Talk:Monty Hall problem

The trifle analogy

Suppose there are two trifles, trifle A being ten times bigger than the trifle B. A ring is hidden in one of them by Mr X. For Mr Y the chance of teh ring being in trifle B is 1/11. Or is it? Does the size really matter? FleischerDan (talk) 14:34, 8 August 2017 (UTC)

Probably belongs on the 'arguments' page (and is an analogy to the first stage only of the MHP) but using the information available to Mr Y, and making the assumption that a ring is smaller than a trifle, it's 1/2. Freddie Orrell 15:05, 8 August 2017 (UTC) Correction: that assumption about size is superfluous since the ring is successfully hidden, and I should not have included it. Freddie Orrell 22:27, 12 January 2018 (UTC)

Referencing

It’s a relatively minor point, but I was intrigued at the referencing system used? I’ve not seen many Wikipedia articles using the Harvard Referencing system, and indeed other mathematics articles use the Vancouver system (eg Fermat’s Last Theorem). I was wondering what the rationale for using the system here was - I have no plans to change the referencing system, but amidst all the other Wikipedia articles it does stick out a little. Mofs (talk) 21:37, 13 March 2018 (UTC)

I suggest that you contact Rick Block with your question as he was instrumental in bringing this article to featured article status. *I believe it was Rick's personal preference but* He may provide additional insight. hydnjo (talk) 19:19, 1 April 2018 (UTC)

*Strikethrough my error. hydnjo (talk) 19:12, 8 April 2018 (UTC)

The article was originally written using Harvard referencing (before I ever saw it). I just continued the existing system. -- Rick Block (talk) 14:59, 5 April 2018 (UTC)

Basic assumptions

I don't have a problem with the mathematics, but they do not correspond to the problem as posed here. The problem merely says that Monty Hall opens the door on this occasion. It is perfectly possible that "Monty Hall" only chooses to open the door when the door you have first chosen conceals the car. We either need to know that Monty Halls decision is independent of your initial choice (including the possibility that he always opens the door), or that he is trying to help you (rather than the budget of the show). The problem badly needs rewording. PhysicistQuery (talk) 00:23, 22 April 2018 (UTC)

Please have a look to "Standard assumptions" in the article. Btw: vos Savant says that the host, who knows the location of the car, OPENS a door and shows a goat. --Gerhardvalentin (talk) 16:10, 22 April 2018 (UTC)

Yes, but the point that people always seem to miss here is that she didn't say (at least in her first article) that he had to do that, or that he would always do that, but only that he did. It actually does matter, for purposes of the Bayesian update. --Trovatore (talk) 21:19, 22 April 2018 (UTC)

MvS presented a hypothetical "one time scenario". The contestant originally selects one door, having 1/3 chance on the car. Then the host, knowing about the location of the car, does open a door with a goat behind. For the contestant, this results in a now conditional probability of 1/3 to win the car by staying, and in a now conditional probability of 2/3 to win the car by switching. --Gerhardvalentin (talk) 22:31, 26 April 2018 (UTC)

Unfortunately she didn't tell us whether he was required to act as he did. If he wasn't required to, then the update of conditional probability depends on what you assume about Monty's motivations. --Trovatore (talk) 22:41, 26 April 2018 (UTC)

Additonal unproven "assumptions" lead astray, so we have to stick on what we definitely know. No additional unproven assumptions (lame clumsy host and so on). MvS said: The host knows what's behind the doors and, after the contestant made his choice, the host opens one of the two unselected doors and shows a goat, offering his still closed door to switch on. For unbelievers and doubters: Selvin, Mueser and Granberg and others say
that the host must always open an unchosen door and must always show a goat and must always offer to switch. The unalterable hypothetical "one-time scenario" is (and always remains) as it is. --Gerhardvalentin (talk) 09:14, 27 April 2018 (UTC)

Selvin, Mueser, and Granberg may have said that he must open an unchosen door and show a goat, but vos Savant did not. At least, not in the first Parade article. --Trovatore (talk) 09:29, 27 April 2018 (UTC)

Yes, but let us stick on what we definitely know meanwhile.
In the unalterable hypothetcal one-time gameshow as per Marilyn vos Savant, the host does/did open an unselected door and does/did show a goat and does/did offer to switch doors. Marilyn vos Savant said so, and - to make it clear that there is no reason to shake this concrete issue - Selvin, Mueser and Granberg and many many others say that it is like that in the given situation, and never different . We find that in the article, that should be cleaned up to make it clearer.

