Talk:Monty Hall problem

Jerrywickey says:

++ A far simpler and more intuitive explanation ++

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

simple clear intuitive explanation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Points well taken.

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

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

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

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

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

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

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

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

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

Thank you Martin, you are absolutely correct, and you clearly articulate the dilemma of the article:

The Monty Hall Paradox or The Monty Hall "Problem" ?

The world famous "paradox", in a given scenario, is absolutely correct in saying:

Just only two still closed doors in a given scenario, one of them containing the car, the other one containing a goat,
but chance to win the car by swapping from the door initially chosen by the guest to the offered still closed door of the host is not 1/2, but exactly 2/3.

This is quite counterintuitive but absolutely correct in a well defined given standard scenario. As said, this is the world-famous paradox.

But the Monty Hall "Problem" is trying to show that, in quite other assumable scenarios, offside the world famous paradox, additional information could be cogitable by some host who is NOT obliged to maintain secrecy regarding the car hiding door, so that – in those quite other scenarios, and derogating from the standard scenario of the world famous paradox

• the chance to win by swapping could be 0 if the host should offer the swap only if the guest should have chosen the prize,
• or the chance could be 1/2 if the host, not knowing the actual location of the car, eliminates the chance on the car in 1/3 and only in a subset of 2/3 just "happens" to coincidentally open a door hiding a goat and not hiding the car,
• or the chance has to be within the fixed range of at least 1/2 to full 1/1, so on average exactly 2/3 though, depending on a completely unknown but supposed asymmetry in the special behavior of the host, if in 1/3 he opens one of his two losing doors, NOT being obliged to maintain secrecy regarding the car hiding door.

As said, all of this is outside the standard scenario of the world famous paradox that the article pretends to present and that should be the point of the article, just in respect of the readers.

The actual article is a confusing mingle-mangle of all of that, unnecessarily presenting just lessons in conditional probability theory, making it very hard for the reader to grasp what the smoke-screen-article in fact is all about. That intolerable unclear mingle-mangle finally should have an end. The structure of the article should be clear and intelligible, and the title of the lemma should read again "The Monty Hall Paradox", as it was before the sanctioned page ownership appeared. Gerhardvalentin (talk) 09:54, 28 May 2012 (UTC)

Um, what? The page was created as Monty Hall problem on Sept 22, 2001 with this edit. It remained at that name until March 23, 2006, when it was moved to Monty Hall Paradox and then quickly moved back on March 26, 2006 (discussion concerning this is in the archives here). It has remained at this name since. -- Rick Block (talk) 19:04, 28 May 2012 (UTC)

It's interesting to note that in Russian, Ukrainian, Polish and Hungarian WPs this is called a "paradox" in the article titles. In all the Western European languages that I can read (tennish) they go for "problem". Can't read the Asian ones :) Malick78 (talk) 21:27, 28 May 2012 (UTC)

I think Gerhard is actually making a point about the irrelevance of the 'conditional' solutions to this famous puzzle rather than quibbling about the name of the article. Martin Hogbin (talk) 21:41, 28 May 2012 (UTC)

Exactly, but not against a conditional solution per se but for a clear structure of the article showing the "clean paradox" that does not need lessons in conditional probability theory. Showing later that quite other scenarios do not affect the "clean paradox". And any conditional formula – if any – just in the form of clear odds. Gerhardvalentin (talk) 22:08, 28 May 2012 (UTC)

JeramieHicks, please read the above also. Gerhardvalentin (talk) 00:35, 3 August 2012 (UTC)

What is the conflict about

This world famous paradox explicitly says

1. that the car and the Goats were placed randomly behind the doors before the show and
2. in any case the Host is to open a door showing a Goat, offering a switch to his second still closed door,
3. and nothing can ever be known about any possible Host's preference, if in 1/3 he should dispose of two Goats, as he then chooses one uniformly at random.
   
   

This well defined scenario of the world famous counterintuitive paradox ensures that, neither by the Guest's initial choice of door nor by the Host's opening of a special loosing door we could have learned anything to allow us to revise the odds on the door originally selected by the Guest, nor compared to the entity of the group of all unchosen doors, as a whole. Absolutely nothing. No additional information whatsoever.

Once more: In this scenario the odds on the door originally selected by the Guest remain unchanged 1/3, as we have learned nothing to allow us to revise the odds on the door originally selected by the Guest.

And this scenario ("in any case the Host is to open a door showing a Goat") also firmly excludes any forgettable Host who in 1/3 of cases disrupts the chance on the car, and will show a Goat only in the subset of 2/3 when, by chance, he just "happens" to show not the Car but a Goat, reducing herewith the chance on the car by swapping from 2/3 to 1/2.

In this scenario the only correct answer forever only can be based on the average probability of 1/3 by staying vs. 2/3 by switching, as no better knowledge will ever be available.

Actually, the article still is not on the famous paradox, but is about teaching conditional probability theory, staring on door numbers.

Once more: The article, based on actual academic sources, should well arranged report about an extremely counterintuitive but nevertheless clear and very exactly defined world famous paradox with its understood basic standard parameters. So, with respect of the readers, should in the first line pay respect to that clear and well defined standard scenario. That should be the point of the article, just in respect of the readers. But actually the article is an obfuscating and confusing mingle-mangle, not even trying to correctly and clearly address the world famous paradox with its distinct scenario and its well-defined and exact certain information content. As said, those basic parameters of the standard version show a clear scenario, with a firm information content, yes they clearly define the exact information content of this paradox. They firmly and emphatically are excluding, from the outset, in the first place a priori and explicitly, and in a very assertive way that, after an unselected door – irrespective of its door number – has been shown to be a loser, any eventual connected "possibly to be gathered additional information" on the chance of the door originally selected by the Guest could ever be supposed to exist. The standard version clearly shows that such additional information has firmly been excluded and never can be given nor can be expected. And, as a consequence, also definitely excluding any eventual connected possibly to be gathered additional information on the odds of the group of all unselected doors altogether, as a whole entity likewise. – But, quite contrary, and quite outside the famous paradox and its firm scenario, the article is still telling in a misty and treacherous way:

"The popular solutions correctly show that the probability of winning for a player who always switches is 2/3, but without additional reasoning this does not necessarily mean the probability of winning by switching is 2/3 given which door the player has chosen and which door the Host opens."

Cheekily and hideously ignoring the clear and unambiguous scenario the paradox is based on. This article actually still is a lesson in teaching conditional probability theory in a classroom, in an educational way eagerly distinguishing "before" and "after", but never about the core of this world famous paradox.

Woolgathering of teachers in maths may belong to an article on Bayes, but not to this counterintuitive paradox. So a clear structure is indispensable, in any case such woolgathering in later sections at the end of the article, titled "All kinds of quite different other scenarios that, outside the paradox, are used in classrooms to teach conditional probability theory".

But never within the main section about this explicit well defined and clear scenario that this world famous paradox is based on. Just to show respect for the readers.

