsections are split along the unconditional/conditional line (or strategic / belief update, same thing) for that reason. Solve the problem first by looking at the overall probability of winning of each strategy, then look at the probability of the car in the two doors situation. -- Coffee2theorems (talk) 14:29, 11 October 2012 (UTC)

Your approach is IMO ridiculously academic for what was intended to be a simple puzzle and it is based on the assumption of a bit of dubious mathematical pedantry. Some people may imagine a specific case that the player has chosen door 1 and the host has opened door 3 then they have to choose, others may imagine a more general situation but, with a reasonable assumptions or a Bayesian perspective it really does not matter which situation they imagine, the answer is the same they will double their chances of winning by switching, whether they play once or many times.

To artificially split the problem right from the start is utterly confusing and unhelpful to the general reader. Martin Hogbin (talk) 15:18, 11 October 2012 (UTC)

We should keep clearly in mind that MvS presented a one-time problem. The contestant was not offered to "play many times". The article should carefully never confound the frequentist's view and "often-times repetitions". Gerhardvalentin (talk) 15:42, 11 October 2012 (UTC)

Coffee2theorems is *not* artificially splitting the problem. There are a number of different ways in which ordinary people can br helped to realize that switching doors is smart. Different ways catch on with different people. All of them are natural, all have advantages and disadvantages. See Kraus and Wang (very helpful overview, written by experts in cognition, not mathematics). To every popular solution there corresponds a formal mathematical version. His approach is not academic. If you know the literature, and want *on this talk page* to discuss how to structure the article, then you should use appropriate language to discuss things. It doesn't mean that the article has to be drenched in the same language.

Vos Savant deliberately names the doors Door 1, Door 2 and so on, so that the newcomer to MHP will adopt the visual picture which blocks their understanding and entices them to the wrong answer (see Kraus and Wang). To realize intuitively that 50-50 must be wrong you first have to discard that visual picture. Think strategically, or group the doors, or forget the door numbers. Then you see that switching is wise. Finally, it is interesting to go back to the original situation, the visual picture, and try to understand what was wrong with it. This is not *academic*. This is achieving full understanding. Maybe some readers are just interested in a slick one-liner why switching doors is rather smart. But some readers will want to understand what was wrong with the intuitive but wrong reasoning: Doors 1 and 2 must have equal probabilities. This is not perverse, not a matter of advanced mathematics. It is not an academic extension! Nothing is extended! It can be explained with simple verbal reasoning and needs only common sense. No need to use expensive words like "conditional" or "symmetry". There are absolutely simple arguments why, when you are in this situation (you chose Door 1 and the host opened Door 3) 1/3 of the time the car will be behind Door 1, 2/3 of the time the car will be behind Door 2: e.g., Devlin's argument (with missing step filled in), or a simple calculation from a table, or the observation that the probabilities can't depend on the specific door numbers named in the case at hand combined with what we know from the one-liners. I think something like this will be enlightening for many readers. If it can be said in a few simple words then it belongs in the non-technical, not specialist part of the article. It reinforces the conclusion of the one-liners, it doesn't weaken them. If a reader like Martin is uninterested or finds it too obvious then he skips it. Richard Gill (talk) 10:49, 12 October 2012 (UTC)

Exactly! Artificially splitting the explanations into "simple" and not "simple" (by presenting only "simple" solutions first) is utterly confusing and unhelpful to the general reader. -- Rick Block (talk) 16:44, 12 October 2012 (UTC)

Oh! I didn't realize anyone would think I'd want to drench the article in statisticese. Nothing of the kind! I'm quite simply trying to make a minor modification to my current proposal to satisfy Martin's concerns about section names. This could be as simple as just changing those section names and nothing else. You can see the text there, it's not in statisticese. If changing just the names isn't enough, then we can change more things incrementally, yes?

The reason I brought up Devlin and objective probabilities is simple: I'm trying to figure out how to make the text more palatable to Martin! AFAICS Martin is afraid that if one of the subsections contains the explanations he likes and the other the explanations he dislikes, then the latter might come to be presented more favorably ("separate but 'equal'", with the associated problems). So I brought up the possibility of including a Devlin-type explanation, which I gather he likes! That should make further discussions more stable as well, as Martin would have favorite explanations on both sides of the natural splitline. Other simple conditional explanations acceptable to Martin would serve the same purpose. Having attractive words like "objectively correct" or "hard guarantee" etc. on the unconditional side of the line might also help; hence the mention of objective probabilities (the conditional subsection is far more naturally interpreted as subjective/Bayesian than objective/frequentist).

Unfortunately, I have no idea what Rick is talking about above. Rick, could you clarify? Are you agreeing or disagreeing with me? With Martin? Both? Neither?? Or Richard? None of the above? -- Coffee2theorems (talk) 19:25, 12 October 2012 (UTC)

I'm agreeing with Richard, supporting your presentation, disagreeing with Martin. -- Rick Block (talk) 05:56, 13 October 2012 (UTC)

Why "always"?

I don't understand the following phrase: "Vos Savant's response was that the contestant should always switch to the other door." The contestant is a person who plays the game only once. So, she can't switch many times. What is to be said by always? --213.102.99.129 (talk) 17:35, 12 October 2012 (UTC)

It is on a one-time show, yes, but anyone is free to "visualize what happens", though. You also could say "in any case". Gerhardvalentin (talk) 18:12, 12 October 2012 (UTC)

As far as I'm concerned, we're talking about simulations. Any averages or repetitions and so on are repeated trials in simulations. We're just trying to get people to understand probabilities by referring to something concrete, i.e. the average rate of something happening in a simulation of many games (as the simulation length tends to infinity, if you're mathematically inclined). The conditional solutions only apply to a one-time game. If you have multiple games, then it gets really really complicated, and AFAIK we don't have any sources for treating that kind of situation. It's only about a one-time show. Only about imaginary (simulated) repetitions. Would it be clearer if we said that explicitly somehow?

"Always" means a general strategy. It means that no matter what door you pick or what door the host opens, you should always switch, no matter what. Not just in the door 1 chosen, door 3 opened case. And overall, when repeated in simulations, that strategy gets you 2/3 winning rate (given K&W conditions). Further, if you pick your first door randomly, then you get 2/3 guaranteed, no matter what. -- Coffee2theorems (talk) 19:44, 12 October 2012 (UTC)

Asymmetric behaviour of the host in opening one of his two doors (note that in 1/3 he has a "choice"):
The argument of Rick Block so far has always been that "if you follow enough repetitions of this game show you will detect some pattern of the host's behavior...." – tending on some "empirical value" to convince that one "must" base on some "q <> 1/2".
You just asked (regarding simulated repetitions) "Would it be clearer if we said that explicitly somehow?", and my answer therefore is yes, we should.

Falk says that, to applying of any host's bias, the host must be biased and you have TO KNOW about that bias. As we never will have such "necessary evidence", we never "need" to base on unknowns, and therefore all we have is the assumption of symmetry, and nothing else.

But of course we "can" and are free to pay respect to the "best case scenario" and to the "worst case scenario" as well: In 2 out of 3 the odds of the two still closed door (contestant:host) are "at least 1:1" (worst case), and in 1 out of 3 they are "0:1" (best case). We know that for sure, and without applying maths. Let's take the extremely biased host: In 1 out of 3 he got two goats, and he can open his "strictly preferred door", odds being "1:0", and in a second one out of 3 he can also open his "strictly preferred door, having the car and one goat behind that special door, but odds being "0:1". So whenever he should open his strictly preferred door (unknown to us!), then probability to win by switching will be at least 1/2. But if he should open his strictly avoided door (also unknown to us!), because the car being behind his strictly preferred door, then probability to win by switching will be max. "1". So we know that we should switch anyway, and the host is very welcome to be biased as much as he can, giving away additional hints we never can nor ever will be interested in, as confirmed per modern reliable sources. Saying that probability theory never is "the key" to making the correct decision and to giving the correct answer, much less conditional probability theory. Exact "conditional probability to win by switching" once and for all is irrelevant.

Any "actual (!) probability" to win by switching, once and forever is completely irrelevant, as all we can ever "know" is that it will be 2/3. But that is not the point. The point is that we are not interested in "actual" conditional probability, not interested in "probability" at all. As per modern sources, the *point* is never to "know any exact actual (!) probability", but the *point* is that we know for sure, just from the outset, that any "staying" forever will diminish the chance to win, as 2/3 is only guaranteed if any "staying" is definitively excluded, just from the outset, and therefore "staying" never will be any option. Never. Just from the outset! That's what we know for sure, and we do not need to know anything else, as per modern reliable sources. We never need to know "actual probabilities", and we do not need any "actual conditional probability". Asking for "actual probability" is a completely superfluous approach that should be discarded, for making the correct decision and to give the correct answer.

Morgan et al.'s view obviously was in the wrong direction and, originally, they indeed went in that wrong direction, complaining about Savant's wording. – Later on, they had to have been corrected, and they had to officially admit their error, had to admit that they were wrong.
Please underline that the question was not on a series of game shows the contestant is in, but just on "a game show". Gerhardvalentin (talk) 22:17, 12 October 2012 (UTC)

What do we know about "any case"? We only know that the contestant picked door No.1, the host opened door No.3 which has a goat, and offered a switch. We don't know anything about the situation the contestant would have picked door No.2, for example, and about the host's behavior in this case. --213.102.99.209 (talk) 23:10, 12 October 2012 (UTC)

So the article, in its actual form, did not help you to decode the paradox? Please help to make the article clear and intelligible. All we really "know" is that staying forever will diminish the chance to win the prize and therefore "staying" once and forever has to be excluded, just from the outset. You only will have a guaranteed 2/3 chance to get the prize if you switch to the door offered, but never otherwise. Actual "conditional probability" being completely irrelevant. Gerhardvalentin (talk) 00:16, 13 October 2012 (UTC)

Try reading the "Extended description of the standard version" section. Does that settle the issue for you? If the lead is confusing, how would you change it? Would simply removing the word "always" from the lead do?

