Talk:Monty Hall problem/Archive 39

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Archive 39

Fighting to re-add a previously supported External Link

Consensus remains to not include links to simulators

Per a discussion in October of 2012 on https://en.wikipedia.org/w/index.php?title=Talk:Monty_Hall_problem&oldid=520111533#Proposal_to_add_a_link_to_the_External_Links_section I found support for adding the following to External Links: ThreeDoors Three Doors Probability Problem (Simulator to go through 100 simulations in a couple of minutes) If you have an issue with this, please state it clearly here. — Preceding unsigned comment added by Reidme (talk • contribs) 18:35, 15 October 2014 (UTC) I dont see how it comes close to meeting the external link criteria: WP:EL - can you provide your analysis of how or why it does? -- TRPoD aka The Red Pen of Doom 18:45, 15 October 2014 (UTC) Did you read the original Talk discussion I have linked to above? In what way does that not address your question, let me know and I will address it from that point.Reidme (talk) 18:58, 15 October 2014 (UTC)Reidme It doesnt address my concern because there is not a thing about how it meets the criteria for links that should be included. WP:EL. Please specify why / what segments you think it meets.-- TRPoD aka The Red Pen of Doom 19:01, 15 October 2014 (UTC) If you can tell me how the NYT Monty Hall Simulator listed in External Links meets the criteria, then I can tell you how my simulator which is faster and much better meets the criteria. If not, it is a resource that Richard Gill, whose papers appear in References four times on the Monty Hall Problem page, stated that he was in favor of it being added because "NYT uses flash, doesn't work on iPad. Wolfram is dull. This one is good. Richard Gill (talk) 06:48, 23 October 2012 (UTC)" You can see that statement in the link I mentioned above.Reidme (talk) 19:27, 15 October 2014 (UTC)Reidme hmmmm, as I read WP:EL i do not see "when there are already a mess of links that shouldnt be included, you can continue to add more." did I miss that somewhere?-- TRPoD aka The Red Pen of Doom 20:05, 15 October 2014 (UTC) I don't see how that's a valuable link; it's a very crude (looking and working) simulation, that looks like it was put together in five minutes. And you are so keen on it why? Your only activity here is to promote this link. You should review WP:SPAMMER, which describes editors who only seem interested in adding links to pages.--JohnBlackburnewordsdeeds 20:16, 15 October 2014 (UTC) Go to a bar and find someone who doesn't agree with you on the probabilities of the immediate problem and then you might see that, although it is crude, it is very functional and quickly demonstrates the truth of the matter. And I further invite you to try it with other simulators. And contrast those outcomes with articulating the rightness of your belief that it is better to switch doors.Reidme (talk) 20:43, 15 October 2014 (UTC)Reidme Also, JohnBlackburne, "looks like it was put together in five minutes"? You are way out of your world of expertise.Reidme (talk) 16:55, 16 October 2014 (UTC)Reidme uhhh, come again? your rant has to do with policies that support including the link because....??? -- TRPoD aka The Red Pen of Doom 20:46, 15 October 2014 (UTC) No it's simply crude, graphically and functionality. It also gives wrong answers. I refer you to the article for a correct treatment. More generally I don't think a simulation serves any purpose, unless it's exceptionally well done. The problem is more a word problem or a thought problem, one understood by reading and thinking through the logic. Any 'simulation' is going to include all the explanation somewhere, and the 'simulation' aspect can only consist of choosing one of two doors, as that's the only real choice. Better to have that all explained in text rather than over multiple fragmentary pages with limited navigation between them.--JohnBlackburnewordsdeeds 21:07, 15 October 2014 (UTC) What wrong answers does it give? On another topic, a simulation need not include "all of the explanation". The simulation serves to prove empirically and quickly the truth of the matter in real time. Go to a bar and test the different things I said earlier.Reidme (talk) 22:18, 15 October 2014 (UTC)Reidme Wikipedia is not an internet directory, and not the place to promote your website or webpage. To support adding a link, you should demonstrate how you feel it meets inclusion criteria listed at WP:EL. --- Barek (talk • contribs) - 21:39, 15 October 2014 (UTC) This is a special case where if you read the immediate article you will see that simulators are crucial.Reidme (talk) 22:18, 15 October 2014 (UTC) I am pretty certain that people survived for decades discussing and understanding the subject without any internet assist and that your particular website has not been identified as the cure all that complete and immediate understanding. -- TRPoD aka The Red Pen of Doom 23:37, 15 October 2014 (UTC) I would refer you to the immediate article, which says, "Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result (Vazsonyi 1999)." Additionally, if you did your research, you would see that someone who knows what they are talking about, having done extensive published academic research on this problem, considered my simulation "good", and that was in comparison to the other simulators listed and was in favor of it being included.--Reidme (talk) 12:21, 17 October 2014 (UTC) To clarify, are you saying that you cannot demonstrate if/how the inclusion criteria of WP:EL is met? --- Barek (talk • contribs) - 22:21, 15 October 2014 (UTC) To clarify, no one is removing the NYT simulator link and no one is removing the Wolfram simulator link. Why not? If my link has been judged by someone who has written peer review papers as better than those two simulators (this is on the record here), certainly my link should be reinstated. Mr. Ollie slammed it down when I originally added it. The I went to talk, and invited him. It was supported in talk.Then a year and a half later he cuts it out again. I would think that we should put it back up at least until some other published expert on the topic weighs in. A simulator would naturally be linked to in this article, as is proven by the links that exist to simulators in this article.Reidme (talk) 17:18, 16 October 2014 (UTC)Reidme The argument that "other stuff exists", particularly when that other stuff should also be removed, is not a convincing argument. As to the other argument of leaving the link; the current consensus is clearly against the link you added. Current consensus takes precedent over a two year old consensus; and since Wikipedia works based on community consensus, the link should not be restored to the article at this time. --- Barek (talk • contribs) - 19:56, 16 October 2014 (UTC) What is the quality of that consensus? An expert who had studied this problem was in favor of it two years ago, his opinion is ignored in favor of whomever happens to show up and have an opinion? That is what we have here.--Reidme (talk) 12:21, 17 October 2014 (UTC) What we have is community consensus, a Wikipedia policy and a fundamental model for how Wikipedia works. --- Barek (talk • contribs) - 16:17, 17 October 2014 (UTC) Note of related discussion: following a reverted attempt to add justification for the link to WP:ELMAYBE, a discussion closely related to the one here was also started at Wikipedia talk:External links#Proposal: Allow for links to simulations and models under "Links to be Considered". --- Barek (talk • contribs) - 20:09, 16 October 2014 (UTC) In fact, most simulators of the problem are wrong. Handy programmers like to show their skill in building fancy simulators, but do not know the exact ins and outs. To simulate the situation only those cases where door 1 has been initially chosen and door 3 has been opened should be accounted for. And moreover, who guarantees the user that the simulator is constructed correctly? So how trustworthy is the result of a simulation? Nijdam (talk) 11:34, 17 October 2014 (UTC) The simulator I made was reviewed in October 2012 by an expert who had published on the immediate topic and was in favor of its inclusion, which is mentioned above.--Reidme (talk) 12:21, 17 October 2014 (UTC) Background and Final Argument I created the simulation that I think should be included under External Links more than 15 years ago in response to having difficulty proving the value of switching to others, after discovering this wonderful problem. Naively I thought explanations would work. But they often do not work and are summarily rejected because the listener fully believes that any explanation is false. I mentioned this problem to an owner of a math software company where I was interviewing for work long ago - and the owner dismissed it out of hand and me as a potential employee. A similar thing happened on another interview. Others were hard to convince as well. So I created my simulator. And, within minutes people would get that switching was better than staying. No long arguments. Click, click, click. 100 simulations with stats to see the results in real time. Three minutes going through 100 simulations. People actually found it fun. And then they were open to explanations, in light of actual experience. Should an encyclopedia include a link to such a simulator? I would think yes. Because it goes to the core of learning about part of this topic. The simulator I created can prove an assumption false in three minutes that would ordinarily take a long time, even with the best attempts to communicate via words and pictures. I think that is a huge advantage. An advantage worthy of being linked to for teachers and others who can use it to learn and teach others. And use in conjunction with the immediate article. Here I seek consensus in a matter where the written rules do not explicitly support the adding of a link to simulators. The rules do not explicitly block posting of such a link, either. It is actually up to consensus. If the current consensus holds, oh well. It has been an interesting experience to originally get it supported and then have it removed and now return to this article's Talk page to push for it one more time. 100 simulations in 3 minutes where you get to pick the doors.--Reidme (talk) 01:43, 18 October 2014 (UTC) you have made what you identified as your "final argument" which as all your previous positions fails to establish a rationale appropriate to WP:EL , but apparently [1] it is not your "final argument" and you are going to continue to beat the dead horse? Please be aware that this article is under Wikipedia:Arbitration Committee/Discretionary sanctions and Tendentious editing such as relentless push to promote your website in contravention of consensus against such inclusion can quickly lead to a topic ban or even a complete block. -- TRPoD aka The Red Pen of Doom 17:00, 18 October 2014 (UTC) Okay. I see that all links to simulators have been removed, except the University of California San Diego link, which is broken link aspects to it. Also, the "Monty Hall at DMOZ" has links that are broken. I am sort of exhausted by this process right now, so others, if they wish, can decide to ignore or delete those. I now see that this is much more of a fair coin than I did when I arrived here a few days ago. I also went to Encyclopedia Britannica and see that they do not have a similar article. Nor do they do external links. When I posted my "Final Argument" I had already accepted whatever consensus would prevail. But now I am comfortable with that consensus.--Reidme (talk) 18:31, 18 October 2014 (UTC)

The little green woman is missing!

We currently have in the article, 'The behavior of the host is key to the 2/3 solution' but not much else to explain this important point. At one time we had vos Savant's comments on the subject and her little green woman example but they seem to have gone missing.

What, in the opinion of editors here, is the best way of explaining that it is only if the host knows where the car is (or always reveals a goat) that it is advantageous to switch. This is one of the most surprising (to most people) features of the paradox. 16:23, 17 October 2014 (UTC)— Preceding unsigned comment added by Martin Hogbin (talk • contribs)

So "most people" mean I choose door 1. Then I open door 3. If there is a goat behind it, I now have a 2/3 chance with door 2.?--Albtal (talk) 18:41, 19 October 2014 (UTC)

Most people mean I choose door 1. The host opens door 3. There is a goat behind it. The distinction being made here is that the probability of the car being behind door 2 is 2/3 only if the host MUST open a door AND the host knows before he opens it that there is a goat behind it (AND the host effectively flips a coin to decide which door to open if the car is behind the initially chosen door, door 1 in this case). If the host opens door 3 without knowing where the car is AND (luckily for the host) it happens to reveal a goat, there is no advantage to switching. This is the "Monty Forgets" or "Monty Falls" variant, discussed by Rosenthal in 2005 (reference in the article) and by vos Savant in 2007 [2]. Note that her little green woman example addresses a slightly different point, which is that the odds of winning by switching knowing the set up are different from the (50/50) odds of randomly picking one of door 1 or door 2 after the host has opened door 3 (in fact, regardless of whether the host knew what was behind door 3 before he opened it). IMO, the clearest way to explain these subtly different problems is with the language and machinery of conditional probability. -- Rick Block (talk) 21:24, 19 October 2014 (UTC)

But this is not at all an answer to the question which I directed to "16:23, 17 October 2014 (UTC)".--Albtal (talk) 09:23, 20 October 2014 (UTC)

Do you think we should not include the 'little green woman' example then? Martin Hogbin (talk) 22:53, 19 October 2014 (UTC)

Oh, the original poster was you? OK then. I'm outta here. -- Rick Block (talk) 04:59, 20 October 2014 (UTC)

That is not very nice and very much against the spirit of WP. It would seem that you are only willing work with others and contribute to this article if you can have it all your way. Over its history there has been a clear majority of users who prefer to organise the article in that same way that I do and this was confirmed by an RfC. Within that consensus structure, I do not see why editors with differing opinions on the subject of conditional probability should not work together.

Having re-read the 'little green woman' example I think it actually not all that clear exactly what point vS was making so I agree that it is best left out. The 'Monty Forgets' variant should, in my opinion be mentioned earlier. Martin Hogbin (talk) 07:51, 20 October 2014 (UTC)

No Martin. It's much more personal than you are making it out to be. I'm perfectly happy to work with anyone on this article, however it's structured, except you. -- Rick Block (talk) 14:32, 20 October 2014 (UTC)

I do not know why that should be, we have unreconcilable differences concerning the MHP but I have no personal animosity towards you. I was not responsible for the sanctions against you during the Arbcom case. Martin Hogbin (talk) 18:35, 21 October 2014 (UTC)

Did I remotely imply I thought you were? You seem to imagine that I think things that I generally don't. And, quite possibly, vice versa. For example, I think you're bringing up arbcom here to discredit me in this conversation (see poisoning the well). You (of course) didn't bring up that I'm an administrator of fairly long standing, or that this article was (once upon a time) a featured article largely due to my efforts and is no longer arguably largely due to your efforts. Suffice it to say I think it's best if I simply do not engage with you (period). I suggest if you want to pursue this further we take it elsewhere. -- Rick Block (talk) 00:04, 22 October 2014 (UTC)

I do not want to pursue this at all, I was just rather puzzled by your rather personal remarks and wondered what had prompted them. I had no intention whatsoever of trying to discredit you in any way and I apologise if I gave that impression.

On thing for sure is that the reason that this article no longer a featured article is not because of the agreed structure. More likely it was because of the very long period of argument in which all editors lost sight of the purpose of the article.

Now we have an agreed structure, there should be no need for argument about what goes where and it should be possible for everyone to work together to improve it and regain its former FA status. All points of view on the subject should be properly represented and discussed within the article provided that we stick to the principle of having the simple stuff first. Martin Hogbin (talk) 07:42, 22 October 2014 (UTC)

Very simple explanation

Actually, the paradox can be explained with a very short and quick-to-grasp phrase. I think it is beneficial to put a variant of this phrase somewhere close to the beginning of this article. The phrase is this: "If you retain your choice, then you win if you have picked the right door in the beginning, and you lose if you have picked a wrong door back then. If you switch your choice, then you win if you have picked a wrong door in the beginning (since both wrong doors have been eliminated in the process), and you lose if you have picked the right door back then". - 89.110.0.42 (talk) 16:28, 5 November 2014 (UTC)

Are discretionary sanctions here still needed?

