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Canonical assumptions

Instead of the text from K & W, I suggest using the more structured assumptions from Rosenhouse p. 156:

1. The doors were initially equiprobable.
2. Monty never opens the door you initially chose, and he reveals a goat with probability 1.
3. Monty chooses his door randomly whenever he has more than one option.

(It's an actual quote). Tijfo098 (talk) 13:48, 10 April 2011 (UTC)

I think that's a good idea, moreover as long as we focus on the mathematical aspects/ambiguity of the problem citing a math source is more appropriate.--Kmhkmh (talk) 15:36, 10 April 2011 (UTC)

I think 1 is rather vague, we all know what it means but it is not clear. In this subject it is best to make things crystal clear. There is no reason that we cannot state the assumptions in our own words provided that they are all unequivocally supported by reliable sources. Martin Hogbin (talk) 15:34, 10 April 2011 (UTC)

An equivalent formulation of a source content would be fine as long as the editors agree on it.--Kmhkmh (talk) 15:39, 10 April 2011 (UTC)

Yes, 1 could use some clarification. I'll see if I can dig a better source (possibly some other page in Rosenhouse). My WP:OR is that (1) means (1b) the player initially picks the door hiding the car with [equal] probability 1/3. Neither the car placement nor the player's first door choice need to be actually random, as long as (1b) is met. Tijfo098 (talk) 16:46, 10 April 2011 (UTC)

Why not take 1 literally? To wit: every door may hide the car with equal probability. This is a neutral statement that applies regardless of whether it is read as (a) a rule of the game (i.e. a constraint on where the car may be placed before the game is played), (b) the state of knowledge of the player prior to making a choice, (c) the strategy used by the player to make his initial choice (d) (perversely, IMHO) as a long-run frequency of car placements over many games, (e) ...? I may be misinterpreting Rosenhouse here, but I suspect that he chose that form of the statement exactly because it elegantly avoids unnecessary considerations of strategy or, worse, interpretations of probability. glopk (talk) 18:57, 10 April 2011 (UTC)

If you say that the doors are intially equally likely to hide the car, and that Monty is equally likely to open either door when he has a choice, you are being strictly neutral with regard to probability interpretation, but you are already biased to solutions which require full probability assumptions "host side". Many people solve MHP without these assumptions at all, they only assume that their own initial choice was random. Please note that "chooses randomly" is ambiguous, I suppose Rosenhouse (who is not a probabilist or a statistician) means "completely randomly". Anyway "chooses (completely) randomly" biases the reader to the frequentist interpretation. For the subjectivist, the uncertainty is in the mind of the player, not in the brain or action of the host. Richard Gill (talk) 07:47, 11 April 2011 (UTC)

PS, stating assumptions is fine, but do not call them THE assumptions. They are a commonly made set of assumptions, but not unversally made. The simple solutions don't even use all of these assumptions, which means that they are stronger (in the sense of being more widely applicable). Richard Gill (talk) 07:51, 11 April 2011 (UTC)

Analyses of Glopk and Rick Block's arguments against Martin Hogbin's proposed structure

Rick Block and Glopk listed their policy-based concerns in their straw poll oppose statements. The reader is invited to go there now so as to see the arguments in context, and to consider voicing an opinion in the straw poll. The current count is five support Martin Hogbin's proposed structure, two oppose, one abstains.

The arguments are as as follows (comments mine):

Responses inline in italics. -- Rick Block (talk) 15:25, 12 April 2011 (UTC)

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Argument: Martin Hogbin's proposed structure violates WP:STRUCTURE:

Argument fails because it either requires the reader to pretend the words "based solely on the apparent POV of the content itself" are not contained in WP:STRUCTURE or it requires an assumption that both Martin Hogbin and Guy Macon are lying when they say that their motive is not based solely on POV of the content, but is instead based upon a desire to conform to WP:TECHNICAL.