On a related note, we're currently talking about the possibility of replacing the beginning of the solution section with this new version (visible when you click on "[show]"). That one also has phrasings like "always stick", "happens 1/3 of the time", "on average", etc. I'm wondering whether that is suitably clear, or if we should, at the beginning of the solution section, say something like:

Imagine simulating the game show many times and think about what happens to simulated players on average in the long run. The results of such a simulation can be used as a guide for making the final decision to switch or stay even on a one-time show. Actual simulations that have been done are covered in the simulation section. This section explains what will happen in any long enough simulation, and why.

Do you suppose saying something like this would be helpful to the reader? -- Coffee2theorems (talk) 08:07, 13 October 2012 (UTC)

I agree with 213.102.99.129. 'Always' is unnecessary. In the circumstances described the player should switch, and will double their chances of winning by doing so, just as vos Savant said. 'Always' is a relic of a pointless and pedantic argument by editors here. Martin Hogbin (talk) 08:10, 13 October 2012 (UTC)

I would prefer the proposal of C2t to mention early enough that it's not about an ongoing series of shows, but on a one-time problem that, in exactly this form, never took place in reality, and any necessary simulation addresses the possible outcome of that one-time problem. Just to help other editors here not to continually be hoping for "enough repetitions" in order to detect some host's pattern of behavior.

And I repeat my proposal to help the readers to see that the usual wrong conclusion "1:1" is correct only in quite another scenario, in quite another kind of game (if the host does not "in any case" intentionally show a goat, but just opens one of his two doors, be it the car in 1/3 or be it a goat in "only 2/3" !). To see that 1:1 applies to such quite other scenario like "Monty Fall" or "Ignornt Monty". But doesn't apply to the scenario of MvS and MHP. Gerhardvalentin (talk) 08:44, 13 October 2012 (UTC)

If we say that we are analyzing a simulation as a guide for a possibly one-time-show, I think it's clear enough that nobody is trying to detect any long-run patterns of host behavior. I mean, sure, you could figure out any peculiarities of a simulated host; maybe it's a simulation with real players, with your friend playing the role of the host, and he always tends to pick a particular door when he can. But how would figuring out your friend's quirks help you with the real Monty? :) No, I don't think that would be a common point of confusion among readers, especially as we mention computer simulations as well.

Would the example paragraph I gave above (starting with "Imagine simulating") work for you? I think that things like "never took place in reality" fit better in the extended problem description section than the solution section. Such things are more about clarifying what the problem means (just like the "host is not allowed to open the car door" clarification etc.), than about solutions as such. -- Coffee2theorems (talk) 09:37, 13 October 2012 (UTC)

Yes, whatever you prefer. All I fear is that editors of this article will continue to still insist on being able to detect any given host's bias, in "observing him" on the long run.
Saying that "q" could actually "differ" from its average of "1/2" (symmetry), and consequently any solution "must" only be a cure-all so-called "correct solution" in applying "q" (possible range from 0 to 1, although any exact actual value of "q" forever must remain unknown!) within Bayes' formula.

Mathematically fully correct, yes, a suitable description of the obvious spectrum, yes, but forever completely unnecessary to the contestant, unnecessary in the situation the contestant is in. And as any "actual true q", diverging from 1/2 forever is and forever will be completely unknown, this never can be of any help in the given actual situation the contestant is in. Such cure-all so-called "solution" is not suited and hence unneeded to make the correct decision, for

1.) "q" is and remains totally unknown (but on the other hand no-one "needs" to know any "actual q"), and
2.) as "q" forever must remain within the range of "0 to 1", with symmetry at "1/2", staying forever has definitely to be excluded. Excluded. Period.

Yes, it beautifully describes the full spectrum, yes, interesting for students in maths, yes, but without any relevance for the decision asked for. Completely inappropriate: undue weight. Show all of that unimportant display, unnecessary for the decision asked for, in later sections. Show in the first line why staying never can be an option and therefore is discarded just from the outset, and why switching is the only "correct decision". See all modern sources. You know that the overall probability to win by switching on average is exactly 2/3, from the worst case scenario (at least "1/2" in two out of three) to the best case scenario (full "1" in one out of three). And you know just from the outset that staying never can be any option at all. But they continue to distinguish "before" and "after". Undue weight.

Yes, conditional probability maths is capable to nicely describe the full spectrum, but forever is unnecessary to make the decision asked for. Undue weight. Improper to help to decode the paradox, improper to meet all views from different angles that are indispensable and most necessary to understand and to decode. Nevertheless they say that this narrow approach of just only presenting a cure-all mathematical "description" of the given spectrum, and never more, is "the key" to "making the correct decision" asked for, but it never was and never can be such "key", but just only a description. But they insist on such description to be the only correct approach, and that any other approach is "incorrect", as addressing the wrong question ("before" versus "after"). Not being able to read what the sources really say, and especially what modern reliable sources say. That's the ongoing conflict of the article. Gerhardvalentin (talk) 11:17, 13 October 2012 (UTC)

Has anyone checked what vos Savant actually said? It seems to me that if we're appearing to quote her words, we should check her words. There is far too much "own research" going on here, too much original thinking, If Marilyn said "always", she said always. And hopefully the context would clarify what she meant. Richard Gill (talk) 14:59, 13 October 2012 (UTC)

Vos Savant's response was: "Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?" [9] She didn't use the term "always", and it seems to me her interpretation is that of a one-time-game. So, simulations and repetitions are inadequate to solve the problem.

My conclusion after reading through the article is: Vos Savant's response is the wrong answer to her question but the "Extended description of the standard version" describes the right question to her answer. --213.102.99.248 (talk) 17:03, 13 October 2012 (UTC)

Good, the words "always" shouldn't be there. But this doesn't mean that we can't use simulations to solve the problem, or shouldn't think about repetitions. For many people, "the chance of winning by switching is 2/3" means that if you imagine the situation being repeated many, many times, then in 2/3 of the times, you win by switching. Simulation experiments are an excellent way to verify claims about probabilities. The player plays just once. But the player can figure out what is a smart way to play, by imagining many repetitions.

The player who always switches in every imaginary repetition, wins 2/3 of the times. Moreover, among those occasions on which the player chose door 1 and the host opened door 3, he also wins 2/3 of the times. And the same for any other initial door and/or door opened by the host. As long as the car is equally often hidden behind each door, and the host equally often opens either door when he has a choice. These are imaginary repetitions and in these repetitions, we do things equally often because they are equally likely for us in the single case we are faced with in reality. Richard Gill (talk) 15:01, 14 October 2012 (UTC)

To answer this question: I don't understand the following phrase: "Vos Savant's response was that the contestant should always switch to the other door." The contestant is a person who plays the game only once. So, she can't switch many times. What is to be said by always? Vos Savant said "always" because this game tactic was very frequently used on the show. Actually it was the show in that the quests were always asked if they wanted to switch to a different choice. Gandydancer (talk) 16:07, 14 October 2012 (UTC)

You do realise that the puzzle has nothing to do with the real TV show, don't you. The real Monty Hall may have asked contestants if they wanted to stick with their choice (it's frequent in this kind of game show - Noel Edmonds does it on Deal or No Deal), but this theoretical problem doesn't represent one of the games on the tv show. Elen of the Roads (talk) 19:54, 14 October 2012 (UTC)

Of course I realize that it is a "brain teaser" puzzle. How rude of you to act as though I must be stupid. Doubting that anyone here has an IQ in excess of hers, I assumed that she did not make a mistake...or that she is misquoted in the article - and, as it turns out, she is. Marilyn never said "always". [10]. Gandydancer (talk) 03:16, 15 October 2012 (UTC)

Gandydancer, you might try a little good faith here. If you read through the many reams of talkpages - including this one - you'll find that people do pop up who believe that the puzzle is describing an actual game on Let's Make a Deal (someone above suggests including the results of the actual show in the article). Your post sounded exactly as if you were under the same impression, so I certainly wasn't intending to be rude or suggest you were stupid in having that impression. Elen of the Roads (talk) 10:49, 15 October 2012 (UTC)

I'd suggest you try a little good faith yourself. I'm not going to look for the editor you mention, however I'd bet that s/he has not been here since 2006 and made hundreds of good edits on many difficult articles. Gandydancer (talk) 14:28, 15 October 2012 (UTC)

"Do you know who I am?" is never a particularly good gambit in my experience. Unless you're the Queen. But I've never seen anything that suggested she was interested in maths problems. --Elen of the Roads (talk) 17:45, 15 October 2012 (UTC)

Some people take the time to check those things before they make a rude "you must really be dumb" post - apparently you are not one of them. Gandydancer (talk) 18:49, 15 October 2012 (UTC)

Oh, I'm sorry ma'am, I presumed the Queen of England had flunkeys to edit Wikipedia for her. I have to point out that you are the one saying that making the perfectly reasonable mistake that it represents an actual game on the show is "really dumb." Perhaps you might like to explain to the guy further up the page why you think he's really dumb for making that mistake. Or perhaps you might stop ascribing motives to me that bear no relation to anything I said. Ma'am. Elen of the Roads (talk) 21:20, 15 October 2012 (UTC)

Gandydancer, vos Savant's famous question evidently was about an imaginary one time show, not addressing any of those various TV series with their quite differing very own special rules. Yes, with their quite differing rules. Evidently nothing to do with any one of those various differing special "real TV shows" in reality. The question is on an imaginary "one time problem". I suppose you know that, and I suppose you realize that the puzzle is on a quite imaginary one time problem ("imagine you are in a game show"...). And I am repeating:
So you surely do realize, just like me, that the puzzle has nothing at all to do with any of those real TV shows, don't you. As to me, yes I do. Regards, Gerhardvalentin (talk) 21:24, 15 October 2012 (UTC)