Rick, we all know that maths-teachers see it from an entirely different view, not addressing the paradox, but addressing lessens in maths. Please help to address the paradox instead, to make this article intelligible and beneficial for the readers. --Gerhardvalentin (talk) 08:51, 31 May 2012 (UTC)

Thank you for your opinion about what the true paradox is. However, I am much less interested in your opinion than in the opinions of reliable sources, whose views must (per WP:NPOV) be represented "fairly, proportionately, and as far as possible without bias". Whether we agree with what reliable sources have to say or not, editing MUST be done from a neutral point of view. Another way to say this is that we must not inject our own views into the article. I'm happy to work toward that goal. Are you? -- Rick Block (talk) 19:25, 31 May 2012 (UTC)

Rick, you have to present what reliable sources really say, and you are not to higgledy-piggledy mix incoherent multifocal perspective. There are sources on different scenarios, and you have to keep different scenarios apart. No source says that in the standard scenario, where the car and the goats were placed randomly behind the doors before the show, and in any case the host is to open a door showing a goat, offering a switch to his second still closed door, and nothing will ever be known about any possible host's preference, if in 1/3 he should dispose of two goats, as he then chooses one uniformly at random, that in this standard scenario only a player that always switches will have a chance to win of 2/3, but that "this does not necessarily mean the probability of winning by switching is 2/3 given which door the player has chosen and which door the Host opens."

To obfuscate by confusingly mixing and not distinguishing what different sources – based on their respective subject – really say, and not to clearly distinguish those various specific views, is not the basic of an article that should be intelligible and beneficial for the reader. We all, and you too should pay respect for the reader. So please do not call that incoherent misty higgledy-piggledy mix of multifocal scenarios the article actually shows (or hides?) to be "what the sources say". Please help to put things together, but correctly structured and intelligible, and not nebulous convoluted. -- Gerhardvalentin (talk) 22:28, 31 May 2012 (UTC)

Gerhard, you are not alone. Even Morgan et al, who created this whole mess, have now retracted their argument that the answer is anything other than 2/3. You are also quite right that there are no sources which state that the conditional solutions are required when the host is known to choose evenly. This nonsense belongs is a section on academic extensions to the problem. Martin Hogbin (talk) 22:50, 31 May 2012 (UTC)

Gerhard - is there some change you're suggesting to the article? If so, what is it? -- Rick Block (talk) 18:37, 1 June 2012 (UTC)

Rick, so you really don't see the hideous state of that article? An article that perkily pretends to present a world famous counterintuitive paradox that, under the strict rules of a very special appropriate and well known standard scenario, provides just only one correct answer to the question

"In this given scenario, is it to your advantage"?

A scenario, that explicitly excludes that the Host offers a swap only if the Guest should have selected the prize, and that explicitly excludes that the host will ever be showing the car, a scenario that explicitly excludes that the Host's special behavior in selecting one of two goats, e.g. in opening of his "strictly avoided door" ever could signalize that the odds on the door offered to swap on could be beyond average? A scenario that explicitly excludes that those odds can ever differ from average? In this given world famous standard-scenario, there only can be one correct answer.

This standard scenario comprising every single one of the following possible constellations likewise:

After the Guest selected door #1, the Host has opened door #2

After the Guest selected door #1, the Host has opened door #3

After the Guest selected door #2, the Host has opened door #1

After the Guest selected door #2, the Host has opened door #3

After the Guest selected door #3, the Host has opened door #1

After the Guest selected door #3, the Host has opened door #2

Every single one of those possible constellations is / are comprised likewise in this standard scenario.

But the article impudently says that this is incomplete or solves the wrong problem, for It does not consider the question: given that the contestant has chosen Door 1 and given that the host has opened Door 3, revealing a goat, what is now the probability that the car is behind Door 2? And the article says The popular solutions correctly show that the probability of winning for a player who always switches is 2/3, but without additional reasoning this does not necessarily mean the probability of winning by switching is 2/3 given which door the player has chosen and which door the host opens.

Yes, I said impudently. Because the obfuscating, nebulous and indistinct article does not distinguish what scenario it is reporting on.

Another comment: Martin has just proposed today to delete this error from the article. He says "In the light of Morgan's retraction I suggest that we remove that error", and I do support that proposal. -- Gerhardvalentin (talk) 10:00, 9 July 2012 (UTC)

There should be a clear and clean split between the world famous paradox and its standard scenario (say 80 percent), and later other sections on quite other, on quite different scenarios that some sources are talking about (say 20 percent).

The article should be 80 percent on the famous counterintuitive paradox, trying to make it intelligible for the readers that the chances are not "1/2 : 1/2" but that they are "1/3 : 2/3". This section could also present a short conditional formula in odds form. And max. 20 percent of the article should be on quite other, on quite DIFFERING scenarios that some out-of-date sources had in mind. Showing that door numbers could only be of relevance in case that some Host's bias could be suspected, otherwise meaningless and just interesting in teaching probability theory.

And all of that Bayes' formula back to the article of Bayes, and just a link to that article. The article is a mess and should clearly present that counterintuitive paradox, and should stop to be indistinct, nebulous and obfuscating. I hope you can help. With respect for the sources and what they really say and address to, and with respect for the readers. -- Gerhardvalentin (talk) 08:58, 2 June 2012 (UTC)

Rather than move the formal derivation to the Bayes' theorem article (the derivation of this problem has no particular significance in the context of that article), how about moving it to the end of the article as it was when the article was a featured article (for example, this version). Would that help? Is there anyone who would be opposed to this? -- Rick Block (talk) 19:01, 2 June 2012 (UTC)

No Rick, the "formal derivation" does not really need Bayes'. Just present a short proof in odds form right at the beginning, or just write the following:

If for example the Guest initially selects Door 1, and the Host opens Door 3, the probability of winning by switching is 2/3:

$P(C_{=2}|H_{=3},S_{=1})={\frac {{\frac {1}{3}}\times 1}{{\frac {1}{3}}\times ({\frac {1}{2}}+1+0)}}={\tfrac {2}{3}}.$

This is valid for any door the Guest should select, and valid for any other door the Host should open.

And you can show that clearly arranged table also:

Say, Guest selected D1	Door opened by Host
Initial arrangement (probability)	Open D1  (probabilty)	Open D2  (probability)	Open D3  (probability)	Joint  probability	Win by  staying	Win by  switching
Car Goat Goat (1/3)	No	Yes (1/2)	No	1/3 x 1/2	Yes (1/6)	No
	No	No	Yes (1/2)	1/3 x 1/2	Yes (1/6)	No
Goat Car Goat (1/3)	No	No	Yes (1)	1/3 x 1	No	Yes (1/3)
Goat Goat Car (1/3)	No	Yes (1)	No	1/3 x 1	No	Yes (1/3)

We observe that the player who switches wins the car 2/3 of the time. We also see that Door 3 is opened by the Host 1/2 = 1/6+1/3 of the time (row 2 plus row 3), as must also be the case by the symmetry of the problem with regard to the door numbers — either Door 2 or Door 3 must be opened and the chance of each must be the same, by symmetry.

  And please do not cite sources that say "incomplete" or that say "that does not mean the probability of winning by switching is 2/3 given which door the player has chosen and which door the Host opens" without mentioning that those sources clearly are ignoring the only correct standard scenario that shows the world famous paradox, and that the paradox is based on, but that those sources report on quite different scenarios, where the world famous paradox never existed. On quite different scenarios, not just on "variants". The article should be clear and intelligible for the readers. Regards, -- Gerhardvalentin (talk) 15:01, 3 June 2012 (UTC)

The Bayes-based formal derivation has been in the article for a very long time, including two versions that were approved by the entire community during featued article reviews [1] [2]. I don't understand how moving it to the end of the article so that it is effectively an appendix does not address your concern. Including this does not mean any formal derivation needs Bayes. -- Rick Block (talk) 16:41, 3 June 2012 (UTC)

Rick, you correctly say ... showing ... the Bayes-based formal derivation ... at the end of the article... does not mean any formal derivation needs Bayes." So why are you of the opinion that it is still a necessity, though?