An uninvolved editor has asked the Arbitration Committee to review whether the discretionary sanctions are still required in this topic area, and to repeal them if they are not. If you have any opinions on this matter, please comment at Wikipedia:Arbitration/Requests/Clarification and Amendment#Amendment request: Ayn Rand, Monty Hall problem, Longevity, Cold fusion 2, Tree shaping, Gibraltar. Thanks, Thryduulf (talk) 11:58, 14 November 2014 (UTC)

I do not think that discretionary sanctions are in any way helpful to the improvement of this article. Martin Hogbin (talk) 14:08, 22 November 2014 (UTC)

Other host behaviors

I believe the “four-stage two-player game-theoretic” version near the bottom of the table is just a more formal description of the original version. Perhaps it should be merged with the top row. DES (talk) 12:25, 22 November 2014 (UTC)

The original version is not necessarily to be construed as a two person game. The host need not be thought of as a player who can win or lose, and hence would have no strategy for winning. See the reference Seymann, R. G. (1991), which says that in vos Savant's treatment of the problem it is quite clear that "the host is to be viewed as nothing more than an agent of chance". ~ Ningauble (talk) 14:07, 22 November 2014 (UTC)

Thank you, I hadn't considered that. DES (talk) 17:32, 23 November 2014 (UTC)

XKCD

I don't know if you folks have seen this before, but I just ran across it for the first time. http://www.xkcd.com/1282/  :) --Guy Macon (talk) 09:05, 27 February 2015 (UTC)

He was lucky in that he won the right goat ;) Martin Hogbin (talk) 10:13, 27 February 2015 (UTC)

Love will find a way... :) --Guy Macon (talk) 10:44, 27 February 2015 (UTC)

The host opening a door is irrelevant

I wonder if this might be considered original research, but I've come to the realisation that the step where the host opens a door is irrelevant to the probability of getting the car:

Say, I have 3 boxes, only one containing the car. I pick one at random. There's 1/3 chance that I picked the box with the car (in which case, I should stick) and 2/3 chance that I didn't (in which case, I should swap).

Whether or not someone opens a box before I choose (without revealing the location of the car, of course) has no bearing on the probabilities.

Is this something worth noting in the article? Thanks, cmɢʟee⎆τaʟκ 19:40, 24 March 2015 (UTC)

Alas, your conclusion that "Whether or not someone opens a box before I choose (without revealing the location of the car, of course) has no bearing on the probabilities" is incorrect. It actually makes switching a better choice (2/3 chance of winning the car) and not switching a worse choice (1/3 chance of winning the car). BTW, you are in good company. Nearly 1,000 people with PhDs (including at least one world-renowned mathematician) came to the same wrong conclusion that you did. --Guy Macon (talk) 20:35, 24 March 2015 (UTC)

No I can see his point, it's the one that's usually referred to as combining the doors. The opening of the door is the mechanism by which the 2/3 option is collapsed into one choice (the remaining door). However the two 1/3 probabilities of the two unchosen doors are collapsed it's the same. SPACKlick (talk) 20:38, 24 March 2015 (UTC)

Perhaps his description of the game before any doors are opened ("There's 1/3 chance that I picked the box with the car (in which case, I should stick) and 2/3 chance that I didn't (in which case, I should swap") misled him. Yes, at that point there is a 2/3 chance he choose wrongly, but swapping at that point (which is not what happens in the real game) would only let him choose one of the remaining two doors, and thus there would still be a 1/3 chance of winning. --Guy Macon (talk) 20:47, 24 March 2015 (UTC)

Yeah it depends on what he's thinking there. If he's thinking a combined doors style version where "Switch" at that point means "get any cars behind those two doors" then he's right but it's been done. If he hasn't thought about the fact there are two doors left then maybe he's been misled. SPACKlick (talk) 21:19, 24 March 2015 (UTC)

I think that the point that cmglee is making is that the hosts opening of door that is known to hide a goat does not change the probability that the originally chosen door hides the car. This has been a much discussed point in the past. Is that fact obvious? Martin Hogbin (talk) 08:36, 25 March 2015 (UTC)

"Whether or not someone opens a box before I choose (without revealing the location of the car, of course) has no bearing on the probabilities" looks like an imprecise misrepresentation of the "intended clean paradox": cmɢʟee, did you mean "before my ultimate decision, DELIBERATELY showing a goat, and not the car" (as per the intended clean paradox), or did you intend to say "before my ultimate decision, COINCIDENTALLY showing a goat, and not the car"? (see University of California San Diego, Monty KNOWS Version and Monty DOES NOT KNOW Version).

The intended clean paradox says "Before my ultimate decision, DELIBERATELY showing a goat, but NEVER the car". It follows that the chance of the first chosen door still remains 1/3, and the chance of both combined unchosen doors still remains 2/3, yes. But this chance of both combined unchosen doors of 2/3 has collapsed to the still closed unchosen door that the host offers now as an alternative for the ultimate final decision. Gerhardvalentin (talk) 11:32, 25 March 2015 (UTC)

A couple ways to clarify the meaning of the odds

To look at it from a different angle, the contestant has a 2 in 3 chance of choosing a door with a goat. That means the remaining pair of doors have only a 1 in 3 chance of containing two goats. When Monty reveals a goat from the remaining two doors, then the other door will have a goat one-third of the time. Therefore, the contestant halves the odds of getting a goat when switching doors, the same of course of doubling the odds of getting a car.

Note: this also hinders the question as to the importance of Monty Hall revealing a goat.

People get hung up on the idea that the contestant is asked to make a second choice among the remaining two options, leading them to think their odds at that point are 1 in 2. There are two doors. One has a goat, and the other has a car, so a random choice should get the car half the time, they think. It's important to point out that the problem revolves around an initial selection and actions based on that selection. There are two doors, but a car needn't appear behind both of them with the same frequency.

The choices appear to be equal, but they are not because Monty Hall acts on the contestant's choice. His behavior relies on the contestant's initial choice. This is not the same as being offered a choice between two equivalent choices. What happens in a seemingly similar case when there are six contestants in a singing contest and one will be selected to go home? If all things are equal, each contestant puts his/her odds of going home at 1 in 6. But as the MC declares one contestant after another to be safe, the remaining contestants no longer rely on those 1 in 6 odds. Had I placed an initial bet that one of the final two contestants remaining would go home, I would no longer view the odds of my winning that bet as 1 in 6, but I'd think I had a fifty-fifty chance at that point. I wouldn't imagine that the other contestant had a 5 in 6 chance of going home that day just because I made an initial random choice of a contestant. The difference in these scenarios is that the judges and the MC have no knowledge of my bet and take no actions based on it; whereas, Monty preserves the game show contestant's choice when he acts. — Preceding unsigned comment added by Summers999 (talk • contribs) 13:21, 20 April 2015 (UTC)

It is always good to approach the problem from different angles. Different ways of looking at the problem work for different people. It does seem though that once you understand why the answer is 2/3 it is very hard to put yourself back into the mind of someone who does not understand the correct solution.

However, we cannot put everybody's preferred solution or explanation into the article because WP is based on what is said in reliable sources. If you can find a good source that gives your explanation then it could be added to the article. Martin Hogbin (talk) 08:19, 21 April 2015 (UTC)

Edit War over Game Theory

There have been several edits back and forth about whether Game Theory is an appropriate descriptor of the problem. I cannot find a reference to support the inclusion but I would not be surprised if such sources existed. Are there any sources to support the inclusion of the term in the initial description? SPACKlick (talk) 13:17, 10 July 2015 (UTC)

There are a few sources that analyze the problem from a game theoretic perspective, as mentioned in the body of the article, but the vast majority do not. As a degenerate case for games of competition (cf. Seymann, R. G. (1991), "Comment on Let's make a deal: The player's dilemma", American Statistician 45: 287–288: "Simply put, and quite clear considering her [vos Savant's] suggestions for simulation procedures in her two later columns, the host is to be viewed as nothing more than an agent of chance.") MHP may be a useful example for explaining game theory, but using game theory to explain MHP is like shooting a mosquito with the proverbial elephant gun. I do not think it needs to be mentioned in the lede, and it should certainly not be used to state the nature of the problem like this in the opening sentence. ~ Ningauble (talk) 15:54, 10 July 2015 (UTC)

Paragraph which explains why chance doubles by switching

Any random choice of 3 things has a 1/3 chance of finding a certain 1. If you are then shown 1 of the other 2 its not, that changes (in your knowledge) the total chance of the remaining 2 to 100%, but it cant change the 1/3 chance that the 1 you already chose is the winner because the 2 could be eliminated cant be that. Since your first choice has 1/3 chance and total remaining chance is 2/3, you have a 2/3 chance of winning by switching. 97.89.100.56 (talk) 16:17, 14 September 2015 (UTC)

This was removed for being an "unsourced addition", but math is factual and needs no source. 97.89.100.56 (talk) 16:17, 14 September 2015 (UTC)

This is a little way beyond the "Basic arithmetic, such as adding numbers, converting units, or calculating a person's age" permissible under WP:CALC. It also seems to be a not particularly clear reframing of the existing "Monty is saying in effect: you can keep your one door or you can have the other two doors" explanation given in the "Simple solutions" section. --McGeddon (talk) 16:23, 14 September 2015 (UTC)

Beyond basic arithmetic? It doesnt use any numbers bigger than 3 (counting 100% as 1/1). — Preceding unsigned comment added by 97.89.100.56 (talk) 20:02, 14 September 2015 (UTC)

There would need to be a source for the methodology presented. The emphasis or otherwise of elements of the puzzle would be OR without it. SPACKlick (talk) 12:53, 15 September 2015 (UTC)

A goat was revealed

The current lede reinforces the assumption that the option to switch doors is always available, but User:LjL has removed the equally important assumption that a goat, not a car, has been revealed. LjL, you wrote that 'we said (we quoted) "opens another door, which has a GOAT"', but that's also true of Monty offering the option to switch ('He then says to you, "Do you WANT TO PICK door No. 2?"'). So they are no different in that respect.

The point of repeating the key assumptions is to reinforce that both are relevant; in fact, if Monty has a chance to reveal the car when opening the door (but didn't), then both remaining doors have the same probabilities. As (Rosenthal, 2005a) explains, three assumptions are needed to get the standard result:

1. the user can always switch,
2. Monty deliberately chooses a door with a goat, and
3. if there are two doors with goats, Monty chooses at random.

The sentence in the lede about assumptions should mention all three, not just the first one; or maybe none of them, and merely state that "The given probabilities make very specific assumptions about how the host and contestant choose their doors; small differences in their behaviors can yield different probabilities." Diego (talk) 09:30, 20 October 2015 (UTC)

That shorter statement would be okay with me. The lede shouldn't make a long treatise that gets repeated in the article itself, and simply stating things that appear, to a reader who's just reading the lede, to have just been stated in the original quoted version of the problem is just confusing. I've seen too many technical articles that are simply unreadable to the layman because Wikipedia editors want every piece of information to be introduced as soon as possible to not leave any room for doubt... achieving the opposite effect. LjL (talk) 12:22, 20 October 2015 (UTC)

I've tweaked the whole paragraph with a larger change in that direction. See if you like it; if not, I'm OK with using the smaller change with the wording I proposed above. Diego (talk) 13:00, 20 October 2015 (UTC)

It seems okay, although the paragraph is a bit longer than I'd like. LjL (talk) 13:07, 20 October 2015 (UTC)

A tricky word game brain teaser indeed

An easy way to understand why switching is better, is this: When any first choice is made, that choice is 1/3 correct. But after a door is opened, the player has additional information and with that information, the player can also calculate the odds of his next choice. If his next choice is to “stay”, then his odds remain at the original 1/3 (because the actual car location, though still unknown, did not change - and in only 1/3 games is the first pick correct). And since the total odds must equal 1, then switching must be attributed to having 2/3 odds of success. It’s the information which is yielded by the removal of a door that allows the player to refine his second choice. What the revealed information does is allow the player to move away from a 2/3 loss. And in this game, the only way to move away from a 2/3 loss position is by going to the only other remaining door. And under the game rules, the “switch-to” door can only be a 1/3 loss position – because the total must equal 1. Personally, I think one of the major contributors to the erroneous framing which most people experience with this brain teaser is the use of the word “switch”. In modern parlance, most people know the word switch from its use associated with a typical on/off light switch. And because that type of switch is a binary proposition (on/off), one tends to think of a direct opposite such as a yes/no framing. However, the process of improving ones odds with information is a process of refinement, not of “switching”. If the question were asked thusly, this brainteaser would not work: “Is it better to recalculate the odds with the information gleaned from the removal of the door and after that recalculation, consider taking the unchosen door instead; or is it better to disregard that information and ignore whether or not that information might be helpful in informing in your second decision?” 98.118.62.140 (talk) 19:00, 26 October 2015 (UTC)

Has nobody noticed vos Savant's problem in the intro is NOT the same problem described under the Standard Restrictions section?

The standard restrictions section includes "The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999)." This is not part of vos Savant's problem. The host does in fact open a door the contestant didn't pick... she she didn't say he's REQUIRED to. This changes the truth table needed to solve the problem, and thus changes the probabilities, because it assumes that had she picked door #2 instead, he couldn't still have opened door #2. This In short, it's a different problem. Her stated answer to her stated problem was incorrect. Her stated answer to the problem with the additional restriction from the below section would have been correct... but that's not the problem she stated. In her stated problem, the odds that the prize is behind either of the remaining doors is 50%, and her answer was wrong. SteubenGlass (talk) 22:07, 10 August 2015 (UTC) SteubenGlass (talk) 22:07, 10 August 2015 (UTC)

A friend has pointed out that the host "must" open a door she didn't pick, otherwise offering her the choice to pick a different door makes no sense. But in vos Savant's version of the problem, it's never specified that the contestant MUST be offered that choice. vos Savant totally botched the setup of the problem, that's why there's so much confusion. SteubenGlass (talk) 22:25, 10 August 2015 (UTC)

Yes, it is well known that vos Savant did not describe the problem very well. Many sources have discussed what they believe the precise problem intended by vos Savant was and the consensus of these sources is that she intended to specify what we have called the 'standard problem'. I think this is all made clear in the article. Martin Hogbin (talk) 08:50, 11 August 2015 (UTC)

Yes, many people have remarked that her formulation of the problem was imprecise. She herself admitted, obliquely, in a follow-up column that the intended conditions of the problem are only implicit in the answer she gave, not in the question as originally stated. She wrote: "So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. ... Anything else is a different question." [emphasis added]

I imagine she thought there was something implicit in saying "Suppose you’re on a game show", and in the general nature of this sort of riddle, that would make her original intent clear. It was evidently not clear enough, or was sufficiently open to nit picking, that later clarification was necessary.