Well Gerhard, your civil reply is much appreciated. As for my edits, until I did the research and found that Marilyn never actually said "always", IMO it actually may have had something to do with the show. I am old enough to remember "Let's Make a Deal" (back then the only show of that sort),Parade Magazine, and Marilyn von Savant...and the actual question and the following Parade articles. Actually, everybody knew about that show and there were a lot of jokes about it...and even a song, "Door Number Three" (I think( by Steve Goodman. Anyway, until I showed that she did not actually say "always", and considering that every word to the puzzle had been questioned, there had to be some answer. Especially so since a quick read of this article did not seem to show that any of the experts were hung up on that word - again, there had to be some reason. So, I thought I'd throw that suggestion out. @ Elen, please pull your claws in as this is getting pretty old. Even if I was a Ma'am, do you really go around calling editors sir and ma'am, or is this just more of your sarcasm? Gandydancer (talk) 22:42, 15 October 2012 (UTC)

Based on the above, I think that your reaction to my original statement may have been based on some 'separated by a common language' problem. Let me assure you again that although I did think that you thought the puzzle was based on the game show, because that's what you appeared to be saying, I do not think that anyone coming fresh to the problem who makes that mistake is 'dumb' and I don't know why you think I am uncivil and Gerhardvalentin (who also thinks you thought the puzzle is based on the tv show) is civil when we have both said exactly the same thing.--Elen of the Roads (talk) 23:46, 15 October 2012 (UTC)

You cannot repeat this game in the same manner without changing the content of the problem. If you think about repetitions you are creating a rule which is not a part of the question. Your (not specified) rule is: the contestants pick a door, the host always opens another door which has a goat, and always offers a switch. But this leads astray, and it generates another problem, not the one-time-game given by vos Savant. --213.102.96.69 (talk) 10:25, 15 October 2012 (UTC)

I believe you are right that as vos Savant's version (which is unfortunately not the only one floating around in the early days) is a one time problem, any explanation should show a one time answer. You can of course calculate odds on a one time event if that is any help. However, the answer to vos Savant's version as she stated it is "it makes no difference", because she only said that Monty opened a door and showed a goat. She never said if he was operating under any particular constraint. If he picks a door at random, the odds of the car being behind either door are 50/50. It is only if there is a rule that he has to open a door and he has to show a goat that the odds change, and it becomes better to switch, because if you chose one of the goats (which the odds are 3/2 (or 2/3-on in bookie-speak) - ie more likely than not that you did), Monty just eliminated the other goat. Elen of the Roads (talk) 11:00, 15 October 2012 (UTC)

Marilyn said "and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat" implying that he *deliberately* opens a goat door, and emphasizing thet you know that he always can. And almost all of the people who wrote to "Parade" understood her question this way (but still disagreed with her answer)d. And as she herself made explict in her next column on the problem. He can do this, and he will do this, whatever door you initially chose. Richard Gill (talk) 12:24, 15 October 2012 (UTC)

I agree that most people probably if they thought about it assumed that this statement was of some significance. But I also think that once you want to explain her answer, you need to highlight the significance in a way that she obviously didn't because if she had it would have tipped people off as to what the answer is (as if, for example, in the "father and son are involved in a car accident" problem, it said "the surgeon, who was a woman." And I think a lot of people who disagreed with her still hadn't realised the significance of Monty knowing what is behind all the doors. Elen of the Roads (talk) 12:42, 15 October 2012 (UTC)

@Elen: Spot-on. Most people evidently do realize that Monty is selectively showing a goat from the two un-chosen doors, but fail to recognize the significance of it. That is what makes it such a wonderful riddle.

Misdirection using selective information is one of the oldest tricks in the book. It is one of the best tricks because people are often fooled even when they know it is selective information. In very simple situations it is obvious (cf. "nothing up my sleeve" at #Proposed text for Solution section above), but people's grasp of selective evidence is so tenuous and weak that even the very slight complication in MHP of a third door, from which a goat is not to be selected, causes whatever insight one might have about selective evidence to evaporate into thin air.

"Nothing has been falsified – except the impression that it gives."
  — Darrell Huff, How to Lie with Statistics (1954), Ch. 5

I agree that this needs to be highlighted, i.e., pointed out with emphasis. Several editors and sources express the view that the fundamental reason MHP seems paradoxical is that conditional probability is poorly understood. It is; but I believe people's blindness to the significance of selective evidence is the more fundamental stumbling block. This is why selective evidence is used by magicians, con men, and polemicists of every stripe. It is what makes honest, hardworking statisticians unwitting victims of sample bias, and is why hardworking, dishonest statisticians get away with exploiting it.

MHP is an object lesson about selective evidence for everyone, especially for those who can't make heads nor tails of P(A|B) = P(B|A) • P(A) / P(B). ~ Ningauble (talk) 16:59, 15 October 2012 (UTC)

I think the first 'explanation' up has to show the effect that Monty has in deliberately showing a goat. Because that's what makes the difference, and that's what all the 'combining doors' answers fails to explain. If Monty had picked that goat by chance (or even deliberately, but without the constraint that he has to open a door and show a goat) then it isn't any better to switch than to stick. I think the article has to make that clear. Vos Savant made that clear in her second bite at it, where she came up with the million goats explanation. Elen of the Roads (talk) 17:51, 15 October 2012 (UTC)

Am I wrong? The question is about a one time problem: *secrecy*. And the host, in intentionally having to reveal a goat, did reveal a goat. This leads to the famous "paradox". Quite other scenarios not necessarily will generate the "paradox". Gerhardvalentin (talk) 21:42, 15 October 2012 (UTC)

No I don't think you're wrong. I think we are saying the same thing two different ways. It is a one time problem. The significance of 'Monty knows what's behind all the doors' was not brought out in the original iteration, but vos Savant's answer - always switch - is only correct if Monty has to open a door and can't show the car. Elen of the Roads (talk) 21:55, 15 October 2012 (UTC)

Thank you. But: is the WP article really only on a narrow dissection of the history of the first weeks and days of that brilliant famous puzzle that has been presented as a tricky story? Is the article really only on misinterpreting what the story tried to tell? Marilyn vos Savant responded early enough. My RfC-comment above was: It is not so much important what is *the* original question and its "possible" ambiguity, but it is on the certainly *intended* paradox [...] — It is on helping people, *by modern reliable sources*, to decode and understand the puzzle. – Am I really wrong? Gerhardvalentin (talk) 22:12, 15 October 2012 (UTC)

I see what you are saying. But we have to live with the puzzle as it was allocuted, which was as a tricky story. The only significance of it being one-time is as 213.102 points out - we don't know the rules. Which is where we need the other sources who have established what rules are necessary for MvS's answer to be right. Elen of the Roads (talk) 23:31, 15 October 2012 (UTC)

@213.102.96.69 You are right, it's on a "one time problem", and the host always can show a loser "on purpose", as per Marilyn vos Savant, regardless what the player has chosen. Yes, the host shows a loser "on purpose", and not just "by chance". And Marilyn vos Savant added: "by that" we've learned nothing to allow us to revise the odds of the guest's first choice. Gerhardvalentin (talk) 13:03, 15 October 2012 (UTC)

Probably vos Savant had noticed that her first reasoning (a million doors) was wrong. Therefore in her next column she proposed a repetition of the game, but didn't realize that such a repetition, in the same manner, would imply a rule which wasn't given in her original wording. So, her solution doesn't fit to her statement of the problem.

I agree that the host *deliberately* opens a goat door. But you don't know previously that he will do this, whatever door you initially chose. You only know afterwards that he has opened a goat door when you have picked door No.1. You have to pay attention to the chronological (and logical) progress of the game. Hence you must not use knowledge before appearing in the wording.

Nevertheless, if one door is shown to be a loser, that information changes the odds of the remaining doors to be the winner, in general. If we have no further information, a rule for example (but vos Savant's text doesn't give any rule), we have no reason to assume that one door is more likely the winner. So, the odds for both the remaining doors to be the winner are changing to 1/2. --213.102.97.173 (talk) 18:44, 15 October 2012 (UTC)

Yes to your last sentence, but no to the rest (I think - bear with me and see if this makes sense). Vos Savant said that Monty knows what's behind all the doors, but she doesn't say anything about why he picks the door that he does. You pick a door, and the odds are 1 in 3 that you have the car. Monty opens a door at random, it has a goat. That tells you nothing about what's behind your door, or what's behind the third door. As you say, the odds for both the remaining doors to be the winner are 50/50.

Now if by saying that Monty knows what's behind all the doors, vos Savant meant that Monty won't show the car (standard behaviour for a game show host, spoils the game if he gives the result away too early), then the effect is quite different. The odds of you having the car is still 1 in 3, and if you have picked the car, you'll still lose if you switch. But if you picked a goat - for which the odds are 2 in 3, then Monty is left with one goat and the car. He can't show the car so he must show the goat. If you switch, you'll get the car. So you have 1 in 3 odds to get the car by sticking (because you picked it to start with) and 2 in 3 odds to get the car if you switch (because the 2 in 3 odds were that you picked a goat).

