Talk:Monty Hall problem/Arguments

There are ONLY 2 choices left: "Impossible"/Confusing to have a present x/3 statistic.

The fact is, when there are only 2 doors left to choose from, there are then ONLY 2 choices left; and it is therefore impossible to have an x/3 statistic among those at that point in time because you no longer have a choice of 3 but only of 2. Thus, the way some of the article is written is incredibly confusing for the 9/10 people who notice this clear and indisputable fact: There are only 2 choices left.

Seeing that even a famed mathematician thought similarly makes it clear that if you are going to use x/3 when you have left only 2 choices, then you'd better be crystal clear about the logic of application, which from reading much of the article isn't that convincing or clear; because using an x/3 stat when there are only 2 choices left makes little sense to most of us.

Also, placing that remaining x/2 likelihood back in time when the INITIAL chance was 1/3, and continuing to use x/3 makes little sense to most of us.

Saying anything new about x/3 when there are only 2 choices left makes little sense to most; so Why be so confusing with the math? we think. The fact is, the REMAINING chance has only 2 choices possible, and so therefore the remaining statistic MUST be in reference to that: It must be stated as x/2.

Hardly a clear justification was made in my reading of the most of the article for using x/3 at all, at the point where there were only 2 choices left. It is that justification that needs to be made, and clearly applied, which would help the hapless reader understand at least some of the gobbledygook; because otherwise, it continues to seem nonsensical. And after all, isn't this conundrum in particular one that needs to be crystal clear to the general reader who was mystified in the first place when reading the "Ask Marylin" article? Misty MH (talk) 23:55, 9 March 2018 (UTC)

I hand you a die with 2 white sides and 4 black. You only have 2 choices. Roll it, what do you think your odds are if you choose white? — Preceding unsigned comment added by Nijdam (talk • contribs) 11:17, 10 March 2018 (UTC)

The formula ---> (#favorable events / #possible events) is applicable only if you have the same information about the options. Remember the division distribute equal amounts. For example, if I have a cake and two persons and I want to give each one the same amount, then I should give 1/2 to each. But certainly that's not the only way to distribute it. I could give 4/5 to one and 1/5 to the other, or give all the cake only to one, etc.

The probability is a measure about how much information we have. The confusion is to think that the options must be equally likely in any case. In this game, yes, we have two options, but do we have reasons to distribute the probability equitably? No. The contestant's door was chosen randomly from three, meaning that it has 1/3 probability to be the correct. On the other hand, the host knows the positions and must leave the car hidden (because he must reveal a goat door), so always the contestant failed at first, the other door the host leaves closed is the correct one, and we know this is more likely to have happened than the opposite.

In the second selection, we don't have a random door vs another random door. Basically, you have an option selected by someone who hits the correct 1/3 of the time and another chosen by someone who chooses the correct in the other 2/3 cases. Which one do you prefer? The important thing here is nor how many options do you have, but that they were chosen by two different people, one with more knowledge than the other. The host's closed door tends to be the correct with more frequency than the contestant's selection. — Preceding unsigned comment added by 190.36.105.224 (talk) 05:28, 13 March 2018 (UTC)

@Misty MH: I guess Leonard Mlodinow, in his book The Drunkard's Walk, shows it best, in underlining the neglected, but obviously constricted role of the host in the MHP. His role leads to conditional probabilities of surprising 1/3 : 2/3. About Mlodinow, Just have a look there.
It's hard for anyone to grasp that the host can act "randomly" only in 1/3 of all cases. He can act randomly only in 1/3, only in case that the guest (by luck) selected the door with the car (i.e. only if the guest actually is in the "lucky guess scenario").
Only in this special case (1/3) the guest should stay and never switch.

But in 2/3 of cases the guest will be in the "wrong guess scenario" (2/3), having selected one of the two wrong doors (two out of three). Because in that "guest's wrong guess scenario", the host's two doors having the car and a goat resp. a goat and the car, the host nevermore can act randomly, because he never shows the car but the second goat only. So his selection which door to open, in openening always the door with the goat, leads to conditional probabilities. He keeps the secret car undisclosed behind his still closed second door. In that 2/3 switching wins the car for sure.

So in 2/3 of all cases, as a "conditional probability", switching doors helps the guest to get the car.

The contestant, in opening a door, does not know the scenario he actually is in. He only knows that - by the host's constraint to never showing the car - by switching he will win the car with double (conditional) probability. Admittedly, on the long run only.