Using "pretend" here is pejorative and this entire comment is simply a false dichotomy. The proposed structure does separate content based solely on the apparent POV of the content - meaning that only, i.e. solely, content promoting "simple" solutions and, thus, the POV that these solutions are adequate to address the problem, appears first. What you effectively are arguing is that WP:TECHNICAL supersedes WP:NPOV, or that it's OK to violate WP:NPOV as long as your intentions are pure, and this is manifestly not the case (from WP:NPOV): "The principles upon which this policy is based cannot be superseded by other policies or guidelines, or by editors' consensus." -- Rick Block (talk) 15:25, 12 April 2011 (UTC)

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Argument: Martin Hogbin's proposed structure violates WP:WEIGHT:

Argument fails because WP:WEIGHT specifically covers giving some views a larger or more detailed description, and says nothing about structure (which WP:STRUCTURE, being directly above WP:WEIGHT, already covered.)

This is a separate objection from the previous one about the section ordering. This one is based on the proposed structure devoting an entire section with several subsections to "simple" solutions (promoting the POV that these solutions are adequate to address the problem), while the predominant approach used in the relevant academic (math) sources, i.e. a solution based on conditional probability, is relegated to a single (later) subsection ("More detailed and comprehensive solutions") of a section titled "Academic criticism of the simple solutions" (!!??) which is then followed by yet another section titled "Criticism of the criticism". Not only is this overweighting the "simple" solutions based on sheer amount of text, it is clearly suggesting ALL conditional probability based solutions are disputed which is the exact opposite of the actual situation. -- Rick Block (talk) 15:25, 12 April 2011 (UTC)

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Argument: Martin Hogbin's proposed structure violates WP:TECHNICAL (oversimplifying clause):

Argument fails because it is not falsifiable. One could respond to any and every application of WP:TECHNICAL by saying that putting the simpler material near the top is oversimplifying.

See the following comment -- Rick Block (talk) 15:25, 12 April 2011 (UTC)

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Argument: WP:TECHNICAL does not apply because Krauss & Wang's conditional probability solution is as easy to understand as the simple solutions. Any disagreement with this assertion is simply an editor's opinion and not supported by any reliable source:

Argument fails because it is not falsifiable and works both ways; those supporting Martin Hogbin's proposed structure could just as easily claim that Krauss & Wang's conditional probability solution is harder to understand than the simple solutions and that any disagreement with this assertion is simply an editor's opinion and not supported by any reliable source.

The argument is NOT that K&W's conditional probability solution is as easy to understand as the simple solutions, but that K&W provide empirical data suggesting that "simple" solutions are not so simple to understand as opposed to the assertion of various editors that they are simple to understand which is based only on these editor's opinions (not emprical data). -- Rick Block (talk) 15:25, 12 April 2011 (UTC)

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Argument: WP:TECHNICAL does not apply because Krauss & Wang's conditional probability solution is as easy to understand as the simple solutions and that this is supported by a reliable source: Krauss & Wang say that the "simple" solutions are "inaccessible" (quotation marks in Rick Block's original comment):

Argument cannot be verified. A search of the Krauss & Wang paper shows no use of the word "inaccessible", and a careful reading of every instance of the words "simple" and "Savant" show no such claim. Given recent history (example; misinterpreting Ohiostandard saying that a table should be included somewhere as saying the table should be included in the simple section), I would have to see an exact quote from Krauss & Wang so I can analyze this claim in detail.

Krauss & Wang describe an experiment where they tested university students who agreed to participate in a study (which alone makes it problemetical to apply the conclusions to Wkipedia readers), but the experiment was not a comparison between a simple solution and a conditional probability solution. It was a comparison between asking the participant whether to switch vs. first asking some leading questions which the authors call "guided intuition." The experiment does mention the word "accessible", so perhaps that is what Rick was referring to.

The claim is not that conditional solutions are necessarily easier to understand (indeed, as Tijo098 points out, in experiment 3 they show data suggesting that a conditional solution using Bayes Rule cannot be applied to similar problems - but this in no way means all explanations rooted in conditional probability are hard to understand), but rather that the cited experiment shows that understanding the simple solutions requires a different mental model of the problem than most people (97% in their study) initially create - and that changing mental models is difficult. Indeed, keeping all mention of conditional probability far, far away from the initial "simple" solution seems to be the precise intent of the proposed structure - i.e. the proposed structure seems to be intended to convince the reader that their original mental model is wrong so that they can then comprehend the simple solution. By doing this the article would be explicitly endorsing the POV expressed by one editor above - "The common mental picture is misleading, if not wrong." The text of a solution section I've drafted (above) presents both "simple" solutions and what is intended to be an accessible conditional solution, and deliberately relates both types of solutions to the mental model K&W's data says most people construct (where the player has picked door 1 and has then seen the host open door 3). I don't have empirical data that says this specific text is more accessible to most people, but I strongly suspect that this approach is. It is a minor variation on the approach used in Grinstead and Snell, but without the pejorative comments about the "simple" solutions not actually addressing the problem. Since it is also obviously a much more NPOV approach I fail to comprehend why there is resistance to this sort of approach. -- Rick Block (talk) 15:25, 12 April 2011 (UTC)