So for vos Savant's answer - you should switch - to be correct, she must have always been intending some rule or rules which were behind Monty's actions. In a way, the puzzle has been rewritten to fit the answer by those who like precision. Elen of the Roads (talk) 21:43, 15 October 2012 (UTC)

MvS promptly responded that the host always can show a loser "on purpose", regardless what the player has chosen.
She promptly responded that the host shows a loser "on purpose", and not just "by chance".
And Marilyn vos Savant added: "by that" we've learned nothing to allow us to revise the odds of the guest's first choice.
And I am adding: She presented a one time problem, audience and host unknown to us. All we know about the host is that he showed (shows) a goat on purpose, as per MvS. Gerhardvalentin (talk) 22:00, 15 October 2012 (UTC)

Ps: The core of the "MHP" is the famous *paradox*. So the basics of the MHP is that the host on purpose was to open a door in order to show a goat, where he can do that in any case, regardless what the player has chosen. I repeat that we should show (as an *eye-opener*) that - if quite outside the MHP! - another host should be known to open a door regardless of what is behind it (in 1 out of 3 the car, and in only 2 out of 3 a goat) the *paradox* will never arise. Not in just only a "variant", but quite outside the "MHP". Gerhardvalentin (talk) 12:59, 25 October 2012 (UTC)

MvS clarified in her later columns: 1) the host has to open a door, 2) the host knows where the car is and must show a goat, 3) the host must make the offer to switch. She didn't clarify how the host chooses which door to open in the case the player originally chooses the car, but in the "standard" problem this is a perfectly (uniform) random choice as well. The setting being a game show implies this is a standard sequence of events, likely to be repeated - but with all of these rules the chance of winning the car by staying (regardless of which door you choose and which door the host opens) is 1/3 and 2/3 if you switch. Gerhard's "it's a one-time offer" notion is simply weird (if he can provide a source for this notion that'd be nice). If it's a one-time offer (and you don't know, for example, whether the host always makes the offer to switch) you don't know whether the host is "evil Monty" (who opens a door and gives you the option to switch only if you originally picked the goatcar - as Monty Hall himself did in response to being interviewed by the NYT following publication of the Parade columns) - in which case if you switch you lose with 100% certainty. The host not only must open a losing door on purpose (this one time), but also MUST NOT have the option to avoid doing this depending on what the player initially picks (i.e. MUST open a door no matter what door the player picks and MUST offer the trade no matter what). In addition, if you want the chances not to depend on which door the host opens, in the case the player happens to initially pick the car the host must choose between the remaining two doors independently of the player's choice (usually expressed as "chooses randomly" in this case). With all of these rules, the chances are the same whether it's a one-time offer or whether you're going to repeat this 100 or 1,000,000 times. The "Variants" section examines what happens if any of these restraints are relaxed. -- Rick Block (talk) 23:03, 15 October 2012 (UTC)

In just only "a game-show" the host is unable to communicate any "bias", and the "standard" duly pays regard to this fact:  *secrecy*, *randomly*.
Rick, it's still the same pattern as for years now. You wrote above that Marilyn vos Savant did say "3) the host must make the offer to switch," and immediately thereafter you have added your ceterum censeo concern "what if you don't know, for example, whether the host always makes the offer to switch". Are you aware that you just depicted the state of this mingle-mangle article. Gerhardvalentin (talk) 13:52, 16 October 2012 (UTC)

It doesn't matter whether it's one time or not, that doesn't change the odds, which are one-time and not cumulative anyway. I think GV's point is something to do with the way it was said. I could be wrong. Elen of the Roads (talk) 00:08, 16 October 2012 (UTC)

Right. The point is not whether it's a one-time event, it's what the rules were by which we arrived at the situation that was described. For example, if the host is not compelled to open a door showing a goat and make the offer to switch (from the outset), then the 1/3:2/3 answer does not follow (whether we do it once, or a million times). -- Rick Block (talk) 02:22, 16 October 2012 (UTC)

Rick, it is still the same pattern as for years now. Confirming the "standard", you wrote above that Marilyn vos Savant said "3) the host must make the offer to switch",  and now you you repeated again your permanent ceterum censeo concern: "if the host is not compelled to open a door showing a goat and make the offer to switch"  Are you aware that you again depicted the state of this mingle-mangle article. Gerhardvalentin (talk) 13:52, 16 October 2012 (UTC)

What is the article about?

So, as per some few editors here, this article is not to show the strange "famous paradox" with its sole only exclusive solution to "switch" to the other door.
Nor to show the scenario(s) where this paradox can and will crystal-clearly arise.
Nor to clearly show "why" the paradox does arise in an appropriate scenario.
Nor to help the reader to decode the paradox, to decode why it just may "look to him as being a paradox", why his uncalled-for leaving out of fundamental items resp. his uncalled-for adding of inappropriate assumptions can blur and can cloud his / our vision.

Will the article continue to show in the first line that there indeed exists a weird litany of "other" scenarios where the paradox never will arise.
Will the article continue to say that it is impossible to "solve" the paradox but by applying just one special narrow method of approach, resulting in some fixed range of conditional probability (at least 1/2 to 1)? You "can" use conditional probability, but the article always said that any other approach does not address the question asked for. Yes, the conflict will continue.

Rick, you said above that Marilyn said
1) the host has to open a door.
2) the host knows where the car is and must show a goat.
3) the host must make the offer to switch.

But don't forget that the story is saying "you are in a game-show". Nothing is said that you are given the opportunity to detect any special host's behaviour by watching two or twenty or more repetitions. No word about offering such opportunity.

So you need not deplore "but we don't know" regarding the host's choice, if he as a choice. Saying "you still can detect" is incorrect, and so – for the guest himself, and for anyone else except God an the host, there just is no "how the host chooses which door to open in the case the player originally chooses the car". (Btw: it is completely irrelevant "how" he does, because the correct solution is "switch", because any staying will diminish your chance to win the prize.)

As the host's method is and forever remains completely unknown, the "standard" plainly and correctly translates this given fact of our immutable total ignorance to "the host strictly observes secrecy regarding the car hiding door", or in other words "he chooses uniformly at random". Have a look to it.
That's the scenario presented concerning just this one ("a") game-show. Period. And you are right to say that in the scenario as per MvS ("a" show) the chance of winning the car by staying (regardless of which door you choose and which door the host opens) is 1/3 and 2/3 if you switch. Please help the article to be clear and intelligible, by a clear structure. Quite diverging scenarios offside the scenario presented should not be mingle-mangled. Gerhardvalentin (talk) 10:38, 16 October 2012 (UTC)

GV - if I tried to explain this in your first language (German....?) you wouldn't understand a word I was saying, so please don't think this is a criticism of your ability to communicate in a second language. BUT I am having difficulty in getting your point. Marilyn vos Savant gave a version of the Monty Hall problem, in which given the information she provided, the correct answer should have been that it makes no difference whether to stick or switch. However, she said that the answer was that you should switch. The only way her answer was correct was if there was a rule governing the action of the game show host which had not been articulated. Her 'million goats' explanation, and her later comments, suggest that the rule she had in mind was that he had to open a door and not show the car (a loser door, as you describe it). This is nothing to do with 'host bias', this is just a rule of the game, and true for one iteration or one hundred.

So I have to ask - do you understand that if the opened door is chosen uniformly at random, the odds are the same whether you stick or switch. If you do, then I think the point you are making must be just about the choice of words used in the solutions. But the only way you can decide what to do is by guessing (if not given the information) what the host is up to. To get to the right answer in the original vos Savant version, you have to guess that because this is a game show and the host knows what is behind all the doors, the host won't show the prize. And you have to guess that the host had to open a door as part of the game regardless of your choice (because if he only opens a door if you picked the car, then there is a disadvantage to switch). Elen of the Roads (talk) 20:48, 16 October 2012 (UTC)

Elen, it is on the discussion here. Just have a look. Millions of words, and binders full of "But what if the host is biased?" – Binders! – Made up especially by three editors, one of them retired years ago, two others still active. Although we know that the paradox is based on a given scenario as per MvS, she constantly is blamed here of not having given the correct answer to her question regarding doors 1, 3 and 2.

As per MvS, the host has to open a door, the host knows where the car is and must show a goat, the host must make the offer to switch, and – as it is just only "a" show – the host is unable to communicate any additional hint on the actual location of the car. This was adopted as the basis of the "standard".

We should present the paradox and help the readers in decoding and resolving it, as per modern sources that say:

• "If all doors are equally likely to hide the car and the host is equally likely to open either door when he has a choice, then the conditional probability of winning by switching, given intial choice of player and given door opened by host, is 2/3".

Showing just in the beginning that "posterior odds = prior odds times likelihood", and then in contrast, as an eye-opener, immediately followed by the contrary "1:1 result" caused by the quite inconsistent *strange scenario* of a forgetful host who is showing the car in 1 out of 3 by just randomly opening one of his two doors, deleting the chance to win by switching in that 1 out of 3, while in the remaining 2 out of 3 (only "goat:car" and "car:goat") the odds of both still closed doors are indeed "1:1" then, as per the most intuitive common appraisal.
Any other various deviant variants, where the paradox never arises should be shown in very later sections. Gerhardvalentin (talk) 09:33, 17 October 2012 (UTC)

She gave the correct answer - the answer to the classically articulated MHP is as you state - and it is not necessary to make any assumption as to whether the game is played once or many times because the answer does not change. The result for any given game is independent of all other games.

It is historically true that when she first articulated the question, she did not say the host knows where the car is and must show a goat, the host must make the offer to switch she only said the host knows what is behind each door and the host shows a goat

I agree with you that there should be no talk of 'biased' or 'forgetful' Monty in these first paragraphs. The classic iteration of the problem has only the rules that you state above. However, I believe - as you do, if I've read you right - that to explain the paradox you must show clearly and simply that it is the rules that create the paradox. Elen of the Roads (talk) 15:14, 17 October 2012 (UTC)

Thank you, yes. But as per some editors here the result for any given game is NOT independent of all other games, because they insist hat one "can detect any host's bias" in opening his door. See their full binders over years of discussion. Although it remains "a" game show, and moreover the host can never be supposed to "show" any bias in repetitions. Giving away hints on the location of the car is completely contrary to his focus of interest. It has nothing to do with the actual question "is it to your advantage". See the standard scenario that pays regard to this fact. But they insisted, referring to MCDD. We just should be aware of that fact.

Once more: We should present the paradox, making way and assisting the reader to decode it. Redundancy welcome for this effort. And, to help the reader to grasp why the paradoxical probability to win by switching pws is 2/3 and not 1/2, we should show the difference very early, of intentionally showing "a goat in any case" – and showing a goat "just only by chance, in only 2 out of 3", where pws is 1/2.