But the really important point is not to obfuscate. To present the clear MHP with it's only correct decision: to swap doors. And to clearly show that correct scenario of symmetry. Only this scenario can give the famous answer to the famous question, based on the chance to win the door by staying in only 1/3, and to double it to 2/3 by swapping. Only the correct scenario forever firmly excludes that the Host could be forgetful and firmly excludes that the Host could be biased. Only the correct scenario gives the correct probability that has forever to be the exact average probability. Show that in this scenario the conditional probability for door #2 and for door#3 is identical if the Guest should have chosen door #1, and that this is true for all permutations. Show it by Bayes' factor or by Bayes' rule in odds-form:

The odds on the car being behind Door 1 given the player chose Door 1 are 1:2 against. If the car is behind Door 1 (the door chosen by the player) the probability the host opens Door 3 is 0.5 (no host-bias). If on the other hand the car is not behind Door 1 (the door chosen by the player) the probability the host opens Door 3 is 0.50, because the car is now equally likely behind Doors 2 and 3, and he is forced to choose the other one. The Bayes' factor or likelihood ratio is therefore 0.5/0.5 = 1, and the posterior odds remain at 1:2 against.

And, in citing sources that address quite different, quite other scenarios, make clear that they do not address the usual standard scenario. Try to help the readers to grasp what that famous paradox is about, and what it isn't about. - Gerhardvalentin (talk) 20:46, 3 June 2012 (UTC)

I'm not saying it's necessary, I'm saying it should not be deleted from the article. The reason I think it should be in the article is because it or something very much like it is in pretty much any serious source about the Monty Hall problem, including Selvin's second letter, Devlin's follow up column, Krauss & Wang (which is fundamentally a psychology article, not a math article!), etc etc etc - so the article would not be complete without it. You apparently don't like using Bayes' Theorem. No one is saying you or anyone else has to. -- Rick Block (talk) 21:26, 3 June 2012 (UTC)

Regrettably you do not answer to what I say, but focusing on the unnecessary and secondary Bayes' formula only. I repeatedly said that the article is obfuscating for the (new) reader, in not distinguishing on what hell of scenarios it is talking about, mixing different scenarios with quite different characteristics without clearly to distinguish what that means. The Bayes' formula is not the main concern, although the article should not unnecessarily be messed into a lesson on conditional probability theory. You can say things clearly, and to show probability by Bayes' rule in odds-form would be much more suitable for this article. But please address to the main aspects I mentioned, are you willing to follow those aspects? What do you approve, and what are you disapproving? Can you help to make the article on this world famous paradox clear and intelligible as it should be, even for the (new) reader? As to me, Bayes is just one small side aspect. Regards, -- Gerhardvalentin (talk) 22:35, 3 June 2012 (UTC)

If you're looking for me to in some way pledge allegiance to The Truth as you see it, you're out of luck. My allegiance is to what reliable sources have to say. I don't care even the teeniest tiniest bit what you think The Truth is, so if this is your goal here just stop. Rather than argue about The Truth, I'm perfectly willing to talk about specific changes you're interested in. You've suggested deleting the section presenting a formal proof using Bayes' Theorem showing that the conditional probability of switching to Door 2 given the player initially picks Door 1 and the host opens Door 3 (which also applies to any other combination of initial player pick and door the host opens) is 2/3. My response has been a suggestion to move this to the end of the article so it is effectively an appendix rather than delete it, since it (or something like it) appears in many reliable sources. If this is not the the main change you're interested in, what is? -- Rick Block (talk) 05:08, 4 June 2012 (UTC)

Rick, you should at least show respect for the sources, they are going from different points of view. The article is confusingly decomposed, never clear what it is just talking about. The editor's duty is to represent the sources in their respective context, intelligible and not confusingly mingling different aspects. Otherwise the article is crap. It is a pity that you do not want to respect the sources. It is the whole article that shows that narcissistic dilemma. Please help to make it clear and intelligible, with respect for the readers. -- Gerhardvalentin (talk) 10:47, 4 June 2012 (UTC)

Again, if there are any specific changes you're interested in by all means bring them up, or just fix it. If we can return to the formal Bayes' Theorem section, would moving this section to the end of the article help address any of your concerns? -- Rick Block (talk) 15:18, 4 June 2012 (UTC)

Okay, the explanation after the first table in "Solutions" reads "A player who stays with the initial choice wins in only one out of three [...] while a player who switches wins in two out of three.", so I changed columns accordingly (not "switching <> staying", but "staying <> switching"), according to that explanation. Okay? -- Gerhardvalentin (talk) 09:09, 5 June 2012 (UTC)

Rick, you have moved the Bayes' Theorem (Bayes-based formal derivation) to the end of the article, saying "Thus, if the player initially selects Door 1, and the host opens Door 3, the probability of winning by switching is [...] 2/3". But unfortunately you forgot to mention there what you have written here on the talk page, just a few lines above:

"... which also applies to any other combination of initial player pick and door the host opens) is 2/3."

Please add these (your) words there, because that (otherwise unnecessary) formal derivation should be cleared away completely, yes replace it by a formal derivation in clear odds-form, showing that door numbers are completely irrelevant for the standard scenario. And please make clear there in the article that antique sources that prominently stick on door numbers are not addressing the (since decades) world famous paradox with its unique famous standard scenario, that never can exist anywhere else, but are talking of quite different stories, not addressing the world famous counterintuitive paradox itself. And you should say there that such differing scenarios are prominently used in maths classes to teach probability theory. Please keep different scenarios clearly apart. Thank you. -- Gerhardvalentin (talk) 01:41, 19 June 2012 (UTC)

Does the source that is referenced (Gill 2002) say this? If so, why do you ask me to do this rather than doing it yourself? If not, then to what source would you attribute this? I simply moved the section without changing it. It is not in any sense of the word "mine". -- Rick Block (talk) 02:15, 19 June 2012 (UTC)

Rick, the counterintuitive "paradox" (not "1/2 vs. 1/2" but counterintuitive "1/3 vs. 2/3") is valid for each and any single game based on its unique famous well defined standard scenario called the "standard problem". This standard scenario explicitly ensures that, after the host has shown a goat, we have learned absolutely nothing to allow us to revise the odds on the door first selected by the guest (Falk) and hence – in this well-defined scenario – we've also learned absolutely nothing to allow us to revise the odds on the entity of the group of all unchosen doors, as a whole. Absolutely nothing. This shows and assures that door numbers are irrelevant for the standard problem.

The invulnerable "standard problem" assures that conditioning on door numbers is an unimportant side-show there, in this unique famous and well defined "standard problem". Nevertheless you can show the chances by switching in a clear mathematical odds-form, just at the beginning.

Please help to show what the famous paradox is about, in a well organized article. Any differing, any deviating scenario that does not concern the standard problem, should be presented clearly delineated, for the benefit of the (average or new) reader. In later and special sections you can present one deviant scenario, allowing the chance for switching to vary from (assumed) 1 at max. to (assumed) 1/2 at min., but never less, so on average exactly 2/3, though, affirming that it is wise to switch in each and any given game, even in that one deviant scenario: "the more biased, the better!". And you can show there another special deviant scenario of the forgetful host, who will show the car instead of a goat in 1/3 of cases, disrupting the chance on the car in that 1/3 of cases, and who will show a goat only in 2/3 of cases, so reducing the chance to win by switching from 2/3 to 1/2.