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Guy Macon (talk) 10:02, 11 April 2011 (UTC)

I agree that understanding MHP requires changing the mental model. People see before them (in their mind) two doors closed, one opened. Very, very subsidiary to this, they do know that they chose one of the two closed doors, and that the host opened the door which was opened... but they forget the past history, and have in their mind only what their eyes see in front of them at the end of the history. This mental picture makes them think that they needn't switch. In order to be convinced that they ought to switch, they need to change their mental picture. The so-called simple solutions are ways to make you revise your mental picture. For instance the very simple idea that two-thirds of the time your initial choice is wrong, requires you to change your mental picture to the picture that you are standing at a door and behind the door is a goat!!! The host opens the other goat door and you switch in order to pick up the car. To strengthen this change in mental picture, imagine there are a hundred doors. You are pretty certainly not standing in front of the car-door. The door which the host leaves unopened almost certainly is the car-door. This is how you convince people that switching is smarter than staying. Which it is: switching wins you the car 2/3 of the time, staying wins you the car 1/3 of the time (overall, ie not split out over possible door numbers). All this true, assuming *only* that your initial choice is right 1/3 of the time. I go to a frequentistic picture ("1/3 of the time") in order to make it more concrete in your mind.

Learning formal probablitity calculus and Bayes' theorem doesn't change mental pictures. It's a dull way to avoid making mental pictures. It provides a mechanical way whereby we can find a solution which is guaranteed to be correct independently of any mental pictures we might have.

The simple solutions, on the other hand, provide an alternative mental picture which jolts us to realising that switching is a smart strategy.

Once we have realised that always switching is far far superior to always staying, further analysis is for most people a waste of time. However, the conditionalists have a point: how do we know that switching is better than staying not just "on average" but also for every one of the six possible situations we might be confronted with? (door initally chosen, door opened by host). If you are interested to know this, then I can tell you four or more quite distinct arguments. All of them require a small level of abstraction or sophistication. Most people are not interested in any of them. If you are a student in a probability class then you need to learn how to get the answer by routine operations which will satifsy the teacher and give you full marks. There is one way which is foolproof, but unenlightening: it is, use Bayes' theorem. However you need to learn some elementary probability calculus for this.

Fortunately, there is a clever way to combine both the insights of the simple solutions and the mechanics of Bayes theorem. That is by thinking of information as coming to us in stages.

A: We choose a door. To be specific: let's suppose it was door 1. Nothing changes in our beliefs about the location of the car, when we make a choice of a door. Our subjective evaluation of the chances are that the car is behind our door, number 1, with probability 1/3, just as they were before we chose that door.

B: The host opens "a" door revealing a goat (but not telling us which door he opened). *Nothing* changes!!!! (He could do this anyway and he was going to do it anyway. He did do it. No chances have changed!)

C: The host tells us it was actually door 3 he opened! This tells us nothing about door 1's chances to hide the car, by symmetry! (Here I am using probability in the subjectivist sense: it's in my own (lack of) knowledge of the situation, not in the physical devices used to make choices). So door 1 must still hide the car with probability 1/3. On the other hand, door 3's chance to hide the car has collapsed to zero, so door 2's chance must have jumped from 1/3 to 2/3.