First of all the article should be about the arising paradox, especially based on modern sources. Any sardonic historical misinterpretations are not part of the paradox, but quite another issue. Show in very later sections that there are diverging variants where the paradox does not arise. Gerhardvalentin (talk) 16:08, 17 October 2012 (UTC)

@Gerhardvalentin: If you are right and the "standard" problem has been identified with vos Savant's text, there must have gone something wrong in the discussion here. It is clarified since 1991 (John Tierney) that her original statement is ambiguous:

• Mr. Gardner said the ambiguity could be eliminated if the host promised ahead of time to open another door and then offer a switch.
• Ms. vos Savant acknowledged that the ambiguity did exist in her original statement.
• He [Monty Hall] picked up a copy of Ms. vos Savant's original column, read it carefully, saw a loophole and then suggested more trials. On the first, the contestant picked Door 1. "That's too bad," Mr. Hall said, opening Door 1. "You've won a goat." "But you didn't open another door yet or give me a chance to switch." "Where does it say I have to let you switch every time? I'm the master of the show. Here, try it again." --213.102.99.58 (talk) 15:42, 17 October 2012 (UTC)

Indeed it is possible to misread Vos Savant's text. The "standard" problem is Vos Savant's text with the clarification added which rules out this possibility of misunderstanding. Vos Savant pointed out that most of her correspondents had understood her intended meanings just fine, but still got the answer wrong! Richard Gill (talk) 10:12, 23 October 2012 (UTC)

May be that most of vos Savant's correspondents had understood her intended meanings, may be not. We don't know, but in the letters, she published, the words referring to those necessary rules are rare. Nobody should be castigated who insists on the fact that her original statement "is not well-formed" (Gardner). All the pupils and students who are able to give reasons for the 50:50 solution should be praised but not blamed for an alleged misreading of vos Savant's text, and on no conditions they should be designated as psychologically persistent persons.

A brainteaser should be clear and complete concerning its suppositions and should not depend on personal interpretation. In this view vos Savant's text is not a brainteaser but a problem which shows that people are able to start a war for nothing, in this case the war against the 50:50 solution.

If you are a teacher or a professor in mathematics and if you like your students then you should give them the "Bertrand Box Paradox" for exercise, but if you don't like your students and if you want to disturb their minds then you should give them vos Savant's original statement and insist on "switching is better in any case, because the probability of winning by switching is 2/3". --213.102.99.21 (talk) 16:27, 23 October 2012 (UTC)

Proposed text for the initial solution section, take 3

I started a new section, as the previous one is getting very long. The following is a new version of the previous proposal, trying to take into account various concerns raised. It's not minimalist anymore, and I'm very unsure which is better, the previous version or this one. I'm putting it up because I simply cannot know what others think without asking. If it's worse, then we can scratch this and build on the previous one.

Proposed text for the initial solution section, take 3

Resolving the paradox The player should switch. Switching doubles the chances of winning. [many citations] Outline Most people initially imagine door 3 as already open, leaving only two possibilities: the car is either behind door 1 or door 2. This usually leads to the incorrect conclusion that the two possibilities are equally likely, and that switching is no better than staying. The problem is actually easier if one does not focus on the two closed doors in this way. It is not necessary to know the chances of the car being behind either closed door to solve the problem. (Krauss and Wang, 2003; Gill 2011; Morgan et al. 1991) The first subsection below covers explanations that take a step back and solve the problem by showing that the best strategy is to ignore the door opened by the host and switch no matter what. The second subsection below focuses on the two closed doors, explaining why the car is not equally likely to be behind either one. The explanations may be understood as analyzes of what happens on average to a simulated player in a sufficiently long simulation. It is not implied in any way that the actual show needs to be repeated multiple times. The optimal strategy If the player is resolved to stay with their initial choice no matter what, they might as well not be offered the option to switch at all. Staying wins when the player initially picks the car door, which happens 1/3 of the time. Staying loses the other 2/3 of the time. A switcher wins where a stayer loses, so switching must win 2/3 of the time. (Adams 1990, Carlton 2005). This may be easier to grasp if the doors are grouped into player's doors (door 1) and host's doors (doors 2 and 3), as illustrated below. The player is basically given the choice of keeping what's behind their door, or taking everything behind the host's two doors. The host is saying in effect: "You can keep your one door or you can have the other two doors, one of which (a goat door) I'll open for you." No matter what the host does, a player who is determined to switch gets all cars behind the host's two doors. (Adams 1990) The situation can also be analyzed by cases, as illustrated below: Case 1: Car behind door 1 Case 2: Car behind door 2 Case 3: Car behind door 3 Player picks door 1 Player picks door 1 Player picks door 1 No matter what the host does, switching loses Host must open door 3; switching wins Host must open door 2; switching wins One case where switching loses Two cases where switching wins Switching wins in two of three equally likely cases. A player who switches no matter what wins 2/3 of the time. (vos Savant 1990b; Krauss and Wang, 2003) Many people find the situation intuitively easier to understand by considering the same problem with a million doors instead of just three. In this case there are 999,999 doors with goats behind them, and the player's chances of picking the car door are one in a million. The host goes down the line of doors, opening each one to show 999,998 goats in total, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times (i.e. the times the player initially picked a goat) the other door will contain the prize. (vos Savant 1990; Krauss and Wang, 2003) It is crucial that the host knows where the car is and is not allowed to reveal it. If he doesn't know, then switching and staying are equally good strategies. Staying still wins 1/3 of the time. However, switching also wins only 1/3 of the time. The reason is that switching wins only when the car is behind one of the host's doors (2/3 of the time) and the host nevertheless happens to open a goat door. That occurs 2/3×1/2 = 1/3 of the time when the host has no idea where the car is. (Devlin 2005) It is not actually necessary to assume that the car is initially equally likely to be anywhere, or that the host is equally likely to pick any allowed door. If the player makes their initial choice using a fair die, then the car is guaranteed to be behind the initial door 1/3 of the time, and behind the other two doors 2/3 of the time. The player's choice is still between taking one door or taking two doors, so the chances of winning by switching are still 2/3. However, with so few assumptions the chances of the car being behind any particular door are completely unknown. What is known is only the probability of winning if the strategy is followed, and that no other strategy has a better guarantee. (Gill 2011) A further look at the two closed doors Once door 3 is open, the probabilities of the car being behind each of the three doors have changed. The probability of the car being behind door 3 has dropped to zero, even though initially it was 1/3. Likewise, the probability of the car being behind door 2 has risen to 2/3. It is only the probability of the car being behind door 1 that is still 1/3. One way to see why this is so is to again group the doors into player's doors (door 1) and host's doors (doors 2 and 3), and argue as follows: Initially, the chances of the car being behind the player's door are 1/3, and the chances of it being behind the host's doors are 2/3. Imagine for a moment that the player notices only that the host opened one of his doors to show a goat, but not which door. As the host is certain to open a goat door no matter what, the information that he did so cannot change the chances of the car being behind the player's door. The chances of the car being behind the host's doors also must remain 2/3. As the car is not behind the opened door, the chances of it being behind the remaining door must be 2/3. Further knowing the opened door number cannot change these probabilities, because if it did, the answer would depend on the arbitrary numbering scheme used for the doors. The same argument does not apply when the host is ignorant of the car's location, because then it is not certain that the host will open a goat door. (Devlin 2003, Gill 2011) Another way is to consider what happens on average to players in a simulation, as illustrated below: Car behind door 2 Car behind door 1 Car behind door 3 Player picks door 1 This happens to two in six players Player picks door 1 This happens to two in six players Player picks door 1 This happens to two in six players Host must open door 3 (two in six) Switching wins Host randomly opens door 3 (one in six) Switching loses Host randomly opens door 2 (one in six) Host must open door 2 (two in six) Three in six players see door 3 opened. Two of them win and one loses by switching. Three in six players see door 2 opened. What happens to them is irrelevant to those who see door 3 opened. The above illustration shows everything that can happen to players who pick door 1. On average, three in six end up in the situation where door 3 is open (shown to the left of the thick line). Two of the three win and one loses by switching; therefore the chances that switching wins in that situation are 2/3. (Chun 1991; Grinstead and Snell 2006; Morgan et al. 1991) One may also reason as follows. There are only two possible scenarios that match the observations (the same ones shown to the left of the thick line): The car is behind door 2. The host had to open door 3 for sure. The car is behind door 1. The host happened to open door 3 by chance. The first scenario is twice as likely as the second, so the car is twice as likely to be behind door 2. (Lucas et al. 2009) The effect of the host's random choice is visualized in the illustration by the thick line.

Comments on the proposed text, take 3

Some of the concerns (many gleaned from general discussions as there weren't many comments on the previous proposal) I was trying to fix:

1. Ningauble and Martin had concerns with section names, so I changed them.
2. Elen and Martin (at least) wished for the ignorant host case to be covered, so I added it.
3. Ningauble preferred the reference to actual simulations to be elided, so I elided it.
4. Ningauble wanted the common confusion to be described before all explanations, so I moved it there.
5. Ningauble was concerned with the text appearing to take a frequentist stance, so I explicitly stated that the averages and suchlike are about simulations, which are used both by Bayesians and frequentists, and hence are completely neutral.
6. Gerhardvalentin was concerned about any appearance of the text to imply the existence of multiple repetitions of the show, so I added a clarification that it is not implied in any way.
7. Martin was concerned that the text would appear to claim that the unconditional solutions do not solve the problem, so I added an explicit claim there that they do solve it.
8. There appears to be some aversion to the word "always" in the air, so I replaced them with "no matter what", which are the words used by Krauss & Wang. Incidentally, they make much of getting people to ignore the host's behavior when he has a choice by using wordings like that ("[it] turns out to be a crucial building block of an intuitive solution").
9. Richard apparently liked the unconditional/conditional split and Krauss & Wang's explanation of what is difficult for people, so I made the text more explicit about it. Hopefully this is also clearer to the reader and other editors. It is rather nice that the same splitline is significant both cognitively and mathematically.
10. Richard likes a solution with objective probabilities and few assumptions and nobody has objected to it, so I added an explanation of it. Perhaps that also makes it clearer that the probabilities of the car being behind the two closed doors are not necessary for solving the problem, which also ought to help with Martin's concerns that those probabilities are portrayed as necessary for solving the problem.
11. Martin was afraid that the conditional solution section would be incomprehensible to some readers, so I added two other explanations there, one from Devlin/Gill and another from Lucas et al.
12. Ningauble was concerned that cases with unequal weights would require some explanation. Lucas et al.'s explanation does that. I also tried to tie it in with the pre-existing illustration.
13. Nijdam was concerned that Devlin's explanation is completely bogus, so I used Gill's version of it.
14. Martin was concerned that Devlin's explanation appears to give the wrong result for the ignorant host case. Gill's version is not as misleading, but I added an extra sentence explicitly saying how it fails for the ignorant host case.
15. My concerns: I tried to clarify the text in various ways, cover the essential approaches, keep it nicely organized and also correct. I also more or less agree with the other concerns listed above.