Unproven assumptions do lead astray, the article shows enough unfounded deviations. And finally the article now presents the critical stage of the advanced show: "Is it of advantage for the contestant to switch doors, yes or no?"

As per Henze, the contestant has no knowledge at all about the scenario he actually is in, and the host does not give him any hint on the scenario the guest actually is in.

The contestant might be in the lucky guess scenario, having selected the door with the car and therefore was better to stay, or – with double probability – he might be in the wrong guess scenario, having selected one of the two goats, where switching doors wins the car for sure. That's the actual situation of the guest, in the unalterable hypothetcal one-time gameshow as per Marilyn vos Savant, and his situation never can be different. No room for inappropriate variants. --Gerhardvalentin (talk) 14:59, 27 April 2018 (UTC)

"Stick to what we know." Yes. The whole point is that from what we "know" — that is, using only what vos Savant said in the first Parade article — the problem is under-specified and can't be answered. You can answer it if you add extra information, such as the requirement that the host must behave in that way. --Trovatore (talk) 18:40, 27 April 2018 (UTC)

Trovatore, you are right again. But the host DOES/DID behave in that way. Exactly knowing the locations of the three objects, the host DID open one his two doors indeed, and in opening it, he DID show a goat and not the car. In case that, by occasion, he didn't have two goats, but only one goat and the car, then he obviously did show the goat deliberately. That's all we know. And now comes the moment that the host offers the switch to his still closed second door. In the famous one-time show, there is no place for deviating variants. We have to accept the given information. No chance to believe that he didn't open one of his two doors, and no chance to believe that he showed the car, and not a goat. The show is almost over, the host already has offered to switch doors. We have to "stick" only on what we know — and we have to accept that, in this case, it is like it is and not otherwise. Regards, --Gerhardvalentin (talk) 19:40, 27 April 2018 (UTC)

I really don't know why you keep repeating what the host did, when it isn't the point. In any case, to get back on topic, I just looked back at the article, and I'm reasonably satisfied with the current state of the introductory section. The second sentence of the third paragraph starts with [u]nder the standard assumptions, without actually saying what those assumptions are, which is a little less than ideal, but perhaps the best that can be done without disrupting the flow. --Trovatore (talk) 19:49, 27 April 2018 (UTC)

Okay. Considering your objection, and the years of dispute, I strived to show that without full particulars, it is trouble-free to recognize vos Savant's implied preconditions that make it easy to solve the puzzle as she intended. See Selvin and many others. But I admit that without that vista, and under deviant preconditions a congruent solution will not be achieved. It just depends on your goals, and the article doesn't make that clear enough. --Gerhardvalentin (talk) 09:32, 28 April 2018 (UTC)

Explaining ourselves better

Three initial configurations of the game. In two of them, the player wins by switching away from the choice made before a door was opened.

Recently in the Arguments subpage, Rick Block made the point that maybe we are not doing our best to explain the problem to someone who still doesn't understand it.

Because of that, I've created a simple diagram based on Rick's images, that conveys in visual form the information of the table in the Simple solutions section. Please comment on ways to improve the image and/or the accompanying text, and I'll add it to that section as a reinforcement to the information already found in the article.

Optionally the diagram could be complemented with this second one, which shows the "little green man" version of the game where the probabilities are aligned with the common misconceptions of the students. I believe showing both diagrams side by side could illustrate the differences with the probabilities in the standard assumptions in a simple, intuitive way. Diego (talk) 06:22, 24 April 2018 (UTC)

A different selection process, where the player chooses after a door has been opened, yields a different probability.