This is a sorry state of affairs because so much attention has been given how sloppy her formulation of the problem was, rather than the much more interesting fact that the intended problem defies common sense: most people get it completely wrong even when it is stated in a completely precise form. Vos Savant surely had no idea of the importance this simple riddle would take on, or she would have been a little more careful about how she presented it. ~ Ningauble (talk) 15:00, 11 August 2015 (UTC)

I still think this needs to be addressed in the lead section. The fact that the answer depends so subtly on conditions of this type is actually one of the most interesting things about the whole paradox. I am not at all convinced that vos Savant had considered this issue in her original column; she may well have assumed that it didn't make any difference. Even at this late date I have run into people who cannot be convinced that it does make any difference. --Trovatore (talk) 03:49, 12 August 2015 (UTC)

The lead section already alludes to problems of interpreting the question when it says "under the standard assumptions". (At one time the article explained those assumptions in the very first section after the lead.) Perhaps slightly more could be said about it there, but it is not really so ambiguous.

"Suppose you're on a game show, and you're given the choice of three doors" and after you pick a door the host says "Ok, you win/lose. Thanks for playing." Not much of a game, is it? Suppose the host opens the door you picked and asks if you want to pick another. Kind of a no-brainer, isn't it? Suppose the host shows you where the car is, and asks whether you want it. What game show ever worked like that? No, let's suppose you're on a realistic game show.

I am quite convinced vos Savant did not consider such scenarios in her original column: the answer she gave in that column makes it clear what scenario she had in mind. The essence of the problem lies in being given a second chance with additional information that is still inconclusive. As vos Savant wrote: "Anything else is a different question."

I agree that quite a few people are convinced that this makes no difference. Many theories have been suggested to explain why people don't get it, and there are probably multiple factors in play. The two that are most interesting to me are confirmation bias (the goat behind door #3 taken as confirming evidence for the car being behind door #1), and the problem that common sense intuitive reasoning is ill equipped for interpreting selective evidence (as if it doesn't matter that Monty knows what's behind the doors).

The real difficulty of the Monty Hall problem lies in the nature of the problem itself, not in the vagueness of saying "suppose you're on a game show". ~ Ningauble (talk) 14:30, 12 August 2015 (UTC)

I don't know how old you are; you may never have watched the actual show. Americans of my generation or older mostly have watched it at some point, and their analysis is likely to have been influenced by that.

My recollection is that Monty appeared to make these offers (to switch) more or less ad lib. You didn't actually know whether it was scripted or not, could have been, or he could have decided on the spot whether or not to make the offer. It was entirely obscure whether his decision to make the offer or not depended on whether you had picked the door with the car.

So I disagree with your assertion that "it is not really so ambiguous" for a "realistic game show", because it is completely ambiguous for the actual game show. --Trovatore (talk) 16:59, 12 August 2015 (UTC)

Well, it is completely different than the actual game show (which I do indeed remember from my youth), where you never got a second shot at a choice among the same three options – only an offer to trade for something else.

Have you read Krauss & Wang (2003), cited in the article? In the section "Are There Possible Effects of Incomplete Information?" at pp. 9–10, they make a good case that it is not ambiguity of the rules that leads people astray, but an inability to appropriately model the scenario for solution.

Some people do get hung up on guessing at Monty's motivations, but Krauss & Wang observe that such guesses would not logically lead to the common answer that the odds are 50:50! (Any lack of impartiality in the host is really not in the spirit of game shows from his era, for those who remember, but harkens back to a much earlier era of crooked games on the radio, which were before my time.)

I am not aware of any reliable source that demonstrates or claims that people get it wrong due to trying to apply rules from the actual game show. ~ Ningauble (talk) 19:14, 12 August 2015 (UTC)

In fact there's a letter from Monty Hall to Steve Selvin, cited in the article, in which he clarifies that on the show the rules were "no trading boxes after selection" (this was 15 years before vos Savant's column) - so applying rules from the game show to the problem simply doesn't work. What was offered was cash in lieu of whatever was behind the selected door, and sometimes a non-selected door was opened (making the contestant think her odds of winning had gone up). This is discussed in the NY Times interview with Monty (also referenced in the article) in which Monty Hall agrees that you should switch if the host must open a door and make the offer to switch doors [there's an implicit assumption that the host knows where the car is and never reveals it], and then goes on to demonstrate (!) that if the host is not compelled to do this you could lose every time by switching. -- Rick Block (talk) 15:43, 13 August 2015 (UTC)

The majority of the people in general may get it wrong even among the standard assumption (according to Krauss/Wang). However much/most of academic discussion of the subject deals exactly with the ambiguity of the original problem, the justification of those standard assumptions and what is happening when they're dropped. So having that in mind it is indeed justified to mention something to that regard in the lead.

Having said that however this however, this is also reopening old (and apparently neverending bitter) edit disputes over the most appropiate descrtiption of the problem and which aspects are supposed to be played down or up and which aspect is the "really important" one and the "true reading" of the problem. So in doubt be prepared for yet another round.--Kmhkmh (talk) 20:02, 13 August 2015 (UTC)

I suggest the word 'always' is removed from the Standard Assumptions. If taken to mean 'every time the MHP is stated' it is redundant, and if 'every time the game show plays the three-door game' it is incorrect. In the citation 1991a, vos Savant uses 'always' only in the context of the six games that exhaust all the MHP possibilities. If the host acts non-randomly, the MHP odds apply in those games where he reveals a non-chosen goat even if in other games he reveals the car, the chosen door or nothing.Freddie Orrell (talk) 06:34, 11 October 2015 (UTC)

I disagree, per the sources that are cited in this section. And it is not correct that "the MHP odds apply in those games where he reveals a non-chosen goat even if in other games he ...". See, for example, the variant Monty Hall himself apparently created on the spot during a 1991 interview with John Tierney from the NY Times (cited in the article). In this "Monty from Hell" variant, switching always loses. If you'd like to discuss this I suggest we move to the /Arguments page. -- Rick Block (talk) 16:16, 11 October 2015 (UTC)

@SteubenGlass: Of course you are absolutely right. Marilyn vos Savant's Monty Hall problem, together with the claim of the 2/3 solution, is a joke. And it persists because publicists and their followers feeled confident that the 2/3 solution holds without any additional assumption. But according to the Wikipedia rules and their brave advocats we have an excellent article: it is just as bad as the so-called reputable sources.--Albtal (talk) 22:41, 6 November 2015 (UTC)

I think we have been here before. Vos Savant was quite clear about what exactly problem she intended to describe and it is the problem with the standard assumptions as stated in this article. Martin Hogbin (talk) 09:43, 7 November 2015 (UTC)

Well, she made it clear in the second article, after the point had been raised. It's not at all clear whether she had thought of it at the time of the first article, or realized it mattered. --Trovatore (talk) 18:23, 7 November 2015 (UTC)

It doesn't matter at all what Marilyn intended. It's only important which problem has been going around the world. The shell game which she proposed to illustrate her 2/3 solution, really has this solution. But her implicit claim, that the shell game corresponds to her Monty Hall problem, is wrong. Marilyn, as opposed to most of her poor followers, knows that the crucial condition has been missing in her original Parade problem. But she also knows that revealing this would collapse her famous Monty Hall paradox. - Oh yes, the Wikipedia article: Write and motivate that in the Parade problem there is no reason to prefer one of the remaining doors. Formulate the most simple problem which has really a 2/3 solution, with the immediate implication that the contestant will win by switching in 2 of 3 cases. That's all.--Albtal (talk) 16:31, 7 November 2015 (UTC)

Are the odds ever 50-50 ?

It's been established that the contestant and Monty have produced a pair of doors containing a car and a goat that do not have the same odds of producing a goat. Yet, on Marilyn vos Savant's website where she discusses this teaser, she perhaps yields too much ground to the academic furor that confronted her. She made this concession:

"Suppose we pause at that point [when two doors remain], and a UFO settles down onto the stage. A little green woman emerges, and the host asks her to point to one of the two unopened doors. The chances that she’ll randomly choose the one with the prize are 1/2, all right. But that’s because she lacks the advantage the original contestant had—the help of the host." [3]

But are the chances 50%? Probabilities don't care much for who does the choosing, and neither the contestant nor the green woman has insider information about the whereabouts of the car behind the doors. The contestant can know that the odds of finding a car behind the switched door are 2/3, but the odds don't change just because someone doesn't know the odds. If I know that four sides of a die are red and two are blue before I paint over the sides, the odds of someone who has never seen the die rolling red are not 1/2 due to her unawareness.

If the odds of randomly selecting a car among the two blind choices are 50-50 for anyone other than the contestant, then they must be 50-50 for everyone. We know that they are not, so the green woman approaches a stacked deck, so to speak. Otherwise, we end up saying that all odds between situations with two outcomes are 50-50 so long as we know nothing about the underlying conditions. That's not correct.

Vos Savant refers to the help of the host as being a missing ingredient, but that help actually settled the odds prior to his final selection, and it must've done so for anyone choosing between the two doors. We are now convinced that this problem is a matter of frequency of appearance of the car behind particular doors after particular actions have settled the odds, so vos Savant needn't have made any concessions to the contrary.

Thoughts? — Preceding unsigned comment added by Summers999 (talk • contribs) 14:24, 25 April 2015 (UTC)

The question is what odds are you talking about? If you pick randomly between two choices the odds of your choice being the correct one are 50-50. Note how I worded this - the odds we're talking about are the odds of "your choice". If we roll your red/blue die the odds are 4/6 that it ends up red and 2/6 that it ends up blue. If you roll it 100 times and I randomly guess red or blue, about 50 times I'll guess red and 50 times I'll guess blue. Of the 50 times I guess red I'll be right about 4/6 of the time, so about 33 times. Of the 50 times I guess blue I'll be right about 2/6 of the time, so about 17 times. Altogether I'm right about 50 times out of my 100 guesses, i.e. 50-50. If I don't know 4 sides are red and 2 are blue, my odds of correctly guessing are 50-50 whatever the "actual" odds may be because I have no choice but to randomly guess. If I always guess red (because it's my favorite color), I'll be right about 4/6 of the time, but now we're not talking about the odds of a random guess but the odds that the die ends up red (which is a different question).

If you want to talk about this further let's move the discussion to the Talk:Monty Hall problem/Archive 39/Arguments page. -- Rick Block (talk) 16:40, 25 April 2015 (UTC)

I'm only referring to a single selection of unequal odds. Namely, if the contestant or a green alien chose one of the remaining doors, it would still be better to choose a particular door. And this is true whether or not someone asked to select knows of the contestant's original choice. Even two of three random selections will yield a car from the alternative door, just as blind rolls of the die I described will yield red twice as often as blue. Summers999 (talk) 21:32, 25 April 2015 (UTC) user:Summers999

Like I said - if you want to talk about this further let's move it to Talk:Monty Hall problem/Archive 39/Arguments. -- Rick Block (talk) 04:09, 26 April 2015 (UTC)

It entirely depends on your information, that's all odds are.

The car is behind a specific door. Stick or switch are each either 1 or 0 probability of getting the car. The host knows the car is fixed so for the host choosing stick or switch the odds are 0 or 1 The contestant doesn't know where the car is but he knows the host could have opened door 2 and didn't the odds are 0.33 0.67 stick and switch. The alien doesn't know which door the host chose not to open and so the odds with that information are 50/50 SPACKlick (talk) 09:42, 1 May 2015 (UTC)

Thank you for stating the truth: Odds truly only do exist on the information level. Without information, odds cannot be calculated. But on the extant level, the level of what actually is, the distribution rate of cars to doors is indeed 1/2 after a door is removed. In other words, it's the information about the removed door which informs the choice to switch, not the odds of where the car is. We are not in fact calculating the odds of where the car is. Rather, we are calculating the odds of how effectively we'll find it. And it's that distinction, a distinction which even most probability experts fail to clarify, which muddies the water regarding this brain teaser. 98.118.62.140 (talk) 05:56, 14 October 2015 (UTC)

That is the normal (Bayesian) meaning of the words 'probability' or 'odds'. Martin Hogbin (talk) 00:19, 10 November 2015 (UTC)

The odds will be 50/50 if and only if the host were to randomly (mindlessly) open a door and this door happens to not be the prize, then your odds will instantly change form .3333 to .5 The main difference in the actual problem is that the host does not mindlessly open a door, he specifically opens the one without the prize, which in effect adds value to the one he did not open--Mapsfly (talk) 04:46, 2 December 2015 (UTC)

This important point is made in the article. Martin Hogbin (talk) 09:55, 2 December 2015 (UTC)

Is this *problem* a "stupid filter"?

Open speculation on the intelligence / psychological state of living persons is an unwise use of anyone's time here. I, JethroBT drop me a line 03:07, 10 January 2016 (UTC)

The odds of choosing the car on the first choice are indeed 1/3 compared to a 2/3 chance of choosing a goat, we all understand that. But the first choice isn't what wins or loses the game. Regardless of whichever door you pick, one of the remaining two doors is discarded from play and what you're left with is objectively a 50/50 chance of winning or losing based upon your second choice of switching or staying. Instead of choosing between 2 goats and a car, you're choosing between 1 goat and a car. The elimination of a goat does not add any information into the problem, because subtraction is objectively not addition. You initially had a 1/3rd chance of blindly choosing the car in the first round because there is only 1 car of 3 possible doors but the first round doesn't win or lose the game and may as well be discarded from the problem entirely when getting down to the statistics, as when the guaranteed second round occurs there is now 1 car behind 2 possible doors, or 1/2 chance. Stick or switch, you *will* win half of the time. It seems there is extremely flawed math behind people who think it as a 2/3 chance for winning by switching, which is expressed by the extremely poorly coded "simulations" one might find by google searching a monty hall problem simulator. Was the person who invented the "proof" of this 2/3 switch win answer even a mathematician or were they known for otherwise pulling people's legs/being a pathological liar? 24.205.110.111 (talk) 21:41, 8 January 2016 (UTC) I looked up Von Savant's wiki page and apparently she wasn't too bright, unable to understand the concept of imaginary numbers, thinking $600 is worth more than $1000, clearly not a mathematician, and also having fabricated the "highest recorded IQ". Good to know, that explains a lot. 24.205.110.111 (talk) 22:11, 8 January 2016 (UTC) Going back to the earliest cited source for this problem within the article, https://www.jstor.org/stable/2683689 , you can find an error in Steve Selvin's math. Von Savant parroted his answer (knowingly or unknowingly), and therefore suffers from the same problem. Here is a table of all possible games in the Monty Hall problem, similar to the table Steve Selvin offered - except this table doesn't incorrectly combine two individual similar but unique results into a single result.24.205.110.111 (talk) 05:23, 9 January 2016 (UTC) Your table has a problem - specifically, each of the rows are not equally probable. For example, the first two rows each have half the probability (1/18) of the next two (1/9). The overall odds of winning by switching are indeed 2/3 - which you can see by adding up the probabilities of the "switch" rows. There are 12 rows, and 6 are rows where switching wins - but they each have probability 1/9, not 1/12. The odds of winning by switching if you pick door 1, the host opens door 3, and you switch to door 2 are also 2/3 (under normal assumptions). If you'd like to discuss this further, let's take it to the /Arguments page. -- Rick Block (talk) 17:01, 9 January 2016 (UTC) Each of the rows are equally probable though, no? For example, lets say it takes 2 minutes to play a monty hall game, and a single monty hall game is a game where [1]The player chooses a single door among 3 doors, [2]A goat is revealed behind door(x) that isn't the chosen door, and [3]The player can choose to stay with their door or switch to the closed door which nets them either a win or a loss. Using these three parts as the assumption of what a basic monty hall game is composed of, and the given that a single game takes 2 minutes to play, Game 1 with the end-result of [A, B, A/win] is an entirely different and equally probable set than Game 2 with the end-result of [A, C, A/win]. Monty can only reveal one goat, and if the chosen Door A is the car behind Door A, he has to reveal either Door B or Door C, but not simultaneously or both within a single game. These two different winning-by-staying situations have the same result, but you can also get the same result by choosing either goat door then switching. It doesn't make sense to combine the "stay-to-win" results [A, B, A/win] and [A, C, A/win] while simultaneously NOT combining the "switch-to-win" results [B, C, A/win] and [C, B, A/win], when both "stay-to-win" games have a total combined playing time of 4 minutes and both "switch-to-win" games also have a total combined playing time of 4 minutes. Do you understand? I've made two javascript simulators to explain: here and here. 24.205.110.111 (talk) 01:51, 10 January 2016 (UTC)

Correct?