<rant> I just don't see a big deal. Two years fighting have gone by because some guys wrote in a low-status statistics teachers' magazine that the intuitive solutions of people like Vos Savant were wrong. They rewrote Vos Savant's question in order to do so. That made her bloody angry, and several smart people defended her. A lot of other folk were delighted to at last have a fun example of application of Bayes' theorem, and copied those guys' rewriting of Vos Savant's question together with those guys' rather boring solution in their elementary textbooks. Yet other people said that the first guys were arrogant bastards and that it was all much easier than they suggested. Just a few other people said that there were a whole load of other ways to approach the problem, all of which actually give insight into the paradox. </rant>

Additional remark: thinking of information as coming in piece by piece is also a change to the mental model of the problem. We solve a problem by simplifying. The immediate (intuitive) reaction of most people to MHP is to simplify it (ignore the history, just look at the stage) and this leads to the wrong answer. The simple solutions simplify it in a different way (ignore the detail of the specific door numbers). One can marry the "full" conditional probability Bayes theorem solution to the simple solutions in a number of ways, for instance as I did above, by adding the detail of the door number (host opened 3 not 2), at the very last step. One can also mention explicitly in the simplifying stage of forming the "good" mental picture, that by symmetry the numbers are irrelevant. Why make a big thing out of it? Richard Gill (talk) 05:27, 13 April 2011 (UTC)

Guy asked if I might explain the difference between the simple solutions and the conditional solutions. The difference is whether you stop at stage B or go on to incorporate also stage C. Most people have no interest in going to the bitter end. For those who do want to, it is an easy final step. That's all there is to say. Richard Gill (talk) 20:38, 12 April 2011 (UTC)

I hope you are aware that this is "nice" polemic that barely manages to avoid openly telling false things (politely put). Starting with "rewriting vos Savants question" (in reality it is Whitaker's question and who was allegedly rewriting/rereading what and why it matters or not is actually at the core of a part of the debate) to "Bayes (or conditional probabilities) not changing the mental picture". Of course they do, depending on how you make the Bayesian argument you place the contestant behind the doors as well (just performing a different calculation from that view) and you shift the focus towards the host's behaviour which is another change of the mental picture. The criticism of vos Savant's approach as incomplete (rather than false) stems not only from Morgan's article in that "low level magazine" way back but it fact it ranges over a variety of different publications all the way to the recent most comprehensive publication on the subject yet (Rosenhouse's book).

I don't want to comment in detail on the rather subjective (and imho just distracting, not helping and to some degree ridiculous) framing of the issue ("boring", "arrogant bastards". "low level", "smart guys"). Such descriptions might be fine and possibly entertaining for polemic or tongue in cheek essay on MHP, however for encyclopedic articles and their discussion it is inappropriate (as far as the article is concerned) and usually not helping (as far as the discussion is concerned, in particular if it is a heavily contested one anyhow). It really irks me to read something like that, in particular if it comes from a smart mathematician, who should know better.--Kmhkmh (talk) 21:49, 12 April 2011 (UTC)

Sorry, we should delete the polemical next to last paragraph I wrote, Preparatory to that, I put it in small print. It was my personal interpretion of the history. But please recall that Whitaker's original question doesn't mention any door numbers at all. I think it supports Vos Savant's intended meaning of her question. Richard Gill (talk) 05:11, 13 April 2011 (UTC)

If you want remove the last postings (including mine) that's fine by me, they are not bringing the discussion forward anyhow. Nevertheless I'd like to point out that Whitaker's original question (as quoted in our article and as described on vos Savant's page) does contain the door numbers (and always had). You can of course still argue that the door numbers might not matter and that they did not matter for Whitaker who was using them merely for illustrative purposes, but then again it is only a "can" and not a "must" and how appropriate the "can" is, was again at the core of some parts of the "academic" debate.--Kmhkmh (talk) 06:11, 13 April 2011 (UTC)