And in another section you can show there the scenario of the sneaky host also, who offers a switch only if the guest, by chance, should have selected the car and not one of the two goats.

And you can show maths there in those later sections, where it belongs, with just assumed differing scenarios outside the "standard problem" that will just lead to assumed probability of 0 to 1, for the apprentice (outside the "standard problem"). Nevertheless this article should not be degraded to a nonrelevant maths lesson in conditional probability theory.

We should stop to disconcert and stop to disorient. A clearly structured outline will help the reader to grasp what quite differing and opposing scenarios are talking about. From the clean and invulnerable "standard problem" with its firm answer/decision to other and quite differing stories. -- Gerhardvalentin (talk) 14:32, 19 June 2012 (UTC)

I agree with much of what Gerhardvalentin says. Glrx (talk) 17:38, 22 June 2012 (UTC)

The never ending conflict could easily be settled: stop fobbing the readers. Gerhardvalentin (talk) 16:53, 11 July 2012 (UTC)

JeramieHicks, please read the above also. Gerhardvalentin (talk) 00:35, 3 August 2012 (UTC)

The article still fobs the reader

No progress, it is still the same. The article pretends to present the world famous "standard version" of full symmetry, citing the famous "extended description of the standard version" by Krauss and Wang (2003), and it shows correct solutions that explain the paradox and give the correct answer to that fully symmetrical "Krauss and Wang STANDARD version": This famous version secures that the technical value of unconditional probability and of conditional probability unavoidably must in any case be fully identical (pws:2/3) and in accordance with the average pws. The host is choosing uniformly at random if (in 1 of 3) he should have two goats, so any host's bias  "q"  is a priori excluded, from the outset. No necessity at all to condition on door numbers, their color or their placement. Everyone is free to do so, but this forever will be completely effectless. This is the charm of that world famous counter-intuitive veridical paradox.

But note: Quite another thing is that a large number of textbooks on probability theory like to pay special attention to undue "still assumed"  host's bias  "q"  for the purpose of teaching and learning conditional probability theory. But this is being quite outside the world famous counter-intuitive standard version of the paradox with its only correct decision to switch to the door offered, based on the only to ever be known average probability.

But the article, instead of helping the reader to grasp what all is about, still likes to inspire the reader with distrust, skepticism, suspicion and disbelief. The owners of the article for years liked to use one special trick: by first of all reporting on sources that criticize quite other scenarios outside and diametrically contrary to the symmetrical standard version presented in the article. You can show conditional probability and the forbidden "q", yes. But where it belongs, in a section that tells about devious scenarios.

Without saying in a "new distinct headline" that they are leaving now the field of the standard version, they like to abruptly and explicitly argue about halve-baked confusion, citing sources saying that any "simple solution", not using conditional probability and not using the variable "q" to express the illegal unfounded and excluded and therefore inapplicable "assumed host's bias", are shaky, incomplete, misleading, bluntly false and not answering the "right" question.  Without a clear headline that they already have left the symmetrical standard version, and not saying that they report now on quite other scenarios, beneficial for teaching conditional probability theory only. The old special trick is still alive.

Rick Block has just edited a section now where all of that applies. Although we have been fighting for a clear structure of the article, he still mixes different scenarios, different versions without a clear headline to distinguish what he just is reporting about. Nopeless. Gerhardvalentin (talk) 16:53, 11 July 2012 (UTC)

JeramieHicks, please read the above also. Gerhardvalentin (talk) 00:35, 3 August 2012 (UTC)

The Final Solution

After literally years of talking, the editors of this page are no closer to agreement than they were a year ago. Nor has there developed any easily-countable everybody-but-the-one-holdout-agrees consensus, as usually happens in these situations.

I have a bold solution that I am absolutely sure will result in a better article with a lot less time and effort expended.

Put an indefinite topic block -- no editing of the MHP page or MHP talk page -- on every single person who has ever edited either. (This of course includes me.) Then turn the article into a stub, ripping out and discarding everything anybody has done. In other words, nuke everything and ban everybody.

Wikipedia has 17,093,564 registered users who have made 543,655,915 edits so far. Plenty of them are capable and willing to create a new page on this topic without all of this drama. Within a few months they will create a new article that is far superior to the one we have now, and they will do it without any major conflicts. Giving the boot to a couple of dozen editors who, collectively, have completely failed to figure out what should be in the article will, in the long run, have a positive effect.

Ban and nuke. It's an idea who's time has come. --Guy Macon (talk) 10:51, 9 July 2012 (UTC)

Although I understand your sentiment, that idea is not really compatible with the concept of 'Wikipedia, the encyclopedia anyone can edit'. Despite fundamental disagreements there has been no incivility or edit warring here nor anything else that would justify the wholesale banning of editors. Martin Hogbin (talk) 14:51, 9 July 2012 (UTC)

I fully agree that it would topic ban those who are not part of the problem (including me!) but in my opinion, this would be a good place to invoke the oft-misused WP:IAR rule. If the "encyclopedia anyone can edit" rule prevents us from improving Wikipedia, ignore it. Clearly what we are doing now is not working, and all attempts to resolve the issue have failed miserably. Topic banning and stubifying will fix the problem. We would, of course, carefully explain that the topic ban is not based upon any bad behavior. --Guy Macon (talk) 17:13, 9 July 2012 (UTC)

Okay! Guy Macon, d'accord! I agree, for obviously there's no way out. Gerhardvalentin (talk) 17:42, 9 July 2012 (UTC)

Since I believe we are on the verge of initiating an RFC, this proposal seems at best premature. -- Rick Block (talk) 02:30, 10 July 2012 (UTC)

I agree, if the dispute gets settled in the next month or so, and will happily report that I jumped the gun, giving up on a years-long process right before success. Please do prove me wrong. Nothing would make me happier. --Guy Macon (talk) 03:57, 10 July 2012 (UTC)

Who will decide which of these editors will be banned? Martin Hogbin (talk) 08:06, 10 July 2012 (UTC)

Under my proposal, nobody will decide. They would all get nice polite notices saying that administrators X and Y are asking everyone who has contributed to walk away and give the topic a fresh start with new editors, and that they are getting the notice even though they have not contributed to the page in quite some time.

An alternative would be to only topic ban those who have contributed in the last six months. What I do not want to do is to require someone to decide that user A is causing the problem and user B is not. Arbcom already did that. It did not resolve the issue.

As I said, I really hope that you will prove me wrong and come up with a solution by the end of July. As Bullwinkle J. Moose so often says; " This Time For Sure!. " --Guy Macon (talk) 17:16, 10 July 2012 (UTC)

I think there is no prospect of agreement, even on an RfC, in the near future. Rick and I agreed the wording but now he wants to add some stuff to it. Martin Hogbin (talk) 18:40, 10 July 2012 (UTC)

Martin and I agreed on the wording of the description of the dispute, but have not agreed on the exact phrasing of the question to be asked following this description. I'm not suggesting we "add some stuff to it" ("it" meaning the agreed description). I have solicited opinions on the phrasing of the question from 3rd parties which have been less than clear, and am currently pursuing other alternatives. -- Rick Block (talk) 19:07, 10 July 2012 (UTC)

Wow. Disagreement about what the disagreement is about. What a shock. Never saw that one coming... --Guy Macon (talk) 19:19, 10 July 2012 (UTC)

Well, it has been an additional twelve days. How is that "doing what we have been doing for the last few years again and again is sure to get results Real Soon Now" plan working out for you? --Guy Macon (talk) 20:04, 21 July 2012 (UTC)