It is a matter of discussion whether vos Savant's response was correct. Even under the standard conditions several authors consider vos Savant's reasoning not correct, although the actual advice of switching is right. Nijdam (talk) 19:46, 10 January 2016 (UTC)

Quite a lot of disussion if I remember correctly. Martin Hogbin (talk) 22:04, 10 January 2016 (UTC)

Actually, I do not see an issue with using 'correct' here. We had, 'Vos Savant's correct response was that the contestant should switch to the other door.' The correct response, as defined in this sentence was that the contestant should switch, which we all agree is correct. The Engish here does not imply that her entire response was correct, only that the advice to switch was correct. Martin Hogbin (talk) 22:09, 10 January 2016 (UTC)

Here is Monty Hall Problem simulator I made. This simulator removes the win-by-stay/win-by-switch choice and assumes the only method is winning-by-staying (the simulator always stays with the first choice). 1/2 of the wins get the Car, 1/4 of the wins get one goat, and 1/4 of the wins get the other goat. It clearly shows that there is a 50% chance of winning the car by staying and that there is no advantage by switching, so Vos Savant was incorrect. 24.205.110.111 (talk) 00:41, 11 January 2016 (UTC)

@24.205.110.111: Please take this up on the /Arguments page as I suggested. Suffice it to say that you're wrong, but this is not the place to argue about it. -- Rick Block (talk) 04:55, 11 January 2016 (UTC)

@Martin Hogbin: Adding "correct" implies Wikipedia is taking a stance here. This is not appropriate, and is in fact counter to a fundamental Wikipedia policy, which is that Wikipedia is neutral with respect to all sources - please see WP:NPOV. What is unambiguously true is that vos Savant's response was that the contestant should switch. Wikipedia certainly can and should say this. Adding "correct" is saying that Wikipedia agrees. Wikipedia does not agree. Nor does Wikipedia disagree. Wikipedia could say "most reliable sources agree" - but Wikipedia cannot, by fundamental policy, say vos Savant's answer is "correct". -- Rick Block (talk) 04:55, 11 January 2016 (UTC)

This is a question about the meaning of English. I am not trying to restart old disputes. My understanding of the sentence, 'Vos Savant's correct response was that the contestant should switch to the other door.' is just that that her words 'you should switch' were correct, it does not imply that her entire response and method of solution (which is not stated here) is correct. Do you understand the English that way?

If you do not understand the sentence that way I would be perfectly happy to make it clearer (so long as we did not use an awkward or hard-to-understand construction or something that implies some sort of mysterty, later to be revealed).

I think it is accepted by all sources that, given any resonable intepretation of the problem, it is to your advantage to switch. I think it would be fair to say that this fact could be considered the result of a routine calculation and part of mainstream maths and science. It is a generally accepted fact that vos Savant was correct in saying that you should switch. Martin Hogbin (talk) 10:32, 11 January 2016 (UTC)

I am not aware of any good quality reliable sources that say you should not switch. It is widely, in fact I would say universally, accepted that vS was correct in advising players to switch. I do know that the rest of her analysis is debated by some but please let us keep clear of that discussion. Martin Hogbin (talk) 10:32, 11 January 2016 (UTC)

Can somebody else chime in here? I do not agree, but I am not going to argue with Martin. -- Rick Block (talk) 12:52, 11 January 2016 (UTC)

Yes please. It seems very simple to me. Vos Savant was correct when she said that you should switch. Nobody disputes this fact. Martin Hogbin (talk) 22:44, 11 January 2016 (UTC)

Well, who's there to decide? Certainly not Wikipedia. Nijdam (talk) 10:53, 12 January 2016 (UTC)

I have no problem with WP using "correct" with vS's interpretation. There were certainly a lot of people who said that she was wrong, and most of those people were confused/wrong rather than delving into subtle issues about the problem statement. Glrx (talk) 20:58, 14 January 2016 (UTC)

Nice to know. Nijdam (talk) 09:28, 15 January 2016 (UTC)

I have no problem referring to the conclusion that "player should switch" as the correct solution. However referring to VS's response as correct probably shouldn't be in WP's voice as it is disputed by some experts. The sentence refers to VS's response as correct rather than, as I would prefer "Correct conclusion". SPACKlick (talk) 12:30, 15 January 2016 (UTC)

The disputed text is, 'Vos Savant's response was that the contestant should switch to the other door'. So what 'her response' was is defined immediately afterwords to be 'that the player should switch doors'. It is purely a matter of English ther is no dispute about the maths.

If sombody wants to change the wording to make it absolutely clear that WP is saying only that the specific response/answer/solution/advice 'you should switch doors' is correct that would be fine with me. This is what most readers are interested in. I assume that everyone agree that switching doors is the correct thing to do? Martin Hogbin (talk) 12:47, 15 January 2016 (UTC)

FourthKind Solution: Monty Hall problem Solved and Confusion Explained

The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

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FourthKind Solution: Monty Hall problem Solved and Confusion Explained

There are 2 games being played here on Wikipedia and the media in general. Let’s call them:

1: Monty Hall Game

2: 1/3-2/3 Model Trap Game

Explanation for Monty Hall Game

The Paradox

The question posed in the Monty Hall problem on Wikipedia: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a)

Standard Assumptions

1. The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999).

2. The host must always open a door to reveal a goat and never the car.

3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

(Note there are 3)

Answer to the Paradox

Question: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a)

Answer: No (with the 3 Standard Assumptions, choose at random)

Not So Simple Solution

If the contestant adheres to all of the 3 Standard Assumptions, the contestant should answer randomly unless the contestant is really good at reading Monty’s “tells”. Toss a coin in your head, use a pocket random generator, etc. The contestant doesn’t even really have to decide anyway. If the contestant can’t decide, guess, that is a random choice itself. A random choice, without reason. It is a 50/50 probability at the real moment of choice (the contestant answering Standard Assumption #3), that is the real moment of consequence when the probabilities become fixed. The moment when win or lose are real possibilities with unchangeable outcomes. Your final answer. There is no consequence in choosing a door initially when the probability for the contestant winning is 0% or losing 0%, the game show producers winning 0% or losing 0%. Why even consider this in the Monty Hall Game? The answer is because it has great consideration in the 1/3-2/3 Model Trap Game. This is the “connection” some mathematicians make between the Monty Hall Game and the 1/3-2/3 Model Trap Game. It is not important to the outcome of the Monty Hall Game probabilities. It is important to the outcome of the 1/3-2/3 Model Trap Game probabilities to explain how the contestant makes an error (losing 2/3 of the time) by “always-staying” in the 1/3-2/3 Model Trap Game. This is why. In the Monty Hall Game the initial door choice is merely used as a “reference point” whether you changed your mind (switch) or not (stay) later in the game. It must be considered in the 1/3-2/3 Model Trap Game. In the 1/3-2/3 Model Trap Game this is a necessary & fixed 4th condition.

After all, the game is not decided when choosing an initial door in the Monty Hall Game but the game is decided at that point in the 1/3-2/3 Model Trap Game. It is 1 of only 2 possible avenues of choice to take in 1/3-2/3 Model Trap Game. In the Monty Hall Game the initial door chosen is merely a reference point. In the 1/3-2/3 Model Trap Game it is described in 1 of only 2 possible answers, always-stay or always-switch, which the player must always choose from (to prove the probability calculations/model). That’s important! That is the difference. The Monty Hall Game has no repeated fixed answer, after all it is a one-time, 2 door, 2 choice, 50/50 event. The reason contestants fear choosing the “wrong door”. Maybe they missed some math advantage? Enter the 1/3-2/3 Model Trap Game.

In the Monty Hall Game the real game from the contestant’s and Monty’s perspective begins after Monty’s 2nd question (via Standard Assumption #3). The game for the contestant (not the viewer) is not after selecting a door of no consequence (sorry Monty). After all the game always continues via Standard Assumption #3. The game never stops after the initial door is chosen by the contestant and the initial door choice has no consequence to the outcome other than being used as a reference point. It is merely used as showmanship for the show (sorry Monty). Down to 2 doors. Win 50% or Lose 50%. Who has the advantage you or Monty. Neither, so the correct answer to the question: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a) is “no”.

The Monty Hall Game For the contestant is a one-time event. The contestant gets 1 choice in their life (assuming Monty doesn’t ask them back). No advantage for Monty, no advantage for the contestant. The choice is 50/50 (2 doors left, 2 possible choices WARNING: switch or stay will result in only a 1/2 advantage).

1/3-2/3 Model Trap Game In the 1/3-2/3 Model Trap Game the game continues for as long as you wish. Play always-stay or always-switch for as long as you want. Answer: yes (with the 4 Standard Assumptions, choose the always-switch answer this will result in a 2/3 advantage and the always-stay answer will result in a 1/3 advantage: WARNING: choosing a random-only answer will result in only a 1/2 advantage) It is not the Monty Hall Game. Even Monty stated as such.

Explanation for 1/3-2/3 Model Trap Game

The Paradox Question: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a) Answer: Yes (with the 4 Standard Assumptions, choose always-switch answer this gives the player a 2/3 probability of winning choosing always-stay answer will result in only a 1/3 probability of winning)

Standard Assumptions

1. The host must always open a door that was not picked by the contestant (Mueser and Granberg 1999).

2. The host must always open a door to reveal a goat and never the car.

3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

4. The contestant must always-stay or always-switch (an improbable event unless the contestant has been on a previous show)

(Note there are 4)

Answer to the Paradox

Question: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a) Answer: yes (with the 4 Standard Assumptions, choose the always-switch answer this will result in a 2/3 advantage: WARNING: choosing a random-only answer will result in only a 1/2 advantage)

Simple Solution

If you adhere to all of the 4 standard assumptions and you always-switch, you win 2/3 of the time. If you always-stay you win 1/3 of the time. If you choose a random answer when asked the important question of whether to stay or switch (cheat) you will win 50% of the time playing this model. How unusual (explanation later)!

Confusion Explained: Mixing the 2 different games

You have noticed by now there are 2 different games competing for referencing solutions. 2 different games competing for rules (Standard Assumptions). 2 different games explaining how to win (Simple Solutions). Which is the real game? FourthKind proposes that the Monty Hall Game is the real Deal (oops, thanks Monty). Monty Hall even claims it is the real game. But that is boring and too simple. 50/50, who needs a mathematician for that?

Comparing the correct answers of the 2 games to the original question posed

The original question posed: Is it to your advantage to switch your choice? (Whitaker, 1990, as quoted by vos Savant 1990a)

Monty Hall Game: no (choose randomly for an even chance)

1/3-2/3 Model Trap Game: yes (“always-switch” choice gives the player a 2/3 probability advantage

Possible Solutions to the Confusion

1: A 4th Standard Assumption such as the one I proposed needs to be added to the 3 Standard Assumptions of the Monty Hall Game before trying to explain the Monty Hall Game using the 1/3-2/3 Model Trap Game.

2: Split the connection of the Monty Hall Game and 1/3-2/3 Model Trap Game entirely (retract the “solutions” to the Monty Hall Game)

3: Start a 1/3-2/3 Model Trap Game page in Wikipedia.

FourthKind FourthKind (talk) 14:02, 16 February 2016 (UTC)

Every time a constant chooses to stick the probability is 2/3 of winning, every time a constant switches the probability is 1/3 of winning. A strategy of randomly choosing to stick, or switch, is statistically less successful than always sticking and more successful than always switching.

Also, a reference for "The contestant must always switch or always stay with their original door choice when offered by the host. (an improbable event unless the contestant has been on a previous show)" would be needed, as Wikipedia aims for verifiability rather than truth - see WP:NOTTRUTH. Jonpatterns (talk) 15:53, 14 February 2016 (UTC)

FourthKind, the Monty Hall problem is infamous for fooling nearly everybody and you are one of those that it has fooled. This page in not for discussing why wrong solutions are wrong or for helping people understand the problem and the correct solution. If you want to do this I sggest that you go back to the arguments page. Martin Hogbin (talk) 16:09, 14 February 2016 (UTC)

Martin Hogbin thank you for commenting on my posting, I truly appreciate it and with all due respect thank you for also helping me prove my points.

You state: "Every time a constant chooses to stick the probability is 2/3 of winning, every time a constant switches the probability is 1/3 of winning"

My Point: I don't see the words "every time" in any of the 3 Standard Assumptions. There needs to be a 4th Standard Assumption for the 1/3-2/3 statistical models to be a valid mathematical argument. This new assumption or method of choosing is not stated before the mathematical arguments are proposed. They are continually referenced as if there are the only 2 possible avenues available to the contestant, not taking into account a 3rd possible avenue of “always-random”. This 3rd avenue is always overlooked and never addressed by any of the 1/3-2/3 models. The 4th Standard Assumption being the one I proposed. The contestant must always switch or always stay with their original door choice when offered by the host.

The Confusion problem with the public: That they are told by some mathematicians that they have a 1/3 chance of winning from the beginning. Before the contestant answers Monty's 2nd question: Do you want to stay or switch? the probability is a 50/50 chance. That is why I believe the public is horrified to be shown such a disadvantage that is often described as not "intuitive".