Sorry Kmhkmh, the last (2010) article by Morgan et al quotes from Whitaker's actual letter to Vos Savant. No mention of any door numbers. Whitaker is comparing Monty Hall to Monty Fall and asks if Marilyn agrees that in the one case you should switch the other not. Vos Savant later insisted that her side remarks "say Door 1" and "say Door 3" were not part of the question, this is mentioned by Rosenhouse in his book. English language semantic ambiguity. Possibly non-native speakers might not be aware of the two possible readings of her words. Anyway, this too is a side issue. Vos Savant asks a question in English with all its ambiguities and vagueness. Different people have converted this into different formal mathematical questions. Fortunately the difference between the two main interpretations is slight. It's merely whether or not you mention explicitly that the actual numbers on the doors are irrelevant to deciding whether to switch or stay. I am assuming here a subjectivist understanding of probability so that the two "equally likely" probability specifications follow from the problem description and do not need to be added as a separate auxiliary assumption. The frequentist cannot solve MHP without adding additional information. Vos Savant doesn't use the word probability or make probability assumptions so she forces the reader to solve the problem with the natural subjectivist (Laplacian) assignent of probabilities. Richard Gill (talk) 06:59, 13 April 2011 (UTC)

Ah ok, you were referring to the until recently unpublished "original" letter being different from the version published in vos Savant's column, I misunderstood that, in that context "vos Savant's question" (=vos Savant's formulation of Whitaker's problem) has different meaning too I guess. I don't think though that this in Rousenhouse's book though, as that was published in 2009. Or if it is nevertheless could you tell me the page please? This version has indeed no door numbers, but is not published version that started the debate and hence I agree somewhat of a side issue.--Kmhkmh (talk) 11:10, 13 April 2011 (UTC)

Yes. Rosenhouse doesn't know about the original Whitaker letter, indeed. But he does refer to Vos Savant's later writing what her intention was with the words "..., say Door 1" and "..., say Door 3". Maybe this is in her own book, I haven't read it yet (and don't have it). Of course her defence of herself came after the attack by Morgan et al., so we don't know what she really did mean at the moment she published her story. Nijdam thinks she was caught out by Morgan and mates. Who can tell?

But, I'm afraid I can't tell you right now where Rosenhouse said that. I'll have to read his whoe book all over again to find the reference, since at the time I did read it, I forgot to note it down.

Anyway, does it matter? The important thing is perhaps how the average reader understands Vos Savant's question, not what her own intention was or what she later claimed it to be. But you'll notice how many authors jump to read into Marilyn's question things which are not there. For instance, Grinstead and Snell write "Other readers complained that Marilyn had not described the problem completely. In particular, the way in which certain decisions were made during a play of the game were not specified. This aspect of the problem will be discussed in Section 4.3. We will assume that the car was put behind a door by rolling a three-sided die which made all three choices equally likely. Monty knows where the car is, and always opens a door with a goat behind it. Finally, we assume that if Monty has a choice of doors (i.e., the contestant has picked the door with the car behind it), he chooses each door with probability 1/2. Marilyn clearly expected her readers to assume that the game was played in this manner." I disagree that Marilyn expected her readers to assume anything like this at all. Even Monty Hall himself writing to Selvin 10 years earlier did not mathematize the problem in this way. And how do Grinstead and Snell know that Marilyn is a frequentist at heart? And if she was, why didn't she tell her readers her frequentist assumptions? (A frequentist can't do anything without making assumptions, while a subjectivist (Laplacian) can and does). Then they go on and write "We begin by describing a simpler, related question" as introduction to the problem which many editors call the simple problem. Then they write "This very simple analysis, though correct, does not quite solve the problem that Craig posed. Craig asked for the conditional probability that you win if you switch, given that you have chosen door 1 and that Monty has chosen door 3." But Craig does no such that thing! He does not ask for a probability, nor does Marilyn! Richard Gill (talk) 12:37, 13 April 2011 (UTC)

Well I agree such comments are annoying and somewhat misleading. On the other you can't quite fault the authors for it, since at least until 2010 (and probably even now) most people did not know Whitaker's original wording but simply based their (maybe somewhat pretentious) judgement on the publication in vos Savant's column (Parade Magazin). Another thing is that even into Whitaker's original wording one can read ambiguities and pursue different solution strategies. Personally I'd agree that the "canonical" MHP and Monty Fall is most likely what he had in mind, but then again we don't know for sure. Morgan's intepretation of Whitaker's words is at least possible though indeed it might appear a bit contrived, Georgii's approach (bayesian angle) however still looks appealling in particular since Whitaker explicity refers to actual Monty rather than formulating an "abstracted" game show riddle. More importantly before ultimately judging Whitaker's original words, I think we would need to see his complete letter rather than having only this snippet by Morgan without knowing whatever else might have been written before or after. I agree that there is no reason to peg vos Savant as frequentist, but there might be none to peg her as a Laplacian or Bayesian either. I agree that solving a problem with no (additional) assumptions (or better minimal assumptions as we still need to exclude Monty Fall) has a certain elegance, but then again making "reasonable" assumptions has always been a basic tool of solving or modelling "practical" problems.--Kmhkmh (talk) 13:29, 13 April 2011 (UTC)