Critizism of the simple solution

I do not agree with the late edit of Martin, I have no idea what he is referring at when pointing to a retraction by the authors. Nijdam (talk) 15:13, 9 July 2012 (UTC)

I removed 'but without additional reasoning this does not necessarily mean the probability of winning by switching is 2/3 given which door the player has chosen and which door the host opens '. This was based on the paper by Morgan but, as you know, the authors said, in response to our letter, 'had we adopted conditions implicit in the problem, the answer is 2/3, period'. That seems pretty clear to me. Martin Hogbin (talk) 21:13, 9 July 2012 (UTC)

They just said "the answer is 2/3, period." Answer to what? To the conditional probability. What else? Nijdam (talk) 10:27, 10 July 2012 (UTC)

This whole section was a mess. I've rewritten it. -- Rick Block (talk) 04:15, 10 July 2012 (UTC)

Why have you missed out any reference to the Morgan quote above 'had we adopted conditions implicit in the problem, the answer is 2/3, period'? What do you suppose Morgan mean by 'conditions implicit in the problem'? Martin Hogbin (talk) 08:09, 10 July 2012 (UTC)

I'm not sure what the point here is. This section does not say anything about the 1/(1+q) answer from the Morgan et al. paper, which is obviously what they're referring to in the quote above. I think a much more relevant quote is this one (from their rejoinder to vos Savant's comments [3]) One of the ideas put forth in our article, and one of the few that directly concerns her responses, is that even if one accepts the restrictions that she places on the reader's question, it is still a conditional probability problem. One may argue that the information necessary to use the conditional solution is not available to the player, or that given natural symmetry conditions, the unconditional approach necessarily leads to the same result, but this does not change the aforementioned fact. -- Rick Block (talk) 16:00, 10 July 2012 (UTC)

• It is inappropriate to criticize a simple solution by changing the problem it purports to solve. Glrx (talk) 15:03, 10 July 2012 (UTC)

Who are you suggesting is doing that? Martin Hogbin (talk) 15:30, 10 July 2012 (UTC)

More specifically, are you suggesting this section is not accurately representing what is said by the many sources to which it refers, or, are you making a personal statement that you disagree with the approach used by these sources? -- Rick Block (talk) 16:00, 10 July 2012 (UTC)

Martin, the article says that Morgan et al. "are" doing it. Yes Glrx, you are right again: the excellent mingle-mangle article, owned for years now, never straightforward clearly did say nor says straightforward and clearly on which scenario it is just actually talking about. The reader never will know: is it actually talking on the standard version, or is it actually talking on quite other conceivable but diametrically opposed versions, completely outside the scenario of the standard version. That's the crux of that owned article, the crux of the perpetual dilemma here. Gerhardvalentin (talk) 16:13, 10 July 2012 (UTC)

As I wrote earlier Critizism of the simple solution is a strange section. Someone in the "literature" came up with the difference between a problem with and without door numbers, and in addition the Wikipedia reader is surprised by the disagreement in the "literature" whether the question he can read at the beginning of the article is the "first or second question" (see my remarks below to door numbers and so on ...). And what is the difference between "switching" and "always switching"? The probabilities for both in the concrete game are the same ... If "always switching" means "whatever door the host opens" it should be made clear. And here my suggestion for a new article: 1/2, 1/2 (or 1/3, 1/3); 1/3, 1/6; 1/3, 1/3*q --Albtal (talk) 16:46, 11 July 2012 (UTC)

This is fully correct, Albtal. Modern academic mathematical literature proofs that the best decision in each and every game is to switch and never to stay, any other decision is throwing you back. This is applicable to the symmetric standard variant, but to any assumed host's bias (whatsoever) as well. The conclusion of this modern academical paper: You cannot do better than to "switch right away in every game, and the host is most welcome to be biased: The more biased, the better!". As said: any staying throws you back. Gerhardvalentin (talk) 17:17, 11 July 2012 (UTC)

Albtal - The difference is between switching whatever door the host opens (effectively, deciding to switch BEFORE seeing which door the host opens or, if you prefer, after the host opens a door but without knowing which door) as opposed to switching with knowledge of which door you picked and which door the host opened. The door numbers are irrelevant in either case (even without numbers there would be a door on the left, a door in the middle, and a door on the right - so the doors are inherently distinguishable to both the contestant and the host). What these sources are saying is that in the latter case the probability of winning by switching (assuming you picked Door 1 and the host opened Door 3) is P(car behind door 2|player picks door 1 and host opens door 3) while in the former case it is P(car behind door 2 or door 3|player picks door 1). These are different probabilities - a solution telling you one is not telling you the other even if they have the same numeric value unless you also show why they must be same. The clearest way to see these probabilities are different is to vary the problem so that they have different values - which is what these sources do, demonstrating both that there indeed is a difference and that solutions determining one don't determine the other. As you say below "Everybody knows that in the MHP the question is for the probability just before the second choice". Indeed. What these sources are saying is the "simple" solutions do not determine this probability and make no attempt to connect the probability they do determine (before the host has opened a door) with what everybody knows the question is asking about, i.e. they're "incomplete" or "misleading". -- Rick Block (talk) 20:35, 11 July 2012 (UTC)

Yes. As I wrote, there is no need for questions, one without and the other with door numbers, but simply ask for the probability of winning by switching

1. before the game starts (or before the host opens a door)

2. after the host has opened a door

One of the "simple" solutions is: Switching wins in two of three cases, so p = 2/3. This is true, before the host has opened a door. This is not a "complete proof" (for 2.), but a key step for understanding the solution, and in no way "misleading". Going further on the simple way just before reaching the goal there is the interesting question: If the probability 2/3 has changed after the host has revealed a goat: Where did it disappear? The only possibility for this is that there are (at least) two cases: One with a probability higher than 2/3, and one with a lower. But where might they come from? The only place for an asymmetry in the game is that the host prefers a door if he has a choice. If he does so I can't know how, and I cannot solve the problem (with exact probabilities). But why should he prefer any door? There is no reason. So the solution is 2/3. (See also the simple solution below: If the solution is independent of the door which the host opens, it is 2/3.)

Another remark: There are no "conditional" and "unconditional" problems or questions. There are only problems and questions, which sometimes may be solved by using conditional probabilities - or not. This whole "conditional salad" seems to come from a source whose authors wrote about the one-million-doors-analogy of Marilyn vos Savant: She then went on to give a dubious analogy to explain the choice.--Albtal (talk) 23:09, 11 July 2012 (UTC)

I should also remark here that it is my opinion that the ongoing argument is only a small side problem within MHP which doesn't really interest anybody but surely should be mentioned for completeness and correctness. And overloading the task setting pedantically will cause a yawn before the reader understands the essential conditions (I cannot prove this exactly, even if applying conditional probabilities). As Martin Hogbin wrote in a discussion with me in May 2012 (I temporarily had the user name Scharzwald) both sides of the ongoing argument don't agree in my view of the main problem. I agree with Marilyn vos Savant that the most significant condition is that the host always opens a loosing door on purpose. And she herself writes in her book (p. 15): When I read the original question as it was sent by my reader, I felt it didn't emphasize enough that the host always opens a door with a goat behind it. I myself would formulate this central condition as follows: The contestant now determines two doors, of which the host has to open one with a goat. For every understanding of the problem which is not equivalent to this has no 2/3-solution. But there is a crucial difference to Marilyn: I would formulate this condition in the question, not in the answer. But so the world is full of strange answers for a question which does not really have a 2/3 solution. I understood Martin Hogbin well. Surely my suggestions for a better article persist. Maybe sometime the majority changes.--Albtal (talk) 09:28, 12 July 2012 (UTC)

Gerhard - no one is saying don't switch. Always switch is the answer Morgan et al. came up with 20 years ago. Nothing "modern" about it. -- Rick Block (talk) 20:35, 11 July 2012 (UTC)

Self-deception? Rick, the simple solutions here in the article are explicitly based on full symmetry, as per the standard version by Krauss and Wang. Presenting the explicitly symmetric standard version where "different numeric values" are banned and are impossible from the outset, by "simultaneously" confusingly using sources that show optional different numeric values though, by irresponsibly ignoring and violently removing the given symmetry??? It is the charade of this messy article to deceive, and to misguide and to mislead. Here are your own words you just wrote above:

"a way to see these probabilities are different is to vary the problem so that they have different values - which is what these sources do".