The confusion being: The problem is explained mathematically with an added condition of "every time" they stay and "every time" they switch. A flaw of reasoning by proving a mathematical model which will ONLY works with this added condition after the 3 Standard Assumptions are explained. This condition incidentally is not in any the 3 Standard Assumptions. Ask yourself who added this extra condition? The answer is the mathematicians who explain the problem using the 1/3-2/3 models. Mathematicians who use this model need this 4th condition to prove Monty has an edge with all the 1/3-2/3 models. There is currently no such condition under the 3 Standard Assumptions. This 4th Assumption needs to be added, or the 1/3-2/3 models to be refined. This is root of the confusion. That is why the public is horrified to be "proven" they were at such a disadvantage. The public is given an explained with a "proof" that is often referred as not "intuitive" and most people fall into a "fool's trap" by staying (which in reality, later in the show (when they both share the “discarded” set) a simple 50/50 chance choice is presented. They can’t figure out why Monty has such an edge (2/3 vs. contestants 1/3). Monty's edge is created by mathematicians and needs a 4th Assumption to be proven accurately and taken seriously. Covertly slipping it in at a later date, unbeknownst to the public undermines the credibility of mathematics in general. Also, mathematicians on YouTube and on the Wikipedia page for the Monty Hall problem do not explain the concept that the "discarded 999,998 choices" in the 1 million-door explanation. Mathematicians fail to explain that these "discarded choices" (also relevant in the 3 doors problem) shown by the host are now a set of "shared choices" by both the contestant and Monty. A set that is shared by both the contestant and Monty now makes the probability a 50/50 one as both sides have 999,999 chances out of 1 million. The 999,998 chances are now exposed to both. The contestant is not stuck with one-in-a-million chance of success as described at the beginning by the mathematician. The 50/50 probability at the end is presented as 1 vs. 1,000,000. Which would you choose? The power of math, or I should say the power of some mathematicians with the 1/3-2/3 models.

Also, I hope the statement: "Wikipedia aims for verifiability rather than truth" is a company slogan and not to be taken seriously. As for me, truth by democracy is not my cup of tea.

Honestly and sincerely thank you again. FourthKind FourthKind (talk) 12:04, 15 February 2016 (UTC)

@FourthKind: It doesn't matter how many times the content plays the game or only once the probability to stick is always 2/3 and switch always 1/3. The success rate over multiple goes is in fact a different question. There are no shared choices, see assumption two '2.The host must always open a door to reveal a goat and never the car.' Or in the many door question 'the host must always open all the remaining doors bar one, only revealing goats'.

Okay, now consider I'm wrong and you are right. Wikipedia will need a reference to back up your claim about the fourth required assumption, such as a peer reviewed paper. Its outside the scope of Wikipedia to do mathematical proofs. The scope of Wikipedia is to write an encyclopedia about proofs that have already been published. Jonpatterns (talk) 13:36, 15 February 2016 (UTC)

}}

Can we please keep this kind of mathematical or philosophical discussion to the arguments page. Martin Hogbin (talk) 14:55, 16 February 2016 (UTC)

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Most statements of the problem, notably the one in Parade Magazine, do not match the rules of the actual game show?

I do not think this is correct. Monty Hall made clear that nothing like the Monty Hall Problem, as it is widely understood, could ever have happened on the game show. Martin Hogbin (talk) 14:34, 29 December 2015 (UTC)

And how does this conflict with "most statements of the problem do not match the rules of the actual game show" (which, BTW, is sourced)? -- Rick Block (talk) 17:35, 31 December 2015 (UTC)

By starting with, 'Most statements of the problem..', the wording suggests that there was a real game that was similar to the MHP. We know from Monty Hall's words that nothing like the well known problem could have happened on the TV show. Martin Hogbin (talk) 16:09, 1 January 2016 (UTC)

It's true that the game show in the Monty Hall Problem has rules different from those of Let's Make a Deal. The real Monty Hall varied his moves, to keep the show confusing for contestants and fun for viewers. In the MHP, the host follows a strict algorithm.

The distinction between the hypothetical and real game shows isn't very important to the MHP. What is important is that the statement of the problem stipulate that the hypothetical host, unlike his real-life counterpart, ALWAYS reveals one of the unchosen doors as a loser. Marilyn Vos Savant, for example, failed to include this in her initial discussion of the MHP. TypoBoy (talk) 20:38, 14 March 2016 (UTC)

What was the date of inception for the "Monty Hall problem" in Wikipedia?

Wikipedia talk pages are for discussion of article improvements, not general discussion

FourthKind FourthKind (talk) 11:56, 21 April 2016 (UTC) February 1975, as it says in the first sentence.SPACKlick (talk) 12:09, 21 April 2016 (UTC) If you're referring to the creation date of this article, you can find it at the history page, all the way down to the oldest edits. It was created on by AxelBoldt at 23:29, 22 September 2001 with a reference from a New York Times article in July 21, 1991. Diego (talk) 12:42, 21 April 2016 (UTC)

The probability is 2/3. Where is the trick?

Non-editorial exchange moved to /Arguments subpage. --Trovatore (talk) 20:46, 10 August 2016 (UTC)

I realized that the odds of winning pass to 2/3 changing the initial choice and I realized my mistake. I explain it because it could help somebody. What the host do opening a door and showing a goat is, in fact, to give to the player the possibility of inverting his initial choice (if he found a goat he’ll get a car and vice versa). For a better explanation I use an extreme version of the game where the initial choice is made between 100 doors with 1 car e 99 goats. If the host asked to the player “Do you want invert the result of your initial choice?” it’s easy to realize that changing is convenient. Coming back to the game with 3 doors, the opening of a door with the goat from the host, even though has the effect of giving to the host the possibility of invert his choice, it’s confused with the random opening of a door behind wich there is a goat. It’s this the event that leads to 50-50 odds (but that clearly is a different game because the host has 1/3 of chance of opening a door with the car and this is impossible in the original game). Alessmaga (talk) 09:29, 11 August 2016 (UTC)

Example door numbers should be removed from the quoted letter at the beginning of the page as they just confuse you

Example door numbers should be removed from the quoted letter at the beginning of the page as they just confuse you. If we are specifying that the player choose door 1 and the host opened 3 and it contained a goat, there is no advantage in switching. both switching and stay give a 50% chance of winning (because if the host opened door n3, it means the car could not have been there in the first place, in this case, so only two possibilities are left). You should just explain that the player picks a door and the host opens another one with a goat and he ask the player if he wants to switch to the remaining door. It is the sum of all the cases together that gives you an advantage by switching. — Preceding unsigned comment added by 2001:b07:644c:a33:70f1:7f9b:d96d:4295 (talk • contribs) 23:21, 26 August 2016 (UTC)

If it's not verbatim, it's not a quote. I think that failing to quote exactly what the question actually said can only increase confusion and doubt about what it means. As vos Savant herself has said, "anything else is a different question". ~ Ningauble (talk) 16:15, 3 September 2016 (UTC)

whats wrong with this?

We have this:

[If]f the contestant picks a goat (2 of 3 doors) the contestant will win the car by switching as the other goat can no longer be picked, while if the contestant picks the car (1 of 3 doors) the contestant will not win the car by switching The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial probability.

and an editor (not me) wanted to add this:

[I]n 2 out of 3 chances, the contestants first guess will be wrong. If the first guess is wrong, and the host shows the other wrong choice, then the remaining door must be the winner. Thus, in 2 out of 3 chances, switching will win.

and has been reverted. But the first passage is poor. "the contestant will win the car by switching" is not true. She might win the car by switching. She will increase her odds by switching. But there's guarantee that she's getting a car. "the other goat can no longer be picked" is not true either. The "other" goat -- by which I assume is meant the non-revealed goat -- is certainly a possibility to be picked. If is meant "the revealed goat" then it should say so.

The second passage seems a lot more easy to understand than the first, although it's not perfect. It's also true, unlike the first passage. But editors are resisting this. Why? Editors are invited to explain. Herostratus (talk) 20:47, 26 September 2016 (UTC)

You have misread the current version. If the contestant picks a goat then they will always win by switching because the other door is necessarily a car. The "Other goat" referred to is the goat Other than the one they previously picked and which was an object earlier in the sentence and as such is the revealed goat. There would be nothing wrong with clarifying by adding "because monty has already revealed it" to that sentence. SPACKlick (talk) 21:03, 26 September 2016 (UTC)

To what "standard" do we refer?

The article currently has a section named "standard assumptions". That's a misnomer; no standards body promulgates a list of statements whose truth can be assumed when analyzing mathematical problems.

This quirky bit of lingo seems to be a sneaky way of excusing people who omit important parts of the problem statement. Marilyn Vos Savant made this error in her initial description of the problem; she described the host opening a non-winning door, but failed, crucially, to state that the host always does this. A reader new to the problem can be forgiven for failing to guess at this rule which, after all, the real Monty Hall did not follow.

This section should be named something more like "stipulations", and shouldn't try to do double duty by both stating the stipulations and subtly claiming that their statement shouldn't be necessary. TypoBoy (talk) 03:10, 15 March 2016 (UTC)

Per the sources in the standard version of this problem all of these are usually assumed even if not stated, you can see from MSV's answer that she assumed the host always opened the door even though she didn't state it. Relation to the original show is irrelevant as the problem isn't based on the show. SPACKlick (talk) 14:49, 15 March 2016 (UTC)

What's not clear is whether she realized at the time that the assumption mattered. (She clearly knew that by the time of the second column, but at the time of the first column, I don't see any evidence beyond her later say-so.) --Trovatore (talk) 17:39, 15 March 2016 (UTC)

All of that may seem correct, but take a closer look. She did not just say that "the host opens just one of the two other doors", but (let's forget about the door #numbers) she wrote indeed: "...and the host, who knows what’s behind the doors, opens another door which has a goat. So IMO indeed it's necessary to take a closer look. Kind regards, Gerhardvalentin (talk) 12:19, 2 May 2016 (UTC)

This section is problematic. The MHP can be solved without the assumption that the host always reveals a goat. For example, the host may sometimes reveal a goat and sometimes not; we are presented with a game in which he does. Is he doing so to help us win the car as we chose a goat first (the Angelic Monty variant), or to discourage us from winning it as we chose the car first (Monty from Hell)? In the absence of any other information, the principle of indifference suggests we have equal chance of either, supporting the 2/3 solution. Anyway, surely the key assumption is that a car is more advantageous than a goat, without which the solution cannot be verified.Freddie Orrell 20:54, 14 June 2016 (UTC) — Preceding unsigned comment added by Freddie Orrell (talk • contribs)

You are wrong. If the host doesn't specifically show a goat, the chance to win by switching to the second closed door is not 2/3. Have a look to university of California, San Diego: "Monty Does Not Know Version", and the "Explanation of the game". If the host did show a goat just by chance, the winning rate is not 2/3, but only 1/2. Gerhardvalentin (talk) 10:48, 21 June 2016 (UTC)

Since I said 'we are presented with a game in which he does', I was not discussing the probabilities associated with not showing a goat. The MHP contains the phrase 'and the host, who knows what's behind the doors', which would be superfluous were he to show a goat just by chance; we may therefore assume he is acting deliberately.Freddie Orrell 09:56, 22 June 2016 (UTC) — Preceding unsigned comment added by Freddie Orrell (talk • contribs)

The article currently claims that "Krauss and Wang (2003:10) conjecture that people make the standard assumptions even if they are not explicitly stated." I can't speak for everyone, but I assumed that the host was trying to win against the guest. So if the host knew what was behind each door, and the host hadn't picked the car in the first round, then the host would pick the car before the second round. And if the host did not know, then the host would choose at random. Each of these interpretations is at odds with the "standard assumptions" and each changes the odds and the best strategy for the guest. They obviously can't be talking about all people, since some of us don't make the "standard assumptions," so which people? 96.255.9.115 (talk) 21:36, 2 January 2017 (UTC)

Explaining switch/stay is 50% not 2/3

Discussion about the math moved to Talk:Monty Hall problem/Arguments#Explaining switch/stay is 50% not 2/3

Discussions about the math should be at the /Arguments page
I would know what would happen if another player in the same Monty Hall game had chosen other door (not the one open by the presenter) and given the opportunity to change door also, which would be the probability for him, and the sum of probabilities for both two players?.88.20.162.142 (talk) 03:42, 4 May 2017 (UTC) Copied to /Arguments. -- Rick Block (talk) 14:51, 4 May 2017 (UTC)

Comment placed on article by IP 92.41.24.7

But isn't there a FOURTH possibility: The car is behind door No. 1 and the host opens door No.2 NOT door No.3.

That isn't a possibility because the table doesn't make reference to which door is opened. It only makes reference to which door is chosen. In 1/3 of cases you will pick door 1. In some proportion (generally 50%) of those the host opens door 3 and in the remainder door 2. But irrespective which door is opened in the 1/3 of cases where you pick the door with the car behind it you win by staying. In the 2/3 of situations where you pick a door with a goat behind it you lose by staying. SPACKlick (talk) 16:03, 2 October 2017 (UTC)

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)

I agree with the point at the head of this section that the math does not correspond to the problem as posed. Specifically, 'standard assumptions' are not part of the problem as posed, nor are they material to the solution. However I have been unsuccessful in arguing this, as can be seen on archived Arguments pages. My explanation is simply this: the MHP always states that the host reveals a goat, but the MHP does not state the host always reveals a goat. The issue is not what we ought to assume about the host's general behavior, but what information we actually possess about the host's general behavior.Freddie Orrell 23:57, 19 August 2018 (UTC) — Preceding unsigned comment added by Freddie Orrell (talk • contribs)

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 way i explain this to anyone. I think anyone remaining who hasn't understood yet can be enlightened in a few lines. Its mainly about the switching. "Under the set up conditions if you always choose to SWITCH the only time you will lose is if you've chosen the correct door the first time." That's all you need. Chances of choosing correct door 1/3. Therefore if u choose to switch chances of losing are 1/3. I can't believe the whole world had an argument about this lol DwayneC-137 (talk) 23:38, 12 November 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)

It is mathematically correct but somewhat densely worded. If there are, say, 100 doors, the initial pick has a 1/100 chance of success and a 99/100 chance of failure. The strategy of switching will always yield the opposite of your initial pick, so in the case of 100 doors the chance of winning via switching is 99/100. There is no mathematical reason it should be 2/3 regardless of the number of doors. 2/3 is just the specific answer you get for (N-1)/(N) when N=3. Alexanderplatz (talk) 12:11, 6 November 2018 (UTC)

Of course it matters. If there are 10 doors with 9 goats, then the probability that you picked the car is 1/10, not 1/3, so the probability that there's car behind the remaining door is 9/10, not 2/3. In any case, this page is not the place to dispute the reliably sourced mathematics in the article. -- Jibal (talk) 22:24, 17 September 2019 (UTC)

Criticism of the simple solutions

The explanation given here for non-random decisions of the TV host is clearly incorrect. I didn't read quoted papers so I can refer only to the explanation as given in this Wikipedia article. Let's analyse following case: when given an opportunity to pick between two losing doors, Monty will open the one on the right. What is the probability of winning the car given the player has picked door 1 and the host has opened door 3? I think it's rather obvious it's not 50% as stated in this Wikipedia article, it's still 1/3.