I agree with you, this fact that in 2/3 the host will never act randomly leads to a conditional probability of 2/3 to win the car by switching is hard to tell, using a few words only. --Gerhardvalentin (talk) 15:15, 22 April 2018 (UTC)

I read this as a criticism of the article, not the math. What Misty MH is saying is that the article is not clear, because it immediately delves into the "simple" solutions which more or less ignore the conundrum most people encounter which is that at the point of the decision there are only 2 doors involved, not 3. In particular, we know vos Savant's explanation from her original column is unconvincing based on the reader reaction she got. -- Rick Block (talk) 14:53, 23 April 2018 (UTC)

Yes, you are right. As to my understanding, Misty MH says that the article should clearly show that any assemblage of both unselected (host's) two doors is more likely to contain the prize. Exactly they have probability of 2/3, compared to any originally selected door of only 1/3.

And it's a fact that the host AVOIDS to show the car, but in any case he shows a goat. Solely in the contestant's lucky guess scenario, the host actually having two goats, he can act randomly. But if the CONTESTANT SELECTED ONE OF THE TWO GOATS (with twice chance) the host avoids to show his car, the "strict condition" is that he will show the SECOND GOAT only. So the 2/3 chance to win by switching doors is a typically "conditional probability". It's that easy.

And the famous mathematician Henze says that strictly speaking "math" is unnecessary to solve the MHP. And also Misty MH says: Why be so confusing with the math?

I say once more: math is rather unnecessary to fully understand vos Savant's MHP. That means that the whole ado about the host's biased behaviour belongs to a completely different article, you can name it "Index of lessons on conditional probability theory, based on the famous MHP"

And Marilyn vos Savant says that only in case that you already do have additional information of a factual existing preference of the host, and its extension, it might be sensual to do maths. But this never will be the case in the MHP, so any of similar considerations do never address the MHP of vos Savant, that definitely excludes any "additional (hidden?) information". No hidden hints are given. Such assumptions have to live their own life completely outside of vos Savant's famous MHP paradox.
(Though mathematically fully correct, M.et al. have disqualified themselves, in inventing ungiven additional assumptions, far outside of vos Savant's MHP, similar to "if wishes were horses, beggars would ride" and similar to "if you know that the biased host is paralyzed, then ...").

Any of such unbased considerations, only basing on ungiven and unproven assumptions (suitable for lessons in probability theory only), are obfuscating for the reader and prevent understanding the core of the famous paradox:

"Two still closed doors, why does the door offered have double chance, compared to the door originally selected?"
Answer: because there is'nt a new simple probability of 50:50, but because there meanwhile emerged a conditional probability of 1/3 : 2/3.

See theoretical physicist Leonard Mlodinow, who worked together with Steven Hawking, his diagram (1/3 : 2/3 on the long run) here:
The article, for years and years, has been confusing. It is time to detoxify it from disconcerting, misleading and truthless "conditional bias" math garbage that clearly belongs to a different article.--Gerhardvalentin (talk) 09:55, 25 April 2018 (UTC)

@Rick Block EXACTLY! You said it much better than I did! :) — Preceding unsigned comment added by Misty MH (talk • contribs)

With a small change to the game, you could be told that you could either have what's in the door you chose, or take what you want from the other two doors - in effect Monty did just that by showing you the door not to bother with - you'd rather swap to the 2/3, right? — Preceding unsigned comment added by Gomez2002 (talk • contribs) 16:07, 19 July 2019 (UTC)

Keep It Simple

If the contestant picks the car and switches, the contestant loses. The contestant has 1/3 chance of picking the car.

If the contestant picks the goat and switches, the contestant wins. The contestant has 2/3 chance of picking the goat.

The contestant doubles the chance of winning by switching.

JAKQ0s (talk) 16:59, 15 November 2018 (UTC)

That's about how I think about it. If you know the host has some algorithm (e.g. always pick the right-most losing door), pick Door 1. If the host picks Door 2, the car is behind Door 3; otherwise the car may be behind Door 2 or Door 1. You have a 2/3 chance of winning at all times. If the selection is random, then you pick a door and the host opens all other doors except one, and never opens the winning door, in which case you have ${\frac {1}{n}}$ chance of having picked the correct door and a ${\frac {n-1}{n}}$ chance of having picked the wrong door, so with more than two doors switching has a large probability of winning. John Moser (talk) 15:38, 17 October 2019 (UTC)

93.106.123.184 (talk) 14:05, 27 May 2020 (UTC) The following question could use a simple answer: Why does sticking with the door you chose originally not count as a new choice in a new situation?

93.106.123.184 (talk) 14:05, 27 May 2020 (UTC)

Where did the article go?

Where did the article go? I clicked on the "Article" tab, and there was nothing. Misty MH (talk) 09:12, 21 March 2019 (UTC)

Weird. I found it, I guess. It's like this is an orphaned Talk page. Is "/Arguments" a normal additional page to a regular Talk page? The article is here https://en.wikipedia.org/wiki/Monty_Hall_problem Misty MH (talk) 09:14, 21 March 2019 (UTC)

@Misty MH: No, this isn't normal. Subpages for talk pages are normal for archives (which will present the same "missing" article tab; mainspace articles can't have subpages either), but not for actual discussion. This is an unusual case because of the nature of the article. See also Talk:0.999.../Arguments. –Deacon Vorbis (carbon • videos) 16:13, 17 October 2019 (UTC)

Law of large numbers proves this (not disproves as originally written)

The below information is accurate so leaving it for prosperity whereas my code was not. However, as a cautionary tale: The code I had originally written was tossing out every attempt where the host picked either the winning door or the contestant's instead of just picking the other door. This is more likely when the contestant picked a nonwinning door. This incorrectly shifted the expected results of the contestant winning to 50%. Now to the original argument.