I think the problem is not so much the images (the one in the Conditional Probability section already shows what I think needs to be shown), but the general approach of the simple solutions - which inherently attempt to switch the reader's mental model away from the situation that exists at the point a door has already been opened to the situation before the door is opened. I think text along these lines might help:

Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. This would be true if the host opens a door randomly, but that is not the case. Before the host opens a door there is a 1/3 probability the car is behind each door. If the car is behind door 1 the host can open either door 2 or door 3, so the probability the car is behind door 1 AND the host opens door 3 is 1/3 * 1/2 = 1/6. If the car is behind door 2 (and the player has picked door 1) the host must open door 3, so the probability the car is behind door 2 AND the host opens door 3 is 1/3 * 1 = 1/3. These are the only cases where the host opens door 3, so if the player has picked door 1 and the host opens door 3 the car is twice as likely to be behind door 2. The key is that if the car is behind door 2 the host must open door 3, but if the car is behind door 1 the host can open either door.

This explanation could be sourced to any of the sources that talk about the conditional probabilities. -- Rick Block (talk) 23:43, 24 April 2018 (UTC)

Hi Rick, I agree adding as clear explanations as possible is a good idea, and I support including your text. But I think there's also a problem with not having enough simple images. To imprint a mental model in the reader, a presentation that requires them to parse several complex concepts will be hard to grasp, and doesn't guarantee that they'll get the essential intuition. I think it's best to start with a really simple image that instantly conveys a mental picture which the reader can point to, and only after that explain the concepts of that image in more detail.

So far, the only image of that kind is the table containing File:Monty closed doors.svg and File:Monty open door chances.svg in the "Simple solutions" section; it conveys the simple intuition of "two doors have 2/3 probability". The table at "Conditional probability by direct calculation" is already too complex for what I have in mind, containing two rows and four columns of images interlieved with long text snippets; that's too complex to provide a "single idea at a glance" intuition.

The images I've suggested try to do the same as the "2/3 probability" images, but for the ideas you mentioned of "a priory probabilities favor 2/3 for switching" and "choosing after the door is opened yields 1/2 probability" (which readers should instantly recognize as their original intuition). I think having both ideas exposed in compressed form should help readers to better understand the text, knowing that it refers to two separate ideas.

I'll integrate the images with your text in the "Simple solutions" section, let's work there to improve their final presentation in the article. Diego (talk) 09:12, 25 April 2018 (UTC)

P.S. Even if the images happen to be not that good as an explanation, having them will allow us -when a reader makes the "two doors, 1/2 probability" argument- to point to the second image and say "oh, you're talking about that image; yet my explanation is about the first image". Diego (talk) 09:39, 25 April 2018 (UTC)

This is another way I like to explain it. Let's keep in mind that the host knows the positions and must reveal a goat from the other two doors the contestant didn't select. From the host's point of view, when he chooses which door he is going to reveal he is also choosing which door he is not going to reveal, so let's say that the one he leaves closed is his selection.

In this way, the game can be seen as if he was another contestant who want to win the car; the contestant chooses one and then the host chooses another, which must be different. Those selections are the two closed doors; the revealed is the door no one selected. But the host is a cheater: he has the advantage of knowing where the car is. Note that if he was the first to select, he would win 100% of the time, but since he is second, he can only win when the car is in one of the two doors the first player didn’t select, which are still the most → 2/3. He always picks the best of the other two. If the game started with 100 doors, 1 car and 99 goats, it would be easier for the host to win, because he just have to wait for the first to select any of the 99 incorrect options, and if that happens he is free to look for the correct.

The analogy works because one of the two closed doors always hides the car. The car is never revealed because both players want to win it. If the contestant didn't choose it, the host is not going to choose a goat; he can afford to select the prize one. EGPRC (talk) 19:59, 8 June 2018 (UTC)

The N doors section makes no sense

It doesn't matter if there are 3 doors with 2 goats or 10 doors with 9 goats, the eliminated doors are moot and can't be part of any valid equation. All that matters is, you pick a door, any door, then he picks a door and he knows where the car is. The resulting odds that the door he picks will be the car will always be 2/3. You either picked the car door to begin with or you didn't, either way the car will be behind one of the doors and Monty knows where the car is and because he does, a 50/50 chance becomes 2/3 for the door he picks because he must leave the car door shut.Ealtram (talk) 11:20, 13 August 2018 (UTC)