Comments? -- Coffee2theorems (talk) 16:02, 16 October 2012 (UTC)

Comment from Richard Gill

I like it! Relaxed style, yet at the same time precise; covers different ways people have looked at the problem in a balanced and unprejudiced way. Richard Gill (talk) 10:06, 23 October 2012 (UTC)

In my opinion, Coffee2Theorems has recently made an excellent proposal for the start of the article. In my opinion it is well-written, encyclopedic, neutral, accessible to a broad (non specialist) audience. Why is no one commenting on his proposal? Richard Gill (talk) Richard Gill (talk) 17:11, 24 October 2012 (UTC)

Comment from Gerhardvalentin

+ Agree Gerhardvalentin (talk) 12:14, 25 October 2012 (UTC)

Comment from Nijdam

I comment step by step.

1. I'm not happy with the sentence in the outline: The problem is actually easier if one does not focus on the two closed doors in this way. It is not necessary to know the chances of the car being behind either closed door to solve the problem. As you said, most people focus on door 3 opened, after door 1 has been chosen. And so they should. The solution should make clear the odds are 1:2 instead of 1:1. That's helpful. Nijdam (talk) 13:49, 25 October 2012 (UTC)

And Kraus and Wang explain why the important step which newcomers have to make in understanding that their initial intuition is wrong, is abandoning their fixation on door number. Less is more! The problem is resolved by forgetting the door numbers. Richard Gill (talk) 14:22, 26 October 2012 (UTC)

Well, if the train to Amsterdam is delayed, you can always take the train to Utrecht to arrive in time. Nijdam (talk) 15:24, 26 October 2012 (UTC)

Comment from 213.102.96.55

Coffee2theorems writes:"If the player makes their initial choice using a fair die, then the car is guaranteed to be behind the initial door 1/3 of the time, and behind the other two doors 2/3 of the time." You must not assume that the player is allowed to use a fair die in the show. And it is not self-evident for the reader that the player would use such a die and chose a door uniform at random. The article should avoid additional assumptions. --213.102.96.55 (talk) 13:50, 25 October 2012 (UTC)

Supporting this comment. The problem does not mention a die, there is no reason to believe that the player has a die, why mention a die. Should just say "ifthe player makes their initial choice at random, then etc etc" --Elen of the Roads (talk) 14:32, 25 October 2012 (UTC)

Instead of "If the player makes their initial choice using a fair die ..." IMO better: "As the distribution of the three objects behind the doors is unknown to the player, ..."  --Gerhardvalentin (talk) 16:05, 25 October 2012 (UTC)

My impression is that the whole passage from "It is not actually necessary to assume..." until "...no other strategy has a better guarantee." is basing on a die used by the player. Why not remove this section? --213.102.97.40 (talk) 17:22, 25 October 2012 (UTC)

(edit conflict) The two alternatives in Gerhard's comment precisely identify the very crux of the difference between frequentist and Bayesian interpretations of probability. IMHO, the issue of deciding which it better is best mooted by striving for a way to express this that is agnostic about interpretations of probability. Elen's suggestion is better in this regard. ~ Ningauble (talk) 19:06, 25 October 2012 (UTC)

If I am the player and if '1' is my lucky number then I would chose the left door surely. Why should I be forced to chose a door randomly? "Suppose you're on a game show, and you're given the choice of three doors." (vos Savant). I have my choice!

On the other hand, if I know the rules ahead in time and understand their consequences then I would avoid the left door and chose another door first. --213.102.98.151 (talk) 10:15, 26 October 2012 (UTC)

How about we consult the sources here, rather than presenting our personal opinions? A whole lot of authoritative writers present solutions which start by supposing that the player chooses a door completely at random. The position of the car is considered as fixed (but unknown), what the host will do if he has a choice is fixed (but unknown). The only probability ingredient is the player's choice. And a player who chooses at random and thereafter switches wins the car with probability 2/3. Von Neumann's minimax theorem shows that this is the best the player can hope for. Conditional probabilities given which door was chosen, which was opened, and whether it was a boy goat or a girl goat are irrelevant, from the strategic point of view. Which, I repeated, is a common point of view in the literature, both popular and academic.The distinguished mathematical economist Barry Nalebuff popularized MHP in the ecomoics/game theory literature before Vos Savant made it famous world-wide. Richard Gill (talk) 14:11, 26 October 2012 (UTC)

PS the smart player knows the rules in advance and therefore does *not* choose his lucky number. If he has any idea where the car is, he should choose the door where it's least likely to be. If he has no idea, he *should* pick a door at random, so that he does not have emotional capital invested in his initial choice. The smart player has read the piterature, knows game theory, and realizes that the optimal strategy is pick at random and switch. Richard Gill (talk) 14:16, 26 October 2012 (UTC)

Consulting the article:"The car and the goats were placed randomly behind the doors before the show." and "If both remaining doors have goats behind them, he [Monty Hall] chooses one [uniformly] at random."(Krauss and Wang). No words about the player's way of choice. --213.102.96.162 (talk) 15:35, 26 October 2012 (UTC)

Step by step

I would like to give the following suggestions for changes in the article:

First step: In the introduction vos Savant's answer could be cited:"Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance." The remaining part of her explanation could be part of a new section labeled "The unconditional solution", for example, because the reader doesn't know her assumptions in this moment, and therefore cannot follow her reasoning.

Second step: The illustration in the section "Solutions" above could have a better place in the section "The unconditional solution". A reader, who has in mind the sentence:"Imagine that you chose Door 1 and the host opens Door 3, which has a goat."(Krauss and Wang), could be disturbed by row 3 of the table with the car behind door 3.

Third step: Vos Savant's assumptions should be pointed out explicitly (as proposed by Rick Block and Elen of the Roads) in the section "The unconditional solution", and it should be explained, that without her assumptions the odds are the same for the remaining doors.

Fourth step: The Extended_description... could be the beginning of a new section "The conditional solution". The conditions are:

1. The contestant picked door 1

2. The host opened the goat door 3

Good night! --213.102.99.51 (talk) 21:21, 17 October 2012 (UTC)

Not sure if I would call it 'the unconditional solution' - it may not be immediately apparent what is meant by 'unconditional'. --Elen of the Roads (talk) 22:49, 19 October 2012 (UTC)

The "unconditional" solution doesn't refer to the special doors, one picked by the contestant and the other opened by the host. The corresponding solution table, presenting only one of the three possible cases (the contestant initially picked door 1), shows the host opening door 3 (row 2) or opening door 2 (row 3). For the "conditional" solution ("Imagine that you chose Door 1 and the host opens Door 3, which has a goat."), row 3 is impossible, and you need another illustration. --213.102.96.29 (talk) 15:16, 20 October 2012 (UTC)

I know what 'unconditional' means in this case. My concern is that since conditional probability hasn't been introduced at this point, people might assume that the normal meaning of unconditional applies. This would misdirect them, since Monty is operating under two conditions (using the normal meaning of the word) - that he has to open a door and that he can't show the car. Elen of the Roads (talk) 15:26, 20 October 2012 (UTC)

The 'rules' of a game are not the same as the 'conditions' occuring during the progress of the game (using the normal meaning of the words). At the beginning of a game you don't know anything but its rules. So, the rules are the game itself and they determine the possible moves of the players. The conditions are the special situations which are created playing the game according to the rules. For example:

Rule: The contestant picks a door.

Move: The contestant picks door No.1.

Condition: The host must not open door No.1

Rule: The host opens another door revealing a goat.

Move: The host opens goat door No.3.

Condition: The car is not behind door No.3.

Rule: The contestant may switch or not.

Move: ?

But if you don't like the term "unconditional" in this context, please remember that I had written above:"...a new section labeled "The unconditional solution", for example..." Feel free to propose another label for this section! --213.102.98.144 (talk) 08:46, 21 October 2012 (UTC)

I am with Elen about not calling such solutions "unconditional", but for a different reason. The word does not mean solutions that do not use or are independent of the methodology of conditional probability, it means something very specific within the methodology of conditional probability. Some writers have characterized vos Savant's solution as unconditional for polemical purposes in the context of addressing the methodology of conditional probability. Whether or not it is appropriate to describe these polemics later in the article, it should not be introduced in the early sections, and Wikipedia should definitely not be taking one side or the other on a matter of dispute.