These are your own words. And you do not hesitate to name the simple solutions of the fully symmetric standard version to be shaky, incomplete, misleading, bluntly false and not answering the "right" question etc. And that is what for years you have been doing here. Unbearable. The fully symmetric version does not need to condition on door characteristics, as per modern sources.

Rick, please distinguish the famous paradox with its strict conditions of full symmetry as per the standard version (Krauss and Wang), keeping it apart from quite other contradictory scenarios that a priori are forbidden by those well defined strict conditions. Please clearly distinguish those contradictory versions and show them after a clear headline saying that the following sections are different to the famous standard version and do no more concern that famous standard version. You can show there conditional probability theorems, also. To show the conditional probability is fine, but its a bold lie that it is academical consent that you MUST use it to "solve" the symmetrical paradox.

Once more: you CAN do it, but modern academic sources agree that you do not need it to solve the symmetrical paradox. I hope you can an will help to muck out the crap. Gerhardvalentin (talk) 22:17, 11 July 2012 (UTC)

Gerhard - please distinguish Truth from TRUTH. I'm merely saying what it is that these sources say. Also note that they are numerous and keep appearing (1991, 1992, 2001, 2005, 2009 - and this is not even a complete list). Whether you, I, or anyone else agrees is completely beside the point. The POV they express certainly belongs in the article. -- Rick Block (talk) 02:08, 12 July 2012 (UTC)

Rick, we have recently established on Sunray's page that very few, if any, reliable sources say that, considering only the case that the host must choose an unchosen goat-hiding door uniformly, the simple solutions are wrong or incomplete. Martin Hogbin (talk) 09:35, 12 July 2012 (UTC)

As anyone following the provided link can plainly see, what we have established is that you think this, I don't, and you are apparently willing to argue tendentiously that these sources don't say what they clearly say. -- Rick Block (talk) 15:13, 12 July 2012 (UTC)

As anyone can plainly see, this article on the world famous counter-intuitive paradox is well-founded on its only correct famous "standard version" of full symmetry, firmly excluding any "asymmetric host's bias" (modern sources proof that you only MAY base on symmetry). The famous "standard version" presented here, does firmly exclude any "closer results" than the average probability to win by switching (pws) of 2/3. Each and every "simple solution" shown here is / are based on that famous standard version only, all "simple solutions" here are based on strict symmetry, so they are correct as per modern sources. In an enduring confusing way, Rick Block insists on citing especially one source that – quite contrary to the standard version presented here – explicitly addresses one quite other scenario of unknown asymmetry and – these are Rick Block's own words above on his impurity source: – they "vary the problem so that they have different values - which is what these sources do" (once more: these are Rick Block's words above). This is impure as it only makes real sense if (contradictory to the standard version that the article is about) one can assume that the host's asymmetric bias (!) can reveal additional information on the whereabouts of the prize, giving "closer results" of pws from 1/2 to full 1. The article should meticulously keep apart the fully symmetric "standard version" that the article is about, from quite other scenarios that a dated source had in mind. And it should clearly be shown that you "CAN" use conditional probability theory, but - as actual sources clearly proof - that this never is "NECESSARY" in any way to make the only correct decision. The article is tattered in indistinctly using sources, and needs to be repaired. Gerhardvalentin (talk) 09:31, 18 July 2012 (UTC)

Extended problem description

I suggest to make the article more transparent by renaming this headline to

"Extended description of the standard version"

Any valid objections? Gerhardvalentin (talk) 18:04, 10 July 2012 (UTC)

Please cite the ISO number, or official designation of another recognized standards setting organization.   ;-)

More seriously, there are two substantial changes introduced in this version and I don't think they are equally "standard". The first change is to stipulate that the host must open an unchosen door revealing a goat. This convention is used so universally that it may be regarded as standard. The second change is to provide explicit frequency distributions (for the host's choice of goat and for initial placement of the car). Although this convention is common in sources that use MHP to explain conditional probabilities, it is by no means prevalent in popular literature that does not employ the same formalism. The latter often use, at least implicitly, a Bayesian perspective rather than a frequentist one. (Cf. Richard Gill's paper,[4] §3)

As I remarked elsewhere, I think it is confusing to interpose a frequentist formulation of the question before giving "simple" answers that do not employ frequency distributions. I think it casts those solutions in a poor light, and I do not think labeling that formulation as "standard" represents a neutral point of view. ~ Ningauble (talk) 17:07, 21 July 2012 (UTC)

P.S. – I would be grateful if anyone could locate additional sources that expressly explain the Bayesian perspective on MHP. An accessible explanation of the logical basis might help to frame the different solutions as complementary approaches, and reduce the article's tendency to give undue weight to apparent controversy. (Gill's paper for Statistica Neerlandica assumes that his audience understands the epistemological difference, and the popular literature that tends to employ Bayesian reasoning tends not to trouble its audience with abstract underpinnings.) ~ Ningauble (talk) 17:14, 21 July 2012 (UTC)

;-) The article already said "A fully unambiguous, mathematically explicit version of the standard problem is ..." — So the headline should say what is to follow ... ;-) Gerhardvalentin (talk) 23:15, 21 July 2012 (UTC)

* Also Henze, Norbert (1997). Stochastik für Einsteiger: Eine Einführung in die faszinierende Welt des Zufalls, 9th edition 2011, pp. 50-51, 105-107, Springer, ISBN 9783834818454, (restricted online copy at Google Books) says:

Der Standhafte gewinnt dann den Hauptgewinn, wenn sich dieser hinter der ursprünglich gewählten Tür befindet, und die Wahrscheinlichkeit hierfür ist 1/3. Ein Wechsler hingegen gewinnt das Auto dann, wenn wenn er zuerst auf eine der beiden "Ziegentüren" zeigt (die Wahrscheinlichkeit hierfür ist 2/3), denn nach dem Öffnen der anderen Ziegentür durch den Moderator führt die Wechselstrategie in diesem Fall automatisch zur "Autotür". Bei allen diesen Betrachtungen ist natürlich entscheidend, dass der Moderator die Autotür geheimhalten muss, aber auch verpflichtet ist, eine Ziegentür zu öffnen. Wer dieser Argumentation nicht traut und lieber praktische Erfahrung sammeln möchte, lasse sich unter der Internet-Adresse  [5]  überraschen.

"For all of these considerations it is of course essential that the host is obliged to maintain secrecy regarding the car hiding door, and is also obliged to open a door with a goat behind."  —  That's what I just added to the section "Other simple solutions based on the standard version of the paradox".