Simple example to compare: on different TV show they hide a treasure in one of 150 countries. Your task is to guess which one. Your guess - France. Your chance is 1/150. It's clear and intuitive that the probability will not rise to 50% if later TV host tells you that the treasure has been hidden in France or in Australia. His "algorithm" for coming up with Australia doesn't really matter. The probability of encountering treasure in France is defined when the treasure location is chosen randomly - with equal probabilities of selecting any among 150 countries for the hiding location. To suggest otherwise is like saying that probability of any lottery game is 50% - you will win or not.

I recommend to remove this kind of "criticism". It may be a quotation from some respected source, but clearly something has been "lost in translation". This question can be easily answered via computer simulation. I've done just that to double check if there is any hidden math-magic here I might have missed. Whether the door opened by TV host is selected (pseudo) randomly or he has the preference for the rightmost door, on the milion cases sample, in 66% repetitions player wins when switching the gate. Algorithm for selecting the door to be revealed by TV host (while observing initial assumptions of not opening the door selected by the player and not revealing the car) doesn't affect the result. — Preceding unsigned comment added by 89.68.171.16 (talk) 20:45, 20 April 2019 (UTC)

The text in the article accurately reflects the cited source. If you'd like to discuss the mathematics of this, I suggest you post to the https://en.wikipedia.org/wiki/Talk:Monty_Hall_problem/Arguments page. -- Rick Block (talk) 00:01, 21 April 2019 (UTC)

Surprisingly, the figure you argue against seems true. If Monty opened the middle door, you can't have picked the right door, it has to be in 3, so you win with prob of 1 by switching. He opens door 3 when you've chosen it or its in 2, both with equal probability. Gomez2002 (talk) 15:59, 19 July 2019 (UTC)

This is incorrect. On average you would select the car door in 1 out of 3 game and a goat door in the other 2 out of 3. If the host does it by random, in those 2 out of 3 that you have chosen goat, in one of them the host would reveal the other goat but in the other one he would reveal the car. So, he only reveals a goat in 2 out of 3 times, in which in one of them you win if you switch and in one you win if you stay. The third game is discarded because the car was revealed and it couldn't continue. In other words you still pick a goat 2/3 of the time at first, but now the condition to win by switching is not only to have picked goat, but also that the host has picked the other goat, which has 2/3 * 1/2 = 1/3 probability, the same as picking the car at first.

It is more obvious that the host behaviour can change the probabilities imagining the case that he only reveals a goat and gives you the opportunity to switch if your choice was correct in the beginning. In this way, you would have 0% chance by switching. EGPRC (talk) 16:08, 18 August 2019 (UTC)

This page is not the place to dispute the reliably sourced (and correct) mathematics. -- Jibal (talk) 22:27, 17 September 2019 (UTC)

Video game trivia & Wikipedia's editing policy

About the video game trivia I attempted to add to the page. The history section of the page initially had other trivial information regarding pop culture references to the Monty Hall problem, which were removed for being unsourced. In any event, I do agree that it doesn't belong in the "history" section, at least, but would it be possible for it to be included on the page in some other capacity? On a related topic, I want to bring up wikipedia's editing policy. Or, more over, the fact that I've had contradictory experiences with admin on Wikipedia. In the past, I have reverted sourced edits, and have had my revert undone with messages akin to, "discuss the removal of sourced information on the talk page". I don't fully comprehend how this is supposed to work, even after reading various pages on the subject. In my experience, it just seems entirely random as to what a user is gonna think of how you approach this kinda thing; the results vary drastically and are completely incongruous between the subjective interpretations of different users. --2A02:C7F:3A2B:3B00:781C:FA00:F28C:3A9E (talk) 15:46, 17 September 2019 (UTC)

Sourcing is necessary, but not sufficient. Just because material is properly sourced doesn't mean it's appropriate for the article in that location, or even at all. This comes down to judgment and there's no single formula for it. And yes, this is subjective; I don't know how you get around that.

As to your other experiences, I wouldn't know. It's quite possible that you were in the right in those. Without seeing specifics I couldn't comment. --Trovatore (talk) 16:20, 17 September 2019 (UTC)

It comes down to having a consensus. If an edit is reverted, that means another editor objects to that edit, and there can be a million reasons why. In this case several editors reverted you, so the edit warring was pretty obvious. You're fortunate an admin didn't see that. No matter how right you are, you must not edit war. BRD is not cyclical. It contains only one B, one R, and one D. (Full stop. Do not return to B.) We are now Discussing and this must be carried on until a consensus is reached. Don't add that content again until such a consensus is reached.

If this content can be sourced to RS, then it might be appropriate for an "other uses" type section, but a consensus will decide that. Keep discussing, and it would be really great if you can find more RS to back this up. -- BullRangifer (talk) 04:01, 18 September 2019 (UTC)

Yeah, but don't tease. If really no one but 2A02 thinks this is worth including at all, then there's no point in making him/her chase down RSs. Does anyone think this fits? I'm not a gamer and can't really judge how interesting this factoid might be to a typical reader of this article. --Trovatore (talk) 04:45, 18 September 2019 (UTC)

I'm not teasing. I think the idea of "interest" to "a typical reader" is a bit of a red herring. This is a factoid of interest to gamers and to those interested in all knowledge about the Monty Hall problem and its various uses. It may be of zero interest to some, but that doesn't make it worthless to everyone else. Someone else might like it. I certainly learned something about just how this problem can be used in other settings.

I'm also curious if other RS have also mentioned it, not that multiple RS are required for this type of inclusion, unlike for matters covered by BLP's WP:PUBLICFIGURE, which does require multiple RS. So far its due weight is low, which is why it would be in such a section and not be mentioned in the lead. -- BullRangifer (talk) 05:43, 18 September 2019 (UTC)

It's interesting and ironic to see editing procedure discussed on the Monty Hall Problem page, where it has led to the acceptance of extraneous assumptions as essential to the statistical methodology (a factoid in its original meaning of unreliable information becoming accepted by repetition). Freddie Orrell 06:55, 20 September 2019 (UTC) — Preceding unsigned comment added by Freddie Orrell (talk • contribs)

Freddie Orrell, is your comment relevant to this thread, or does it belong in its own (or a different) thread? BTW, please sign your comments. -- BullRangifer (talk) 14:28, 20 September 2019 (UTC)

BullRangifer It's an observation meant in gentle support of the initial post in this thread. The actual subject is already well-covered in other threads. I followed the instruction to 'sign your posts by typing four tildes'. The system appended my username, the date and time, and the words 'Preceding unsigned comment ...'. This indicates a failure of software usability rather than lack of etiquette on my part. Freddie Orrell 16:30, 20 September 2019 (UTC) — Preceding unsigned comment added by Freddie Orrell (talk • contribs)

Following the tangent here, you seem to have found a corner case; apparently you have your preferences set to provide your signature without links, and it looks like the bot doesn't recognize that as a signature. Your easiest workaround would be to set your signature pref back to the default. By chance, do you have "Freddie Orrell" in the "Signature" box under Preferences->User Profile, but you have a check mark in the box below it, "[t]reat the above as wiki markup"? If so, you can just uncheck that box, and things will go back to normal. --Trovatore (talk) 16:42, 20 September 2019 (UTC)

Trovatore I had visited the SineBot FAQ, where Wikipedia mistakes its own tab 'Preferences' for 'my preferences'. The terms username, signature and nickname were used without clarity about their difference, amid a jumble of jargon which left me none the wiser. However, I have gratefully followed your kind suggestion and we will see the results here: Freddie Orrell (talk) 17:40, 20 September 2019 (UTC)

Proposal to merge with Three Prisoners problem

The following discussion is closed. Please do not modify it. Subsequent comments should be made in a new section. A summary of the conclusions reached follows.

There is no consensus to merge the articles; in fact, the arguments presented lean against merging. No such user (talk) 07:42, 4 October 2019 (UTC)

---

I propose to merge Three Prisoners problem into Monty Hall problem. These are mathematically equivalent problems. The only difference is the wording. The explanation of the problem and its solution are exactly the same. There is no reason to have two articles. Alexei Kopylov (talk) 22:50, 22 April 2019 (UTC)

• Merge per nominator. One can copy the statement of Three Prisoners problem into Monty Hall problem, and there is no need to repeat the explanation twice in slightly different contexts. Wikisaurus (talk) 00:00, 23 April 2019 (UTC)

The Monty Hall Problem and Three Prisoners Problem each has a different history and importance in popular culture independent of any mathematical similarity. The MHP, following its appearance in 'Parade', was the subject of an intense and controversial public discussion which continues to this day. This is the reason for having two articles, and why they should not be merged. Contrary to your comment, it is actually the context, not the explanation, which is materially significant here. Freddie Orrell 10:05, 2 May 2019 (UTC)

These are not similar problems, but exactly the same problem. The history of each of the wordings is a part of the history of one problem and should be in one article. Alexei Kopylov (talk) 19:17, 2 May 2019 (UTC)

Okay, the Monty Hall Problem and Three Prisoners Problem each has a different history and importance in popular culture independent of their being mathematically the same. The context, not the explanation, is the reason for separate articles.Freddie Orrell 19:02, 18 May 2019 (UTC)

"These are not similar problems, but exactly the same problem." -- You are using words incorrectly. They have the same mathematical analysis but they are prima facie not the same problem. In any case, they are not and should be the same article, for reasons that have been thoroughly explained. -- Jibal (talk) 22:41, 17 September 2019 (UTC)

• Agree with nom's reasons to Merge, essentially because it's the same problem. I'd say its incarnation as the Monty Hall problem seems to be much more well-known and written about, so it would make sense to keep the main article here. But any of the mathematical explanation at the other is completely redundant. All that's really needed is to add to the history. –Deacon Vorbis (carbon • videos) 20:34, 2 May 2019 (UTC)
• Keep as is, a similar problem but two different awareness pools. Many readers have heard of both, and the use of different names and concepts differentiate these topics. Randy Kryn (talk) 10:03, 21 May 2019 (UTC)
• Keep as is: their origins are very different, and the actual question is different ("Is A correct or C correct?" vs "Should I stay or switch?") The article is also already too long (but that's another matter). cmɢʟee⎆τaʟκ 22:15, 8 June 2019 (UTC)
• Do not merge. Many problems can be isomorphic, but that does not mean they need to be merged. Three prisoners can point the reader here for more details. Glrx (talk) 17:11, 29 June 2019 (UTC)
• Comment I took a brief look, and I'm actually not convinced by the claim that it's the same problem. The three-prisoners problem seems to be well enough specified to have a determinate answer; the conditions for what the warden is to do are precise. For at least the original version of Monty Hall (meaning the first piece by vos Savant in Parade), the conditions are not so well specified, and the answer depends on Monty's assumed constraints/strategy. --Trovatore (talk) 20:03, 17 September 2019 (UTC)
• Do not merge: Having the same mathematical analysis is just one aspect of these pages, while everything else is different, and they are situated very differently in re the culture. And contrary to the nom, they are not "exactly the same problem" -- the characters and the situation are different, and even the questions are different ("Should you switch?" vs. "Who is right?" ... notably, MH as asked does not include C's [correct] reasoning). That a mathematical isomorphism exists between the two problems is not sufficient reason to merge the articles. (But certainly each should mention the other.) -- Jibal (talk) 22:41, 17 September 2019 (UTC)

---

The above discussion is closed. Please do not modify it. Subsequent comments should be made in a new section.

Differences between the Selvin and vos Savant statements of the problem

The article seems to imply that vos Savant's version of the problem is essentially the same as the original Selvin letter (Selvin1975a), but it is not. Look up the link. Besides superficial elements (boxes, one with car keys, rather than doors), there is a key difference: in Selvin's version, the host doesn't offer the switch. Instead, he keeps offering money to buy the contestant's box, and then the contestant proposes the switch. Although not explicitly stated, the implied question is: is this a good idea? Selvin then goes through the steps to show that the chance of winning goes up to 2/3 by switching.

So the requirement that we make the assumption that the host will always offer the switch doesn't apply to the original version.

Of course, the way Selvin defines the problem, there could be room for trying to second-guess the host. The host offers money for the contestant's box, and raises the offer several times, but even the last offer is way less than 1/3 the value of the car. So the contestant could guess that the chosen box doesn't have the keys based on the low offers. Selvin doesn't bring this up, but just calculates the probabilities on the implicit assumption that the host's behavior would be the same no matter where the prize is. KevinBTheobald (talk) 17:31, 4 November 2019 (UTC)

Popular culture

I'm aware pop culture references (see above section regarding use in a video game) have been removed, but I can't help thinking that a well-sourced "in popular culture" section might be a good addition to the article. The Brooklyn 99 feature was more than a passing mention: it was a major plot point, and bound to send people scurrying here for more detail. Likewise the usage in 21. Not familiar with the game in question, but I'd be happy to take a swing at it while keeping WP:IPC in mind. Basie (talk) 20:02, 4 November 2019 (UTC)

perhaps a simple straightforward explanation for the layman

I don't know where else to put this but it seems to me possibly the simplest way to explain this quandary to the laymen is as follows, or something similar to, as written. please feel free to use or edit for use in this article in any way that could be helpful.

"the contestant is presented with three doors, two of the doors contain a goat, only one contains the prize(a car). The contestant must choose only one door. The contestant is going to make the wrong first choice 2/3 of the time and the correct choice 1/3 of the time (if a choice is made randomly). Once the contestant makes his choice the host opens a random WRONG door because he is not allowed to reveal the correct one to the contestant until the end.

Now because the contestant has likely already chosen the wrong door with his/her initial choice (as he/she would if he/she chose randomly, 1/3 likely-hood) and because the host cannot reveal the correct door to the contestant, but since the host has also confirmed that one of the doors is WRONG the likelihood of winning the prize increases to 2/3 by switching to the only door that has not either been already chosen (which was probably/statistically a wrong choice in the first place) or revealed by the host (which is 100% the wrong choice as revealed).

Because it can be assumed that the first choice was wrong (a statistically good bet) and the host reveals another wrong door (with 100% certainty) that leaves only the last door (not chosen) as "likely" to be the correct door

If the contestant chooses to keep the first door that they selected their chance of choosing the correct door never rises above 1/3."

if that makes sense.

2603:6000:9E00:30DE:2CBB:CE09:8893:36E5 (talk) 05:52, 14 August 2020 (UTC)

It also becomes more intuitable when you add more doors; a situation that you'd never see in an actual game show setting but has the same statistical basis (using the same assumption that the host eliminates every door but one and then asking you to choose to stay or switch). If you have ten doors, and nine of them have goats behind them, you have a mere 1/10 chance of choosing the correct door initially. If the host opens 8 doors, revealing 8 goats, there's still only a 1/10 chance you chose correctly on the first try, but a very good likelihood that the door that survived the host's eliminations has the car behind it. As you emphasize, the problem rests on the fact that the initial choice made by the contestant has a 1/X likelihood (X being the total number of doors), whereas the door the host proffers will always have an X-1/X chance of being right (so 2/3 or 9/10 or 999/1000 etc.). — Preceding unsigned comment added by 99.242.214.203 (talk) 02:51, 11 September 2020 (UTC)

How do the odds change...