My theory:

I dont beleive everyone else, This is my theory: alright, now lets say each O is a door-

door #'s: 1 2 3

doors: O O O

probability: 1/3 1/3 1/3

okay each door has a 1/3 chance; but the two on the left have 66% chance.correct, So now say we pick one for example, door 3; awesome possum, so we say that it has a 33% chance of it being a car. So now we open door 2 and it's a goat. Now some think that door one still has a 66% chance and door 3 has 33% chance, but thats not true because now we only have 2 numbers; therefore it's 50/50 chance between the two and I was looking at this and they were using larger numbers to try and explain it, this is what I think:

door #'s: 1 2 3 4 5 6 7 8 9 10

doors: O O O O O O O O O O

So each door has a 10% chance now say we predict door 9; at this point, you only have a 1 in 10 chance of getting it right. now next step: lets say we open all the doors except for door 7 and door 9. Now you would think it's in door 7, BUT thats not true now we just have two doors and TWO doors only not 10 but Two:

door #'s: 7 9

Doors: O O

okay and now they both have a 50/50 chance. PROBLEM? Well then,This is what I think, my friend tells me I'm crazy but I just don't know. —Preceding unsigned comment added by 68.225.14.69 (talk) 06:06, 13 April 2011 (UTC)

You are supposing just any door got opened and it just happened by chance to show a goat. Then the chances of the other doors jump to 50%, 50%. But the point of MHP is that the guy who opens a door knows where the car is hidden and always opens a door different from your and with a goat behind it. He deliberately shows you a goat, not coincidentally. The chance your initial door hides a car doesn't change since the host is certain to open a door and show a goat, whether or not your door hides a car. And "which" of the two doors he opens is 50-50 whether or not your door hides a car, so this little piece of news doesn't tell anything about the chances your initially chosen door hides the car, either. Of course, the chance that the car is behind the opened door collapses to to zero, and the the chance that the car is behind the other closed door jumps up to 2/3. The host didn't open it. Maybe, because there is a car behind it. In fact 2/3 of the time your initial choice of door was hiding a goat and so it is twice as likely that the host's choice is forced than that it is free. Richard Gill (talk) 06:47, 13 April 2011 (UTC)

BTW there is an "arguments" page and a "FAQ" where this point is mentioned. It's a common misunderstanding of the MHP. Richard Gill (talk) 07:04, 13 April 2011 (UTC)

And BTW, the couple of sentences I have just written solve MHP in plain words, *completely*, incorporating the difference between the simple and the conditional solutions. That's the parenthetical addition "and "which" of the two doors he opens is 50-50 whether or not your door hides a car, so this little piece of news doesn't tell anything about the chances your initially chosen door hides the car, either", which is just as parenthetical as Vos Savant's "say, Door 1", and "say, Door 3". Richard Gill (talk) 07:15, 13 April 2011 (UTC)

Yet another editor who needs to have it explained to them why the answer is 2/3 and not 1/2. This is what nearly everyone wants to know and what the article should make clear at the start. Martin Hogbin (talk) 17:54, 13 April 2011 (UTC)

Indeed. Another editor thinking of a specific (conditional) example, who has clearly read the "simple" solution and finds it completely and totally unconvincing.