Krauss&Wang defined the only senseful valid scenario, and Norbert Henze (Stochastik für Einsteiger: Eine Einführung in die faszinierende Welt des Zufalls) even is utterly ignoring any other improper scenario. He says

of course you can refine/narrow down the scenario, e.g. if the host has two goats to choose from, he could open the door with the smaller number with probability q

and he hurries to instantly add:

Similar examples can be found in educational books on stochastics. Their sole purpose is to schematically practice conditional probability arithmetic.

But he does NOT say that this could be of any relevance for the world famous paradox. —  So it's up to you to decide what the world famous paradox (1/3 : 2/3, and not 1/2 : 1/2) indeed is, and what a mess this article still is. Gerhardvalentin (talk) 23:15, 21 July 2012 (UTC)

Apart from apparently suggesting that Wikipedia is the official standards setting organization, I don't see how the foregoing remarks have any bearing on my objection to labeling a frequentist formulation of the problem as "standard". ~ Ningauble (talk) 13:20, 22 July 2012 (UTC)

As a 'strong Bayesian' I too would love to see a solution based on a Bayesian interpretation of probability but, unfortunately, it seems that most serious commentators prefer what looks like a frequentist (or realist as Richard might call it) approach to the problem. Martin Hogbin (talk) 17:29, 22 July 2012 (UTC)

In Barbeau's literature overview, cited in the article, he says the "standard analysis" is based on both assumptions (host always opens an unselected door to reveal a goat, choosing randomly if both conceal goats). Selvin, in his second letter (not his first as Richard inaccurately claims in his paper), says the same thing - making these not just the "standard" assumptions but also the "original" assumptions. Lacking a source claiming that most popular sources (who don't say what they're doing) are taking a "Bayesian" perspective, I think the most reasonable approach is to include both of these as part of the "standard" problem. Doing this conveniently makes the answer, whatever your interpretation of probability and whether you interpret the question to be about a strategy of switching vs. staying or about a particular case where the player knows which door she picked and which door the host opened, 2/3 chance of winning by switching. Distinguishing among these is clearly something the article should do at some point - but forcing all answers to be the same gives us no reason to prefer one answer over another, which rather than not being neutral seems precisely neutral. -- Rick Block (talk) 00:41, 23 July 2012 (UTC)

If I understand it correctly, Ninguable's point is that all the analyses of the problem seem to take a frequentist approach to the problem in which it is necessary to assume distributions for those that are unknown. Many people find a Bayesian approach, in which probability is defined as a state of knowledge, to be more natural. As I said, years ago, right at the start of this discussion, using a Bayesian approach we are obliged to take probability to be evenly distributed between possibilities about which we have absolutely no knowledge. The Bayesian approach thus gives the same result as the frequentist (realist, probability theory) result where uniform distributions are assumed for the initial car placement, the player's initial choice, and the host's choice of legal door but there is no need for these assumptions to be explicitly stated because they are an integral part of the Bayesian perspective. Martin Hogbin (talk) 08:53, 23 July 2012 (UTC)

More specifically, my point is that those who address the problem using frequentist principles have redefined the question in terms of frequency distributions in order to facilitate thier analysis.

Taking this as the canonical problem can lead to a vague and unfounded impression that logical solutions using inferences that do not refer to frequency distributions are somehow lacking, and hence to the POV that only conditional probability gives the correct answer to life, the universe, and everything. However, life does not always hand us neat frequency distributions on a silver platter: "Imagine you are on a game show..." and the host goes off-script with an unprecedented scenario. ~ Ningauble (talk) 18:32, 23 July 2012 (UTC)

Why delete my contribution?

I added a few days ago this explanation:

Alternatively one can change the game slightly to understand the solution. Suppose that there are two players: the original contestant who does not change his first choice after Monty shows the goat; and a second contestant who comes in at this stage to choose the remaining door. Obviously because there are two players and only two doors which might contain the car, between them the two players have a 100% chance of winning the car. The first player has the 1/3 chance he started with: nothing has changed -- so the second player has a 2/3 chance of winning. This is the same probability of the single player winning the car if he changes his choice.

This a simple explanation in the vein of many solutions to mathematics or logic problems. It is very clear and I have used it dozens of times to explain the MH problem to those who disbelieve it. I don't want to get into an edit war but it is a little high-handed (Glrx) to remove it with the comment that it adds nothing.Cross Reference (talk) 03:01, 19 July 2012 (UTC)

If the "first" player has a 1/3 chance with his first choice after Monty shows the goat, he has a 2/3 chance with the other remaining door. You don't need a second player.

But more important: You write The first player has the 1/3 chance he started with: nothing has changed. If so we hadn't a MHP at all. Indeed your argument seems to be based on the widespread error that the probability for the first door cannot change if the host opens another door with a goat. But the 1/3 chance for the first door only "stays" if the host is obligated by the rules of the game to open a not selected door with a goat. And it is exactly this obligation which leads to probabilities of 1/3 and 2/3 instead of 1/2 and 1/2; not that he opens the door with a goat for some other reason. And it is exactly this obligation which was missing in the original problem setting. As I wrote earlier the crucial condition in the problem set should be formulated as: The contestant now determines two doors, of which the host has to open one with a goat. And to those who disbelieve the 2/3 solution you can now say: If I choose door 1 I will win the game by switching if the car is behind door 2 and if it is behind door 3. For if the host opens door 2 I shall take door 3, and if he opens door 3 I shall take door 2. This is a first important step to understand the 2/3-solution; and you can demonstrate this easily with three playing cards.

If the host is free in his decision but acts exactly like in the original problem set (that is opening an unchosen door with a goat and offering a switch) the contestant cannot know whether the host wants to help him or if he wants to disabuse him of the right door. So the answer p=1/2 for each of the remaining doors is reasonable. --Albtal (talk) 09:16, 19 July 2012 (UTC)

@Cross Reference. You "Added another simple solution".[6] At least one other editor questioned the addition because Ningauble tagged it with citation needed and commented "I don't see how this elaboration is any clearer than the preceding paragraphs".[7] I also questioned its addition, so I removed it. It is appropriate to debate adding the explanation back on this talk page. WP:BRD

To me, your explanation jars by saying "change the game slightly" and then tightly associates the chosen and remaining doors with the old and new contestants. The explanation adds more complexity with the additional contestant, but that contestant does not add clarity. It was the longest paragraph in the section.

To be clear, I did not say the addition "adds nothing" (and Ningauble did not say that either). My summary states "Changing game explanation adds little".[8] There is something to your explanation. It prohibits the initial contestant from switching. At the start, we know the initial contestant had a 1/3 chance, so there's a notion that no matter what Monty does with the remaining two doors, a stick-no-matter-what contestant will end up with the car 1/3 of the time, so a switch-no-matter-what strategy will win 2/3 of the time. We can reach that result without changing the game or adding a second contestant -- which is essentially the explanation in the first paragraph of Monty Hall problem#Other simple solutions and the thrust of the entire section.