... if you'd rather have a goat? SnappingTurtle (talk) 20:34, 9 June 2020 (UTC)

Help me here

Are people changing or adjusting the statistic in the middle of the operation? — Preceding unsigned comment added by 2604:6000:9FC0:1A:1CC2:D5F:358A:6954 (talk) 19:13, 2 September 2020 (UTC)

Converting to shortened footnotes

I am in the process of converting this article from the deprecated inline Harvard referencing to shortened footnotes. Please be patient. Peaceray (talk) 04:56, 17 October 2020 (UTC)

It's a pity. --Trovatore (talk) 20:03, 18 October 2020 (UTC)

Completed. Please review to ensure that I made no mistakes. Peaceray (talk) 02:53, 19 October 2020 (UTC)

This "article" is another waste of electrons

The only "assumptions" were CYA operations by the pseudo-genius and her celebrities. Let's Make a Deal with Monty Hall, like all post-scandal game show, was played under strictly enforced rules. Monty Hall had no opportunity to "trick" or "help" a contestant by revealing one of the booby prizes.

The pseudo-genius is wrong. This "article" is garbage. 172.58.27.220 (talk) 00:09, 25 November 2020 (UTC)

Do you have verification from a reliable source that this is so, or are you presenting original research that is your opinion? Peaceray (talk) 04:35, 25 November 2020 (UTC)

Simple solutions image

I'm sorry, but the first image under Simple solutions section says "Three initial configurations of the game" and yet it shows four? The last row has two distinct configurations, one in which Monty opens door #2 and another in which Monty opens door #3. I agree that the outcome is the same, but the configuration definitely is not. If we apply the same reasoning, then the first two configurations are also the same, we picked door #2 and no matter what door Monty opens, we switch to the remaining choice. In other words, why are 2 distinct choices Monty has and different from 2 distinct choices the contestant has? StojadinovicP (talk) 09:42, 11 April 2021 (UTC)

Since the host only has those two choices when the player's door has the car, which has probability 1/3, then each of those two possible revelations occurs 1/3 * 1/2 = 1/6 of the time. The image in the section "Conditional probability by direct calculation" illustrates it. EGPRC (talk) 21:23, 3 June 2021 (UTC)

Contradictory

I moved the following section to here, because it's contradictory to what was given in section before. Savant stated that there's a two-thirds of a chance when switching, so that would rather mean 33% chance that the goat is behind their first chosen door. Mikael Häggström (talk) 00:02, 3 March 2021 (UTC)

"The simplest explanation for the effectiveness of the strategy concerns the placement of the goats: when the player first makes their choice, there is a 66% chance that a goat is behind their chosen door. This probability does not change after the host reveals the location of the other goat. Since it is always more likely the player has chosen a goat to begin with, and the host will always give away the location of the other one, there is a 66% chance that switching was within the player's best interest all along (as opposed to the less likely 33% chance the player had chosen the car)."

This edit seems incorrect. Savant's statement means two-thirds chance OF THE CAR when switching, therefore one third chance of a goat when switching. That leaves two thirds (66%) chance of a goat behind the first chosen door. In summary: switching = 2/3 car, 1/3 goat; not switching = 1/3 car, 2/3 goat. The figures for switching change after revealing one of the goats (from 'either of two single doors' to 'one of a set of two doors'), but those for the first chosen door do not change. Without checking, I wonder if that accounts for the wording you have identified? Freddie Orrell (talk) 10:09, 17 March 2021 (UTC)

In the absence of any reply so far, I have undone the stated move. The rationale for it seemed incorrect. In particular, there is 66% chance a goat is behind the first-chosen door at all times. I have left the wording as it was, though others may feel it could be improved. Freddie Orrell (talk) 13:32, 5 April 2021 (UTC)

This game can be played in 2 ways:

- Pick a car, you win. That is: stay with your choice.

- Pick a goat, you win. That is: take the offer to switch.

86.83.108.100 (talk) 15:49, 4 November 2021 (UTC)

Not even a statistical problem

It is interesting that this article like every discussion of the Monty Hall problem treats it as a problem in statistics when nothing about the question as originally posed implies that it is. The only answer to 'Is it to your advantage to switch your choice?' is, in the context of the question, 'That depends if I originally choose a door with a goat or a car behind it'. The question describes a game show and doesn't in any way imply that the contestant will get more than attempt. If the question had asked for the best strategy to maximise winnings over many cycles of the game then the probabilistic approach makes sense. However if you only get one attempt then knowing that the probability of a car being behind the closed door is 2/3 is of no help when there is actually a goat behind it. You'll still lose.

This article should at least mention what, if anything, 'probability' means for a single, isolated event.

130.246.148.29 (talk) 17:34, 13 December 2021 (UTC)

You appear to confuse the number of possible outcomes of an event with the number of events. In your case, one event but several possible outcomes. Using your words, knowing the probability ‘that there “is actually a goat behind it”’ is, in fact, what helps. See ‘Probability of a single event’ online. Freddie Orrell (talk) 23:15, 25 December 2021 (UTC)

The problem is ill-posed

The initial discussion should clarify that addtitional assumptions were made in the solution to the problem, and which were explicit. This should not be left until later in the article, as the implications of the initial text are otherwise misleading. Indeed, Savant's original answer is not clear about the full assumptions: she states that the host will always avoid the car, but this does not necessarally imply always opening a door. Readers making different assumptions could justifiably have interpreted her mathematical statements as erroneous. IMO the present text unfairly lambasts the earlier critics.
As a side point, the solution would require no additional assumptions if the original stated that the host did not know the location of the car - opeining the door reveals the information regardless of the host's knowledge (at the cost of crystallising the incorrect decision in 1/3 of cases). Part of the current text is incorrect in this regard, though the value is correctly stated at one point.
Similarly, if the host always opens a door to reveal a goat, it does not matter at what point the contestant makes the decision to switch; the time of making the decision only matters if it can affect the host's actions. Here again some of the text appears misleading PhysicistQuery (talk) 23:26, 16 January 2022 (UTC)

The biggest assumption whcih makes this problem unsolvable, is that it is not known to us if the host would offer the contestant an option of switching doors no matter which door the contestant originally selected. Just as important, is that the contestant does not know the answer to that question. If the contestant had been told from the beginning that the offer was forthcoming, he could do the math that von Savant did. But as worded, it is not clear if the host is trying to trick the contestant. Did the game show "Lets Make a Deal" always offer this door switch option, so the contestant could at least use history of the show as a guide in his decision? — Preceding unsigned comment added by 69.204.92.210 (talk) 14:33, 18 January 2022 (UTC)

Slight of hand, not statistical proof of human intuitive failure.

Edit: I had an aha moment and get it now. But part of the problem is the reliance on statistics and high-level probability given that we are being spoofed by the info. Given that the first choice has a 1/3 chance of selecting the correct door, the host has a 2/3 chance of being forced to "dodge" the correct door. So then the player ends up with a 1/3 chance of switching away from correct against a 2/3 chance of switching to the correct door.

Latex

Ew looks ugly someone should change the formating to use LaTeX — Preceding unsigned comment added by 2604:3D09:1580:9600:ECAA:144E:36B1:E1C8 (talk) 16:33, 10 April 2022 (UTC)

I think that's wise. In this era of Covid one should always wear gloves. EEng 16:24, 10 July 2022 (UTC)

"Monty Fall" host behavior variant

Firstly, there are 3 sources given, but [53] doesn't mention this variant at all (it also doesn't mention the 2nd variant in the table, "The host always reveals a goat and always offers a switch. If he has a choice, he chooses the leftmost goat with probability p"). As such, I've removed this source from the two variants. Also, [62] simply asserts that the probability for this variant is 1/2 in its introduction, but never explains why. However I've left it in place for now.

[34] does explain the variant via the "Proportionality Principle", but I believe there is a hidden assumption being made, namely that the contestant doesn't know ahead of time that Monty will always fall on a goat door. If the contestant has the same information as the reader and knows Monty is guaranteed (e.g., via divine intervention) to fall on a goat door that wasn't initially chosen, then the variant should be identical to the original problem, only instead of Monty choosing which door to open, it's a divine being forcing him to fall onto a certain door.

As stated, the variant is confusing because Monty cannot both fall randomly (ie. have equal probability of falling on any of the 3 doors), and be guaranteed to always fall on a non-initial goat door. It seems artificially constructed specifically to prove the point that modifying assumptions about host behavior also modifies outcome probabilities. However, the other 10 variants in the table already do this, and have the added benefit of being realistically testable. I am unsure whether this particular variant actually helps in understanding, and am considering removing it from the table. However, I would like to get some other opinions first 30103db (talk) 15:46, 19 July 2022 (UTC)

Update to direct calculation of Monty Hall problem

EDIT: I have just updated the proof to no longer rely on mathematical prose, to include a citation, to improve the organizational structure, and to elaborate on issues that might be unclear to the reader (e.g., why P(Hi|Xj) = 1/2). These improvements address many of the issues raised in the comments.

Hi there. I originally posted updates to a proof for the direct calculation section of the Monty Hall problem page, but I accidentally typed "Monty Python problem", leading a user to revert my changes under the assumption they were not serious. I learned this on the talk page of that user, where others said I was fine to reimplement my changes so long as I fixed my typo. But when I did so, my changes were again reverted because people assumed I was just ignoring suggestions since I did not post on this page. As a result, I am being made to post here to ensure the consensus agrees that my changes are fine before reimplementing them. This is a bit of a headache since this originates from a miscommunication rather than an actual problem with my proof, but that is alright.

My changes accomplish three things. They:

• Improve the intuitiveness of the proof by deriving solutions from claims that more closely correspond with the Monty Hall problem rules
• Prove the probability for all three doors by deriving a probability equation that applies for all three doors, rather than just the second
• Remove redundant text and improve clarity by making explanations more concise (also adding in a section on conditional probability rules)

Here are my changes:

---

Direct calculation

In this proof, we will directly compute the probability of the car being behind each door using Bayesian statistics.

Let us label each of our doors as Door 1, Door 2, and Door 3. Next, let us define three functions. Xi denotes that you initially choose Door i, where i represents any door from 1 to 3. Similarly, Ci denotes the car is behind Door i. And Hi denotes the host opens Door i.

Let us define j and k as numbers, like i, from 1 to 3. i, j, and k will represent different numbers, meaning, for example, that Xi and Cj will represent that you chose Door i which is not the door with the car, Door j.

With these rules in mind, we can define our probabilities as the following:

${\begin{aligned}P(Ci)&={\tfrac {1}{3}}\\P(Hi|Ci)&=0\\P(Hi|Xj)&={\tfrac {1}{2}}\\P(Hi|Cj,Xj)&={\tfrac {1}{2}}\\P(Hi|Cj,Xk)&=1\\P(Ci)P(Xi)&=P(Ci,Xi)\end{aligned}}$

• P(Ci) = 1/3 represents the probability that the car is behind one of the three doors.
• P(Hi|Ci) = 0 represents the impossibility of the host opening the door with the car.
• P(Hi|Xj) = 1/2 represents how there are two remaining doors for the host to choose from once you have chosen a door. It is important to note this expression does not care which door the car is behind, so we do not have to worry about how the possible doors containing the car affect the host's door choice.
• P(Hi|Cj,Xj) = 1/2 represents how, given you chose the door with the car behind it, there are two remaining doors the host can open.
• P(Hi|Cj,Xk) = 1 represents how, given you chose a door without the car behind it, there is one remaining door the host can open.
• P(Ci)P(Xi) = P(Ci,Xi) since our rules do not specify there being conditional dependencies between P(Ci) and P(Xi), meaning that Ci and Xi are independent events.

From Bayes' theorem, we know that P(A,B) = P(A|B)P(B). We can extend this logic to three events using the chain rule: P(A,B,C) = P(A|B)P(B|C)P(C).

The conditional probability, we know that P(A|B) = P(A,B)/P(B). We can extend this logic to three events using the chain rule: P(A|B,C) = P(A,B,C)/P(B,C).

Next, let us derive a formula to calculate the probability that a given door contains a car just in case X1 and H3:

${\begin{aligned}P(Ci|X1,H3)={\frac {P(Ci,X1,H3)}{P(X1,H3)}}={\frac {P(H3|Ci,X1)P(Ci,X1)}{P(H3|X1)P(X1)}}={\frac {P(H3|Ci,X1)P(Ci)P(X1)}{P(H3|X1)P(X1)}}={\frac {P(H3|Ci,X1)P(Ci)}{P(H3|X1)}}={\frac {P(H3|Ci,X1){\tfrac {1}{3}}}{\tfrac {1}{2}}}={\tfrac {2}{3}}P(H3|Ci,X1)\end{aligned}}$

We have done the bulk of the work, proving P(Ci|X1,H3) = 2/3 P(H3|Ci,X1). So the probability of the car being behind a door given you chose Door 1 and the host chooses Door 3 equals 2/3 * P(H3|Ci,X1). We are in luck, since we know the probabilities of P(H3|Ci,X1) for each possible Ci! All that remains is for us to substitute Ci with each of the three possible car-behind-door events and plug numbers into our equations.

To calculate P(C1|X1,H3) (car behind Door 1):

${\begin{aligned}{\tfrac {2}{3}}P(H3|C1,X1)&={\tfrac {2}{3}}{\tfrac {1}{2}}={\tfrac {2}{6}}={\tfrac {1}{3}}\end{aligned}}$

For P(C2|X1,H3) (car behind Door 2):

${\begin{aligned}{\tfrac {2}{3}}P(H3|C2,X1)={\tfrac {2}{3}}1={\tfrac {2}{3}}\end{aligned}}$

For P(C3|X1,H3) (car behind Door 3):

${\begin{aligned}{\tfrac {2}{3}}P(H3|C3,X1)={\tfrac {2}{3}}{\frac {P(H3,C3,X1)}{P(C3,X1)}}={\tfrac {2}{3}}{\frac {P(H3|C3)P(C3|X1)P(X1)}{P(C3,X1)}}={\tfrac {2}{3}}{\frac {0P(C3|X1)P(X1)}{P(C3,X1)}}=0\end{aligned}}$

We have now shown that, given you initially chose the first door and the third door is opened by the host, there is a 1/3 chance of the first door containing the car, a 2/3 chance of the second door containing the car, and a zero chance of the third door containing the car, thereby solving the Monty Hall problem using Bayesian statistics.