Regarding step 3: If you are referring to the stipulation that Monty must reveal a goat behind one of the unchosen doors then please note that this is not peculiar to Vos Savant's treatment of the problem, it is common to most treatments. Any sources that do not use this stipulation are really addressing variant problems. Because this issue is such a recurring source of confusion, I am beginning to think it should be pointed out in the lede, immediately after the very first statement of the problem. (Cf. discussion of iterations of the problem definition at Talk:Monty Hall problem/Archive 30#Does this version work.) ~ Ningauble (talk) 20:45, 20 October 2012 (UTC)

I think that it should be distinguished between the answer given by vos Savant, not referring to the specific doors named in the problem text, and the solution with conditional probability given by Krauss and Wang. Vos Svant's answer, her solution table and the simple solutions should be explained in a particular section before presenting the "Extended description...", because her answer doesn't fit to the sentence:"Imagine that you chose Door 1 and the host opens Door 3, which has a goat." --213.102.98.53 (talk) 09:25, 22 October 2012 (UTC)

ASS(Another simple solution)

You have 2/3rds chance of picking a goat. If you pick a goat and switch you win. The reason to change doors is because it effectively gives you another chance to pick a goat at the start(not at the end). If you do not switch then it is 1/3rd chance to win for the obvious reasons. Or alternatively, not switching has 1/3 chance of winning ==> swithcing must have a 2/3rds chance. By casting the problem in terms of switching and not switching instead of door 1, door 2, door 3, makes the problem much easier. Since the permutations of the doors do not change the results it is best not to include them in the analysis. — Preceding unsigned comment added by 65.65.56.243 (talk) 09:31, 19 October 2012 (UTC)

This is no solution at all because you don't take into account the rules of the game for your reasoning. If the host only opens a door if you picked the car, then there is a disadvantage to switch.

The permutations of the doors may change the results, the permutations of the door numbers do not change the results, but we have to know what we are speaking about. The door numbers are a useful enumeration for knowing door 2 being another door than door 3, for example. --213.102.97.190 (talk) 12:41, 19 October 2012 (UTC)

Total symmetry proof

The door numbers are irrelevant only before the beginning of the game. If one door has been chosen by the player then the number of this door is identified with the (content of the) specific door. A subsequent renumbering of the doors would imply another arrangement of the (content of the) doors themselves. If you think that the door numbers in a specific case are irrelevant to deciding whether to switch or stay, see the following example:

Assuming that the distribution of car and goats behind the doors is randomly but fixed, the player, who picks door No.1, picks the left door in common sense, and the host, who opens goat door No.3, opens the right door in common sense, basing on the permutation (1,2,3). Another permutation (3,1,2) would lead to the situation that the player picks the middle door No.1 and the host opens the left door No.3 which has a goat. With the car behind the middle door, switching in the first case would give the car, but switching in the second case the goat. So, on one side we have a contradiction referring to the chosen door (left vs. middle) and on the other side the Indicator random variable 'I' depends on which permutation has been realised. --213.102.98.78 (talk) 11:46, 22 October 2012 (UTC)

Suppose the location of the car, the player's initial choice, and the choice of the host which door to open (when he has a choice) are all uniform at random. Full symmetry. Then the probability or conditional probability of any event is not changed by renumbering the doors. In particular, Prob(switching gives car | player chose door x and host opened door y) does not depend on x and y. Therefore, the values of the numbers x and y are irrelevant for making the choice whether or not to switch. Whether or not the remaining closed door hides the car is independent of the actual numbers on the door chosen by the player and the door opened by the host. Richard Gill (talk) 17:06, 22 October 2012 (UTC)

Please read what the article says:"Now, however we renumber the doors, whether or not switching gives the car does not change, 'I' is invariant." This is not true for any permutation. --213.102.99.174 (talk) 17:30, 22 October 2012 (UTC)

The article says that the event that switching gives the car does not change on renumbering the doors. Notation. Let (1,1,2) denote the outcome (omega, in probability theory language) car is at door 1, player chooses 1, host opens 2; let (1,2) denote car is at door 1, player dhooses 2 ( host necessarily opens 3). The event "switching would give car" consists of the outcomes (1,2), (1,3), (2,1), (2,3), (3,1), (3,2). Replace in that set the numbers 1,2,3 by any permuatation thereof, and the set remains the same.

At the point when the player is offered the choice of stay or switch, the three doors each have a different manifest (visible) role: door chosen by player, door opened by host, remaining door left closed by host. Two of those three doors also have a hidden role: which of the two hides the car. That's one part of the problem. The three doors also have numbers (1, 2, 3: as viewed from left to right by the audience). With the probability assumptions giving total symmetry in problem description (except for the specific naming "say, Door1, say, Door 3"), ie first three choices completely random as far as the game rules allow -- location of car, door of player, door opened by host -- the relationship between manifest and hidden roles of the three doors is statistically independent of the numbering of the doors (ie which door as identified by roles is which door as identified by number). Thus to decide whether to switch or stay, the numbers are irrelevant. As Kraus and Wang (psychologists) emphasize, it's the specific numbering of doors which builds the visual picture of the problem which traps people into giving the wrong answer. Under the motto "less is more" they point out that a succesful strategy in problem-solving in general consists in realizing what is irrelevant and discarding it, thereby facilitating a new, better, mental picture. I think that many amateurs (people not trained in probability theory) instinctively can and do realise that the door numbers are irrelevant and hence - given this fact - intuitively know that the simple solution is perfectly adequate. My guess is that Vos Savant herself belongs to this category of people. A quick and flexible mind, but not formally trained as a mathematician. Time and time again, wikipedia editors come up spontaneously with the same informal solution. Such people can't be blamed for not knowing the professional mathematician's code words for describing a situation like this. Symmetry and invariance under a group of transformations which leave a problem invariant is one of the most elegant way a mathematician can reduce a problem to essentials, thereby making it easier to solve. Less is more! Personally, I like it that one of the mathematician's most powerful professional tools comes up here in this popular brain teaser as also being a general tool in problem solving, cognition. Richard Gill (talk) 10:36, 23 October 2012 (UTC)

I hope you include change of goat revealed in your group of transformations which leave a problem invariant. Martin Hogbin (talk) 17:06, 23 October 2012 (UTC)

The point is to simplify a problem by showing that certain features of the problem description are irrelevant. There is no point in adding superfluous detail only to thereafter to come up with a good reason why these extra details are not important. Anyone who wants to simulate MHP starts by choosing a door number for the car, choosing a door number for the player, and so on. It's a nontrivial fact that they could have done the following: decide with probabilities 2/3 and 1/3 whether the other closed door hides the car; then, independently of this, assign door numbers (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2) or (3,2,1) to (door chosen by player, door left closed, door opened by host) each with probability 1/6. However, if you are aware of this fact, it means that the door numbers of any specific case are irrelevant to someone who has to decide whether to switch or stay. Martin: your intuitive understanding of MHP is splendid, but you seem unable to discriminate between what is given in advance and what can be / has to be derived from what is given, I take it that you "buy" the usual K&W assumptions. The irrelevance of door numbers is now a fact which can be deduced. And which needs to be deduced if you want to show, not only that "always switching" is smart, but also that there is nothing better. Richard Gill (talk) 17:50, 23 October 2012 (UTC)

I have no idea where you get, 'Anyone who wants to simulate MHP starts by choosing a door number...', from, or what its relevance to my point is.

You say above, 'I think that many amateurs (people not trained in probability theory) instinctively can and do realise that the door numbers are irrelevant...'. If I wanted to be as patronising as you I could say, 'It seems that you instinctively realise that the goat revealed is irrelevant...'.

By the way, I think it would be better to continue this discussion in my user space as you started to. Martin Hogbin (talk) 23:56, 23 October 2012 (UTC)

I am not trying to be patronizing. I want the article to be based on intuition and common sense, Reduce dogmatism and formalism. Keep MHP fun. I do not want the article deformed by the way so-called professionals think about probability problems. Who have become so accustomed to one particular formalusm that they can't even communicate with ordinary folk any more. Richard Gill (talk) 08:15, 24 October 2012 (UTC)

I am with you there. Common sense tells us that the door numbers (and goat ID) are unimportant. It was also the clear intention of vos Savant that we should completely disregard these things. So let us do that and be done with it. The simple solutions (ones that do not consider door numbers, such as that opened by the host) are complete, correct, and answer the question exactly as asked. Nothing more is needed, no K&W, no symmetry argument, no special assumptions. That is what I, and many others, have been saying for years.

The problem is you seem to be suggesting that 'mathematicians' are, in some way, not satisfied with the simple solutions, possibly for reasons that mere mortals do not understand. That seems to be where we disagree. If so, I would be happy to discuss that with you elsewhere. Martin Hogbin (talk) 09:11, 24 October 2012 (UTC)

I would prefer not to see a side discussion in userspace. It is hard enough to keep up with the ever-growing wall of text as it is. BTW, in my opinion Dr. Gill is on the right track here. --Guy Macon (talk) 09:15, 24 October 2012 (UTC)

Guy, I was only trying to stop clogging up this page with yet more discussion on the same subject. You were the one who complained about the quantity of text produced on this page. I am perfectly happy to keep all discussion here if you prefer. Martin Hogbin (talk) 17:59, 25 October 2012 (UTC)

Martin I am not trying to suggest that mathematicans collectively are not satisfied with simple solutions. There are several categories of mathematicians. There is the pedantic formalist type who only knows one way to solve MHP and thinks everything else us wrong. There is a more imaginative type who sees value in informal popular solutions and moreover sees that they correspond to other more creative mathematicians' solutions. Solutions based on fundamental structural insights, rather than on blind calculation by first principles (within the conventional formalism - which ordinary people don't know anything about). Richard Gill (talk) 10:21, 24 October 2012 (UTC)

I agree with that exactly. The only question I have therefore is why have you supported pandering to the 'pedantic formalist type who only knows one way to solve MHP and thinks everything else us wrong' rather than the 'simplist' view which appeals to both the general reader and the 'more imaginative type who sees value in informal popular solutions and moreover sees that they correspond to other more creative mathematicians' solutions'? As I have said right from the start, the simple solutions are most easily understandable by the general reader and perfectly correct (or at least defensible) to experts.

What are we doing complicating the puzzle with the so called 'conditional' solutions? These are an academic, and eventually rather pointless sideshow. Martin Hogbin (talk) 17:59, 25 October 2012 (UTC)

The important thing is whether a solution is intuitive and accessible. There are solutions which are intuitive and accessible, whiche help the reader gain insight, while at the same time being mathematically/academically equivalent to a conditional probability solution. EG: Devlin combining doors argument fixed. Initially the odds on your door hiding the car are 2:1 against. The host opens a door (you don't know which) revealing a boy goat with blue eyes and a pink ribbon around its neck. The odds on your door hiding the car remain 2:1 against because he was going to open a door and reveal a goat anyway. The specific infirmation about the colour/sex/age/religion... of the goat is irrelevant. Finally you are told that the door that was opened was door 3. The odds on your door hiding the car are stilll 2:1 against since whether or not the car is behind your door, it's equally likely for you that door 2 or 3 gets opened.