Many people are only convinced by doing some kind of simulation themselves. They'll simulate the simple solution. And quickly become convinced. I'm not sure if anyone is convinced by formula manipulations. These showed to that famous pure mathematician Erdos that the naive answer is wrong but he still was not convinced. Of course, he wasn't a probabilist or a statistician, he was a pure mathematician and number theorist. Some people will never be convinced. Richard Gill (talk) 17:02, 14 April 2011 (UTC)

If the original poster is checking in here, can you try reading the "Solution" section suggested above (repeated here so you don't have to go looking for it) and let us know what you think? Thank you very much. -- Rick Block (talk) 18:51, 13 April 2011 (UTC)

Proposed text/structure Rick offers to satisfy both NPOV and WP:Technical

Solution Different sources present solutions to the problem using a variety of approaches. Simplest approach The player initially has a 1/3 chance of picking the car. The host always opens a door revealing a goat, so if the player doesn't switch the player has a 1/3 chance of winning the car. Similarly, the player has a 2/3 chance of initially picking a goat and if the player switches after the host has revealed the other goat the player has a 2/3 chance of winning the car. (some appropriate reference, perhaps Grinstead and Snell) What this solution is saying is that if 900 contestants all switch, regardless of which door they initially pick and which door the host opens about 600 would win the car. Assuming each specific case is like any other, this means a player who initially picks Door 1 and sees the host open Door 3 wins the car with a 1/3 chance by not switching and with a 2/3 chance by switching. Enumeration of all cases where the player picks Door 1 If the player has picked, say, Door 1, there are three equally likely cases. Door 1 Door 2 Door 3 result if switching Car Goat Goat Goat Goat Car Goat Car Goat Goat Car Car A player who switches ends up with a goat in only one of these cases but ends up with the car in two, so the probability of winning the car by switching is 2/3. (some appropriate reference, perhaps vos Savant) What this solution is saying is that if 900 contestants are on the show and roughly 1/3 pick Door 1 and they all switch, of these 300 players about 200 would win the car. Assuming the cases where the host opens Door 2 or Door 3 when the player picks Door 1 are the same, this means a player who initially picks Door 1 and sees the host open Door 3 wins the car with a 1/3 chance by not switching and with a 2/3 chance by switching. The probability of winning by switching given the player picks Door 1 and the host opens Door 3 Tree showing the probability of every possible outcome if the player initially picks Door 1 This is a more complicated type of solution involving conditional probability. The difference between this approach and the previous one can be expressed as whether the player must decide to switch before the host opens a door or is allowed to decide after seeing which door the host opens (Gillman 1992). The probabilities in all cases where the player has initially picked Door 1 can be determined by referring to the figure below or to an equivalent decision tree as shown to the right (Chun 1991; Grinstead and Snell 2006:137-138 presents an expanded tree showing all initial player picks). Given the player has picked Door 1 but before the host opens a door, the player has a 1/3 chance of having selected the car. Referring to either the figure or the tree, in the cases the host then opens Door 3, switching wins with probability 1/3 if the car is behind Door 2 but loses only with probability 1/6 if the car is behind Door 1. The sum of these probabilities is 1/2, meaning the host opens Door 3 only 1/2 of the time. The conditional probability of winning by switching for players who pick Door 1 and see the host open Door 3 is computed by dividing the total probability of winning in the case the host opens Door 3 (1/3) by the probability of all cases where the host opens Door 3 (1/2), therefore this probability is (1/3)/(1/2)=2/3. Although this is the same answer as the simpler solutions for the unambiguous problem statement as presented above, in some variations of the problem the conditional probability may differ from the average probability and the probability given only that the player initially picks Door 1, see Variants below. Some proponents of solutions using conditional probability consider the simpler solutions to be incomplete, since the simpler solutions do not explicitly use the constraint in the problem statement that the host must choose which door to open randomly if both hide goats (multiple references, e.g. Morgan et al., Gillman, ...). What this type of solution is saying is that if 900 contestants are on the show and roughly 1/3 pick Door 1, of these 300 players about 150 will see the host open Door 3. If they all switch, about 100 would win the car. A formal proof that the conditional probability of winning by switching is 2/3 is presented below, see Bayesian analysis. Car hidden behind Door 3 Car hidden behind Door 1 Car hidden behind Door 2 Player initially picks Door 1 Host must open Door 2 Host randomly opens either goat door Host must open Door 3 Probability 1/3 Probability 1/6 Probability 1/6 Probability 1/3 Switching wins Switching loses Switching loses Switching wins If the host has opened Door 3, these cases have not happened If the host has opened Door 3, switching wins twice as often as staying ( Preceding offered by Rick Block at 23:00, 5 April 2011 UTC )