---

With that in mind, I would greatly appreciate if people could reply to this post telling me whether they think that my changes are alright. Thank you for your time! GabeTucker (talk) 06:28, 14 January 2023 (UTC)

• David Eppstein? (Hate to put this on you, but I'm just too busy for a few days.) EEng 11:01, 15 January 2023 (UTC)

This complicates the explanation for no clear benefit. The older, simpler calculation is preferable. - MrOllie (talk) 02:52, 16 January 2023 (UTC)

"This complicates the explanation for no clear benefit" here is a list of the clear benefits my proof provides:

• My calculation is 23 lines, compared with the 21 lines of the original equation. So although it is negligibly longer, this allows us to more clearly convey the steps to arrive at our conclusions.
• The original calculation shows:

${\begin{aligned}::P(A,B,C)&=P(A|B,C)P(B,C)\\::&=P(B|A,C)P(A,C)\\::&=P(C|A,B)P(A,B)\\::&=P(A,B|C)P(C)\\::&=P(A,C|B)P(B)\\::&=P(B,C|A)P(A)::\end{aligned}}$

The majority of these lines play no role in the derivation and the explanation of Bayes' rule can be greatly simplified, like in my proof.

• The original calculation only calculates P(H3|C2,X1) = 2/3. This does not prove that P(H3|C1,X1) = 1/3, nor that P(H3|C3,X1) = 0.
• The original calculation shows the following premises:

${\begin{aligned}::P(H3|C1,X1)&={\tfrac {1}{2}}\\::P(H3|C2,X1)&=1\\::P(H3|C3,X1)&=0\\::P(Ci)&={\tfrac {1}{3}}\\::P(Ci,Xi)&=P(Ci)P(Xi)\\::P(H3|X1)&={\tfrac {1}{2}}::\end{aligned}}$

However, it does not justify any of these premises intuitively except for P(H3|X1) = 1/2, which it takes an entire two lines to do. My proof concisely justifies all of these premises in just three lines.

• Even so, most of these premises hold true for H3 and H3 alone. One intuitively understands the probability of a door being chosen given a car is behind it is 0. However, the premises only apply this rule to H3, which makes the explanation follow less directly from the rules defined using English (i.e., the rules state the host will open a door without a car behind it, not that door 3 is opened given door 1 has a car). Moreover, it does not provide intuition for turning the Monty Hall problem rules into general, stronger rules that cover all possibilities. Understanding how to model the core of where something comes from is healthy practice when trying to gain intuitions surrounding a complex topic.
• The original proof does not define the conditional probability identity, which is used in the derivation.

"The older, simpler calculation is preferable" This is incredibly vague. Can you please give any amount of specifics?

GabeTucker (talk) 03:25, 16 January 2023 (UTC)

@GabeTucker:

One should not write things like this:

P(C1|X1,H3)

Instead, that should look like this:

P(C₁ | X₁, H₃)

Notice that

• The letters are italicized but the digits and the parentheses and the vertical slash are not. This is codified in WP:MOSMATH. The point is to follow LaTeX style as closely as possible. One should also not italicize things like max, cos, log, sup, det, etc.
• Actual subscripts are used. (Doesn't that seem easier on the eyes to you?)
• Horizontal space precedes and follows the vertical slash.

Likewise the following

$P(H3|C1,X1)$

should instead look like this:

$P(H_{3}\mid C_{1},X_{1})$

Here again, actual subscripts are used. The vertical slash is coded as \mid.

Michael Hardy (talk) 20:06, 3 February 2023 (UTC)

Thank you for letting me know—I will remember this for the future, and I will integrate this into my solution should people decide my solution is worth implementing. GabeTucker (talk) 22:37, 20 March 2023 (UTC)

References

RfC about the proof

Are the changes proposed in the above proof reasonable for the direct calculation section of the Monty Hall problem page?

GabeTucker (talk) 11:39, 19 January 2023 (UTC)

EDIT: I have just updated the proof to no longer rely on mathematical prose, to include a citation, to improve the organizational structure, and to elaborate on issues that might be unclear to the reader (e.g., why P(Hi|Xj) = 1/2). These improvements address many of the issues raised in the comments.

Hi GabeTucker. I removed the RfC tag. Can we work together to craft a "brief and neutral" opening question? Better yet, could we post a neutrally-worded request for input at WP:WikiProject Statistics and WP:WikiProject Math, where we're more likely to get input from editors familiar with this topic area? Firefangledfeathers (talk / contribs) 17:32, 19 January 2023 (UTC)

@Firefangledfeathers Sure, I've neutralized and shortened my question. Are we good to readd the RfC tag? And how might I go about asking on the pages you provided? Thanks for the help. GabeTucker (talk) 17:46, 19 January 2023 (UTC)

I think that's much better. You don't need to link the article, as the central RfC listings will direct interested editors to this page. For the WikiProjects, you'd just start a new talk page section at WT:WPSTAT and/or WT:WPM and say something like "There's a dispute at Talk:Monty Hall problem#Update to direct calculation of Monty Hall problem about how best to present a proof. We're hoping interested editors can help us decide between two options." To be clear, I'd prefer to do this instead of opening an RfC, which I view to be one of the most expensive dispute resolution options in terms of community time, but you could do both if your determined to open an RfC. Firefangledfeathers (talk / contribs) 17:52, 19 January 2023 (UTC)

Thanks so much for the help! I'll post in one of those talk pages in lieu of opening an RfC. Again, I appreciate the help. GabeTucker (talk) 18:26, 19 January 2023 (UTC)

Both the current version and the proposed changes go against the general ethos that we are writing an encyclopedia, not a textbook, and aren't really in accord with the style we should adopt for mathematics prose. XOR'easter (talk) 22:00, 19 January 2023 (UTC)

Thanks for letting me know, I hadn't realized this was an issue. If I could spend some time on restructuring my changes such that they read more intuitively and less like a textbook/whiteboard lecture, would my changes be acceptable as an improvement to the current version? GabeTucker (talk) 23:05, 19 January 2023 (UTC)

Potentially; it's hard to say without a specific text to read, of course. And to be compliant with policy, we need at least one reference to a reliable source that works through the problem in this way and explains the virtues of doing so; we can't just assert our own opinions about what is most clarifying. XOR'easter (talk) 16:47, 21 January 2023 (UTC)

Hi there—thanks for the reply. How's it looking now that I updated it? GabeTucker (talk) 21:02, 23 January 2023 (UTC)

Looking at your changes, they don't seem to have addressed the issue of prose style. The text is still full of "we" and "you" colloquialisms, and it still contains unencyclopedic remarks like "We are in luck". XOR'easter (talk) 03:05, 26 January 2023 (UTC)

• (Summoned by bot) First, I see that the RfC tag was removed. That should not have been done. Please reinstate it. Second, someone should please state the issues in a manner understandable to the mathematically illiterate. Coretheapple (talk) 16:31, 21 January 2023 (UTC)

  Hi Coretheapple. I removed the RfC tag since the RfC was started with a non-brief, non-neutral statement, and the OP and I discussed alternatives for dispute resolution, including re-starting the RfC with a better opener. We decided to go with some WikiProject posts, which have (as far as I can tell) garnered the involvement of some experienced math/probability content area editors. I feel ok about my actions, and won't reinstate the RfC tag, but I also won't edit war over it if someone really feels an RfC is needed. Firefangledfeathers (talk / contribs) 19:06, 21 January 2023 (UTC)

  Bots often summon me to very poorly drafted RfCs. If you decide to go ahead with one, if you can't come to a consensus, I'd suggest that it be made clearer if possible. Coretheapple (talk) 21:34, 21 January 2023 (UTC)

  Will do! Firefangledfeathers (talk / contribs) 03:55, 22 January 2023 (UTC)

  Hi there—thanks for the reply. How's it looking now that I updated it? GabeTucker (talk) 21:03, 23 January 2023 (UTC)

Updated what? Please revise the beginning of this RfC. Remember that some of us can barely balance our checkbooks. Coretheapple (talk) 22:38, 23 January 2023 (UTC)

I updated the proof to no longer rely on mathematical prose, to include a citation, to improve the organizational structure, and to elaborate on issues that might be unclear to the reader (e.g., why P(Hi|Xj) = 1/2). GabeTucker (talk) 16:20, 24 January 2023 (UTC)

• Neither On further reflection, we should not include either option, as they are both unsourced / WP:OR. The section should just be removed entirely. - MrOllie (talk) 21:29, 23 January 2023 (UTC)

  Hi User:MrOllie—please take a look at my comment to User:Firefangledfeathers right below and let me know whether this source will be acceptable. GabeTucker (talk) 23:42, 23 January 2023 (UTC)

• Neither, mostly per comments by XOR and MrOllie. This is a problem about which we have an abundance of reliable sources. If this line of explanation isn't common, or at least present, in that body of sources, we shouldn't be spending so much article real estate on it. I'm open to changing my mind if such sources are cited. Firefangledfeathers (talk / contribs) 21:57, 23 January 2023 (UTC)

  Hi Firefangledfeathers—I based my answer on the general structure of the proof from a paper that I cited in my update: Gillman, Leonard (1 January 1992). "The Car and the Goats". The American Mathematical Monthly. 99 (1): 3–7. You can access it here: http://scipp.ucsc.edu/~haber/ph112/goats.pdf. I made some changes in order to improve concision and clarity since his proof was 6 pages long and had some more advanced statistics than most might understand upon a brief reading. But overall, it follows the same structure (sections 3, 5, and 6 are the most important for our purposes). With that in mind, is this update acceptable? GabeTucker (talk) 23:41, 23 January 2023 (UTC)

  Taking someone else's argument and then modifying it in the name of "concision and clarity" isn't what Wikipedia is suited for. XOR'easter (talk) 03:07, 26 January 2023 (UTC)

  Fair enough.

  I'll rewrite the entire argument along the lines of one of the mainstream proofs once I get the chance (without using prose style). Thanks for the input. GabeTucker (talk) 05:13, 26 January 2023 (UTC)

  Honestly, I don't see why the section exists. What benefit does it bring? Why is it not redundant with all the text above it? XOR'easter (talk) 21:24, 29 January 2023 (UTC)

  The Monty Hall problem is an article about statistics and probability. So computing a statistical solution to a problem about statistics and probability makes a lot of sense, at least on the surface. GabeTucker (talk) 06:30, 30 January 2023 (UTC)

  Isn't that what all the text above it does, multiple times? What does this section offer, besides more notation? XOR'easter (talk) 12:30, 30 January 2023 (UTC)

  I think the notation offers mathematical precision and clarity that a text description can't capture in the same way. At least, understanding it in this way helped me to better understand the topic. GabeTucker (talk) 11:29, 31 January 2023 (UTC)

  But it doesn't say anything that hasn't already been said. Whatever precision it offers is no better than what one would get from reading the prose carefully. The list of formulae is just a transcription of the scratch-work one would do in a notebook during such a careful reading. XOR'easter (talk) 15:28, 31 January 2023 (UTC)

  Is that really so? After looking through the article, I struggle to find a place that would have a one-to-one mapping, or even any rough mapping, with the provided Bayesian proof. GabeTucker (talk) 08:02, 1 February 2023 (UTC)

• Comment (Summoned by bot) I would respectfully suggest that all mathematical formulas not be in the article unless sourced. Coretheapple (talk) 19:57, 25 January 2023 (UTC)

• Comment I might be a little late to the subject, but I think the issue raised by GabeTucker for this article is really true: it lacks a formal demonstration written using standard notation. There is a reason that this notation exists, and it is precisely for clarity. The fact that you can somehow work out the formulas by knowing probability theory and transcribing from the article text does not mean that such a contribution would not improve greatly the article. We do not explain Matrix multiplication by only using text, even if it could be done. Likewise, WP:NOR does not preclude from reformulation. So I did not have the time yet to analyse the contribution in depth, but I think it should not be dismissed on the basis of the arguments that have been brought forward so far. Cochonfou (talk) 21:53, 14 March 2023 (UTC)

  Hi Cochonfou,

  I appreciate your comment, and I agree with your points regarding clarity and how WP:NOR does not preclude reformulation.

  An additional issue I've been encountering since the onset of this thread is the following: there are no reliable sources proving the Monty Hall Problem. The only sources proving the Monty Hall Problem do so informally and inconcisely, and are mostly done by non-academics writing blogs. So me writing a Bayesian proof for the Monty Hall Problem based on original research is just as credible as citing one of these blogs since in both cases, there are no academic credentials solving this proof. Examples:

  Top 3 sources after searching:

  • "Greek Data Guy": https://towardsdatascience.com/solving-the-monty-hall-problem-with-bayes-theorem-893289953e16#:~:text=Bayes%20Theorem%20%2B%20Monty%20Hall&text=Let's%20assume%20we%20pick%20door,correct%20if%20Monty%20opened%20B%20.

  This solution does not provide a Bayesian proof.

  • Julia Daikawa: https://medium.com/@judaikawa/the-monty-hall-problem-and-the-bayes-theorem-4415e50e233f

  This does provide a Bayesian proof for the Monty Hall Problem, but it does so A) using vague notation and B) without clearly establishing our premises. The source is non-academic, and it does not provide a proof for all three conditions.

  • Brilliant: https://brilliant.org/wiki/monty-hall-problem/

  This does not use standard notation, nor does it prove all three conditions. Moreover, it does not have a listed author, and its only source is non-academic and does not include the proof they provided.

  Academic sources:

  • MIT: http://web.mit.edu/rsi/www/2014/files/MiniSamples/MontyHall.Old.Bad/montymain.pdf

  This is a computer science student simulating the Monty Hall Problem without proving it.

  • Whitman College: https://www.whitman.edu/documents/Academics/Mathematics/2019/Lopez-Martinez-Keef.pdf

  This is an unpublished paper by an undergraduate student (non-credible source) proving the Monty Hall Problem.

  • Mathematics Magazine: https://www.researchgate.net/publication/233565559_The_Monty_Hall_Problem_Reconsidered

  This article proves alternative formulations of the Monty Hall Problem than the one in this Wikipedia article.

  I think the reason is pretty simple: no serious academics are going to be spending time to prove a relatively simple statistics problem, and no research article is going to publish this since it is, at this point, self-evident. There is nothing interesting to be said about a simple Bayesian explanation of the Monty Hall Problem by serious statisticians. So it becomes impossible for us to base our formulation of the Monty Hall Problem off a credible source. As a result of this, should we have no standard notation? On Cochonfou's point: clearly not, since standard notation provides clarity that text does not offer.

  Moreover, WP:NOR says there cannot be non-original "facts, allegations, and ideas". This seems to be a statement primarily about things in the world, like discoveries. In addition, this proof is not a point of view, so it does not break WP:NOR's neutrality rule. And, most importantly, it is verifiable, as it is a simple mathematical proof. Simple mathematical proofs do not seem like things that need credible sources when you are referencing other (cited) Wikipedia articles that establish the rules you are using in said proof.

  GabeTucker (talk) 23:37, 20 March 2023 (UTC)