Why do you want to withhold this solution from the wikipedia reader? Already the discussants of Morgan et al. pointed out that symmetry means the difference between simple and conditional solutions is essentially a matter of taste.

Do you agree that the chance of winning by switching is 2/3, whatever door the player picks and whichever door the host opens? Do you agree that it is not difficult to explain why this is the case? Why do you want to keep this fact hidden from the reader of wikipedia? There is nothing academic or formalistic about saying "the chance of winning by switching is 2/3, whatever door the player picks and whichever door the host opens". Solutions which deliver this, in particular which add the "whatever", come to a more powerful conclusion than solutions which don't deliver this extra. Some readers will appreciate the extra mileage. Others can ignore it. A matter of taste. See Bell's discussion of Morgan et al. Richard Gill (talk) 14:32, 26 October 2012 (UTC)

Proposal to add a link to the External Links section

I propose allowing the following to be added to the External Links section:

→ Three Doors Probability Problem (Simulator to go through 100 simulations in a couple of minutes) ←

I have twice introduced the link and had my edits undone. I believe the link has strong merit because of its utility and is relevant to anyone who is struggling with, and studying, the Monty Hall problem. I link it as a valuable resource of functionality at a level that is not duplicated in the other links already there.

Though I did create the content on the pages (over a decade ago), I think it stands on its own merit and its self-evident utility in facilitating learning through experience. I have looked at the other simulators and I don't see why anyone would see the NYT version as even being equal, let alone providing an experience superior to the one my simulator offers. What the web application I have linked to does is allow people to quickly go through a bunch of Monty Hall problem-like games in real time while stats are kept for them. 100 games. They can switch or stay or change for every round. No waiting for the next game to load, since all 100 are loaded and visible at the onset.

I am currently in a Model Thinking class on coursera.org where more than a few people did not get the logic that is so obvious to you and me. Graphs mean nothing. Logic means nothing. If you have known this problem a long time then you know what I mean. They have to smash their heads against something that doesn't allow them to hold onto their erroneous constructs. And it has to be accessible enough to keep them engaged.

According to my calculations, if you run through the problem for 100 games, there is a 95% chance you will wind up with more than 57 wins if you switch every time. And a 99.7% chance that you will get 53 or more out of 100. But with 10 games, there is about 32% chance that you will get 5 or less wins by switching every time. That is based on my recent learning of standard deviations, so you can correct me if I am incorrect here. If it is correct, that means that if someone goes the NYT version for 10 tries (I found it to be very slow), they may well find it is 50/50, based on their limited experience, and be burned out from the transaction time. Simply put, the feedback from my simulation is instantanous with no waiting for a new game to load.

So why not allow the link above to be added to the External Links Section? Anyone in favor of this, please respond. And anyone opposed, please articulate why.

Thank you,

Reidme (talk) 18:59, 22 October 2012 (UTC)

In favour. NYT uses flash, doesn't work on iPad. Wolfram is dull. This one is good. Richard Gill (talk) 06:48, 23 October 2012 (UTC)

PS The probability an always-switcher wins 59 or more of 100 games is 95.66%. The probability he wins 53 or more of 100 is 99.83%. the probability he wins 5 or less of 10 games is 21.31%. (Binomial probabilities computed with "R"). Why report approximations, when you can easily report the exact probabilities? There are webpages which compute exact binomial probabilities for you, as well as state of the art free statistical software. Richard Gill (talk) 17:48, 24 October 2012 (UTC)

Richard, Excellent! Will correct my simulator page accordingly and include links to two binomial probability calculator pages. Reidme (talk) 02:10, 26 October 2012 (UTC)

Thank you, Richard. I posted the standard-deviation based binomial probabilities for this on the wiki page for the Model Thinking at coursera.org. The SD work I did changed my understanding of what we are dealing with when people don't quickly get it. Do you think we should inject this information into the immediate Article page? Reidme (talk) 17:35, 23 October 2012 (UTC)

By SD based I suppose you mean that you used the mean n p and the standard deviation sqrt ( n p (1-p)) of the binomial distribution, and the Gaussian (normal) approximation, to compute tail probabilities approximately. I agree with you that just playing 10 times isn't enough, but 100 times is. So your webpage is a lot better in several respects than that of NY Times. Also better than Wolfram's (more interactive, accessible,.... ). Richard Gill (talk) 17:59, 23 October 2012 (UTC)

@Reid: I should however change it in such a way as to comply with the problem. I.e. the simulation should only count cases with the same chosen and opened doors. Nijdam (talk) 18:38, 23 October 2012 (UTC)

@Nijdam, The first indented paragraph in the Article ends by posing the question, "Is it to your advantage to switch your choice?" So by my reckoning, that is what the simulator should be addressing. Nonetheless, it does count all cases and infers the advantage/disadvantage for every completed game if one had switched every time, even if they didn't. Reidme (talk) 20:45, 23 October 2012 (UTC)

Richard, Yes, that is what I mean by SD. But I think I will leave it to anyone else who may want to get the underlying point across if they wish. Thank you for your support! Reidme (talk) 20:45, 23 October 2012 (UTC)

@Reidme, Nijdam thinks your simulation does not address the question at all. He thinks that the question is, should you switch on those occasions when you chose Door 1 and the host opened Door 3. Supposing the website visitor makes the possible initial choices equally often, only one sixth of your simulation are relevant, according to his point if view. He would like six times as many trials and separate statistics on six different cases.

@Nijdam, a simulation in which the player always chooses Door 1 and the host always opens Door 3 is highly artificial. It would require "throwing away" all the times when the host wants to open Door 2. And the player would be forbidden to make other choices himself. You could ask for separate statistics for each of the six cases "door chosen by player, door opened by host". But then the experiment should be a whole lot larger to ensure reilable statistics for all 6 cases. I don't think this will result in a popular or useful website. If you don't believe me, why don't you design one yourself? At last, a popular visual interactive introduction to the conditional approach? Richard Gill (talk) 07:55, 24 October 2012 (UTC)

I know this all, but it should be made clear what situation the simulation simulates. And till now my experience is that most people think such simple simulation are correct simulations of the (standard) MHP. And, Reid, the question is not whether switching in general is advantageous, but whether it is in YOUR advantage to switch, referring to the player who is addressed. And they know which door they have chosen and which door is showing a goat. Nijdam (talk) 10:26, 24 October 2012 (UTC)

Nijdam, the simulation allows you to select whatever door you like. No matter which door you selected, and no matter what door has been opened by the host, the result will always be identical. No matter what door you did select and no matter what door the host opened. You never are able to say that – for any "actual game" – the chance to win by switching definitely "IS" below 2/3.

• you select door 1 and host opens door 2 showing a goat: staying wins in 1 out of 3 whereas switching wins in 2 out of 3
• you select door 1 and host opens door 3 showing a goat: staying wins in 1 out of 3 whereas switching wins in 2 out of 3
• you select door 2 and host opens door 1 showing a goat: staying wins in 1 out of 3 whereas switching wins in 2 out of 3
• you select door 2 and host opens door 3 showing a goat: staying wins in 1 out of 3 whereas switching wins in 2 out of 3
• you select door 3 and host opens door 1 showing a goat: staying wins in 1 out of 3 whereas switching wins in 2 out of 3
• you select door 3 and host opens door 2 showing a goat: staying wins in 1 out of 3 whereas switching wins in 2 out of 3

You can descry this without any "additional reasoning". And remember: It is just on ONE game show, so no need for unfounded assumptions on uncertain unknown empirical values. It is an illusion that "one still could know better" for the actual game, and it's insincere to pretend "one still could know better" for the actual game, by means of maths. You never can nor will know better.

Let us stop chasing after "before and after the host opened a door" in that one-time game show, although knowing that sources say that the number of the door opened does not give you any "new" information. Millions of words and binders full of "but what if the host ...",  in just disliking the literature. Let us stop confounding the famous "paradox" with school class textbooks on teaching and learning conditional probability theory. Gerhardvalentin (talk) 12:09, 24 October 2012 (UTC)

Gerhard, Nijdam knows that the chance of winning by switching is 2/3 in all six situations. His point is that Reidme's simulation does not demonstrate this. Nijdam thinks that the only correct way to solve MHP is to mathematically prove this fact. And that a simulation of MHP has to demonstrate this too. Not everyone agrees. Fortunately, what you or I or Nijdam think is irrelevant. What's important is what is found in the literature. Wikipedia is an encyclopedia of generally accepted facts, not an oracle containing only The Truth. If Nijdam wants to promote an idiosyncratic understanding of MHP, he had better publish articles about it in the academic literature, and get journalists to report his breakthrough in popular literature as well. After about 10 years, when every popular writer writes about MHP in Nijdam's fashion, wikipedia will be able to follow.

I think that distinctions which are important in a university probability class, are not necessarily of much interest to an amateur. The first thing to get across to newcomers to MHP is that switching is advantageous. The simple solutions do that. The rest can be considered an academic refinement. Important for experts. Important in a probability class. In the article, this distinction can be discussed later.

However, fortunately there are simple ways to present what is in effect a conditional solution. So the reader can be presented with a range of different solutions, making different assumptions, getting correspondingly weaker or stronger conclusions. The general reader will find for themselves what solution most appeals to them (a matter of taste). Most importantly, they will learn that there is not One And Only One correct way to think about MHP. An informal verbal question about an imaginary game show can be converted into a formal logical or mathematical problem in many different ways. The literature reflects this fact of life. The wikipedia article has to reflect the literature.Richard Gill (talk) 17:41, 24 October 2012 (UTC)