We have 3 doors.

1 with a car behind it and 2 with goats - Assumed randomly assigned

Contestant randomly selects door - Also assumed random

We have to assume that the host cannot pick the door with the car AND cannot pick the contestant's door but other than that they randomly pick an available goat door (ultimately meaningless as it'll always come down to prize door and a goat door)

So at this stage we have 4 possible states...

1. Contestant picked the right door. Host picks goat door A. Contestant chose correctly

2. Contestant picked the right door. Host picks goat door B. Contestant chose correctly.

3. Contestant picked goat door A. Host must pick goat door B. Contestant chose incorrectly.

4. Contestant picked goat door B. Host must pick goat door A. Contestant chose incorrectly.

Host opens a goat door...

At this point the host HAS opened a door and it's known to be a goat and the problem has completely changed. We're left with the new states of...

1. Contestant picked the right door. If they stay they win but if they swap they lose.

2. Contestant picked the goat door. If they stay they lose but if they swap they win.

This can be programmatically shown by this pseudo-code.

Randomly pick 1 of 3 doors to have the prize.

Randomly pick 1 of 3 doors to be the contestant's pick.

Randomly pick 1 of 3 doors to have the host pick a door to throw out. If it's the prize door or contestant's pick just randomly pick again until it's not.

If the prize door is the contestant's pick we add 1 to the win column. Else we add one to the lose column.

Run it for as long as you want, assuming 1 million+ times to be at least semiaccurate, then then unsurprisingly find out that the tally is approximately 50/50. It cannot be argued against because we did the entire Monty Hall Scenario. We have a random prize door. We had a contestant pick a random door. We had the host pick a random goat door. We then compare if the contestant already picked the right door or not. If they always swap doors or not doesn't impact their chances of winning in the slightest.

If we don't do exactly what the pseudo-code does we don't fit the description of the Monty Hall Problem but rather a situation where we're doing math for before the host opens a goat door but after the host has announced the door has a goat behind it. — Preceding unsigned comment added by 184.166.97.144 (talk) 03:07, 28 June 2020 (UTC)

Nope, it's definitely 2/3 for switching and 1/3 for staying. If you implemented your description of the scenario correctly, a simulation should verify that. See the main article for a more detailed explanation. –Deacon Vorbis (carbon • videos) 03:26, 28 June 2020 (UTC)

Ran the simulation. Worked exactly as I said it did. 50/50 — Preceding unsigned comment added by 184.166.97.144 (talk) 03:38, 28 June 2020 (UTC)

Please indent your replies and sign your posts with 4 tildes; see Help:Talk for more info; thanks.

Then you made a mistake in implementing it; there are references in the article that describe doing exactly this and that it supports the analysis showing a 2/3 win chance for switching. –Deacon Vorbis (carbon • videos) 03:49, 28 June 2020 (UTC)

I gave exactly the code I used. Random numbers for everything. If you see a problem with the pseudo-code point out exactly where I made an incorrect assumption.184.166.97.144 (talk) 06:52, 28 June 2020 (UTC)

No, you gave a very very rough description of an algorithm, certainly not rising to the level of pseudocode, let alone actual code. What you described seemed more or less in line with the usual version of the problem, so if you got a different result, you must have erred in its implementation. Without seeing that, it's impossible to say how. –Deacon Vorbis (carbon • videos) 12:45, 28 June 2020 (UTC)

3 random ints [0,2]. They stand in for prize door, chosen door, and the door the host shows to have a goat. Ensure that the host hasn't picked either the prize door or the contestant's door but rather a free goat door, and then just a comparison if the prize door is the chosen door. It's that easy.184.166.97.144 (talk) 19:19, 28 June 2020 (UTC)

But if you're simply checking to see if the player's choice matches the car as you say, then you're simply checking to see if those two random numbers were equal (the value of the host pick is irrelevant), which is 1/3, just as it should be. Again, if you're getting 1/2, then you're implementing something that doesn't match what you've described, because what you've described plainly would result in a 1/3 win chance. –Deacon Vorbis (carbon • videos) 23:18, 28 June 2020 (UTC)

Found the mistake. It didn't help that the the error shifted it to ~50% at large numbers which exactly fit my misconception but it did. I apologize for the inconvenience.184.166.97.144 (talk) 01:00, 30 June 2020 (UTC)