Glrx (talk) 16:05, 19 July 2012 (UTC)

@Glrx -- sorry for misquoting you. The explanation (it's is not really another solution) is in the vein of other questions such as the well known one about the pilgrim who goes up a mountain. There is a single winding path up the slope, and he leaves approximately at dawn and arrives at the top more or less at sundown. He never deviates from the narrow path, but he does take time on the way up for eating, resting and playing his flute. Next day he comes back down the hill, again stopping randomly and and arriving at the bottom as the sun goes down. Was there any time where he was at the same point on the path to the nearest millisecond of time as he was the day before? That can be proven by some reasonably heavy duty maths, but the easiest way to satisfy yourself is to imagine him doing it in both directions on the same day and meeting himself at some point. Change the question and it become obvious. The solution (2/3 if you switch, 1/3 if you stick) is obvious to me after a little thought (and I have modeled it in Excel to 10,000 lines) but I have rarely successfully explained it to the doubters until I started using the second player concept.Cross Reference (talk) 02:33, 20 July 2012 (UTC)

This is conceptually similar to what I suggested a while ago (see discussion now archived at [9]). Explicitly changing the problem makes the 1/3 vs. 2/3 answer easier to understand. What I suggested was meticulously sourced, but even so at least Martin and Glrx didn't like it. Are there other opinions about this? -- Rick Block (talk) 15:37, 21 July 2012 (UTC)

Very good and illustrative example, Cross Reference. The crux here is that, when showing illustrative examples, some editors say that "this is changing the problem". Gerhardvalentin (talk) 18:11, 21 July 2012 (UTC)

No Rick, you are disarranging again. This article is about the famous symmetrical standard version of the paradox,  that ALL simple solutions here are based on. You have to observe that it is inappropriate to criticize them by changing the problem, in drudgingly observing which one of his two doors the host just has opened. And to call it "changing the problem" if you talk of the original problem again. In forgetting what the article is about. Yes, in talking of the symmetrical version, you call that "explicitly changing the problem". The article is, in the first line, on the "symmetrical standard version" of this counter-intuitive paradox. In this standard version, the probability to win by switching is 2/3, full stop.  2/3 "before", and  2/3 "after".

In this standard version, the "difference" of "simple" probability and of "conditional" probability is irrelevant. So please help the reader to grasp, without still trying to confuse him/her. Quite different scenarios of a host, who will only offer to switch in case the guest should first have selected the car, are to be shown quite "outside" the famous counter-intuitive standard paradox, in a later and clearly separated section. As well as the host, who in some certain scale will forget the location of the objects and subsequently will risk to be showing the car instead of his goat, and by that ruining the chance to win by switching to a certain extent. As well as the forever only to be "assumed" host who shows, by exceptionally opening the door that he usually is "assumed" to strictly avoid to open, that the probability to win by switching will be "1" in that case, being subject matter of maths teachers in probability theory. Every "simple solution" presented in THIS article is based on the symmetric standard version, so no need to "explicitly changing the problem". Once more: All "simple solutions" here are based on the famous symmetric standard version. And please stop to inappropriately criticize a simple solution here, by changing the problem it purports to solve. Please stop to disarrange. Once more: please help the reader to grasp, without still confusing him/her. Gerhardvalentin (talk) 18:11, 21 July 2012 (UTC)

Time for The Final Solution?— Preceding unsigned comment added by Guy Macon (talk • contribs) 01:27, 22 July 2012 (UTC)

With the help of Norbert Henze ("For all of these considerations it is essential of course that the location of the car behind the doors has to be kept secret by the host, and that he is also obliged to open a door showing a goat", or: "For all of these considerations it is of course essential that the host is obliged to maintain secrecy regarding the car hiding door, and is also obliged to open a door with a goat behind." and: "Of course you can refine/narrow down the scenario, e.g. if the host has two goats to choose from, he could open the door with the smaller number with probability q. Similar examples can be found in educational books on stochastics. Their sole purpose is to schematically practice conditional probability arithmetic."), and with the help of other modern sources, and with the help of Rick Block and of Martin, of Glrx, Guy Macon, of Kmhkmh and Ningauble and others it should be possible to get to a far better version without further health warnings aimed in the wrong direction, and without lessons in practicing conditional probability theory, within the irrevocable fixed period, I guess ;-)   Gerhardvalentin (talk) 10:45, 22 July 2012 (UTC)

I support you and wish you luck Gerhard but the article is still dominated by editors who cannot see the simple fact that you (and yet one more new editor above - Albtal) plainly can, namely that in the symmetrical case the host's choice of unchosen goat-hiding door is unimportant and need form no part of the solution. Martin Hogbin (talk) 15:05, 23 July 2012 (UTC)

Meaning?

What is the meaning of: "Other simple solutions based on the standard version of the paradox", and why is the so called "combining doors solution", which is wrong, as Devlin himself admits, still in the article? Nijdam (talk) 08:55, 23 July 2012 (UTC)

All "simple solutions" are based on symmetry. After citing the mathematically explicit Krauss and Wang version of the standard problem (symmetry), the article says "Solutions" and presents as one first solution the table by vos Savant (Parade) saying that staying has a chance to win the car of 1/3, while switching has a chance of 2/3.

After vos Savant's solution, before presenting Carlton, Henze and many others, there is a new headline titled "Other simple solutions ...".

And as to the entities of 1, of 2 and of 3 doors: The entity comprising only one single door has 1/3 chance to hide the prize, while the entity of all three doors (for example door#1 + door#2 + door#3 altogether) have 3/3 chance ("1") to hide the prize. Gerhardvalentin (talk) 13:37, 23 July 2012 (UTC)

Nijdam, can you show me where Devlin admits that he was wrong. Martin Hogbin (talk) 15:06, 23 July 2012 (UTC)

+1: [citation needed] ~ Ningauble (talk) 17:53, 23 July 2012 (UTC)

We should not be calling MPH a paradox in the section heading, because a so-called "veridical paradox" is not a genuine logical paradox. It is not a case of two sound lines of argument leading to incompatible conclusions, it is simply a case of common sense being wrong. (There would be a real paradox if Bayesian and frequentist reasoning gave conflicting conclusions; but they don't, and it is not merely a coincidence, as some here seem to believe, that they both give the same odds.) The whole "based on..." clause is superfluous: context should indicate what question the solutions answer. ~ Ningauble (talk) 17:53, 23 July 2012 (UTC)

+1, Ningauble. The context should show it. Yes! But for years, after presenting the Krauss+Wang definition of symmetry as "the basics", followed by simple solutions obviously based on those "basics", especially Nijdam ("that has to be proven, you need a proof") insisted in citing health warnings in an opaque manner, saying that simple solutions do not answer the question put, that they solve the wrong problem, that they are incomplete etc. etc., in a very common way, WITHOUT mentioning that those sources either explicitly address asymmetry (changing the problem) – or just are mathematical cure-all approaches that try to solve both, the symmetrical AND the asymmetrical variants also. But, as more and more modern sources are emphasizing, regarding the paradox it is indispensable ("OF COURSE"!) to base on symmetry, whereas "assumed" deviations by "assumed" host's bias, giving "assumed" results, are a matter of learning maths, their purpose is to schematically exercise conditional probability, as per numerous convenient textbooks, serving exactly that educational purpose. And, at least in the past, Rick had permanently pointed up that this educational material for maths classes is quite numerous, indeed. Yes, it is, but nevertheless the clear and distinct structure of the article should help the reader to grasp what the paragraph he just reads actually is talking about. Gerhardvalentin (talk) 18:52, 23 July 2012 (UTC)

Request for explanation

Let's suppose I'm out of the room during the initial phase of the event, and walk in at the last moment. All I see are two closed doors, and presumably one prize. Why should I assume that knowing any history prior to this moment would change the "two doors, one prize" moment? The article didn't clearly explain, at least to me, why the history should have any significance. I guess I (incorrectly) think of it like the gambler's fallacy... prior events having no impact on the current ("two doors, one prize") moment. JeramieHicks (talk) 22:06, 1 August 2012 (UTC)

If the show proceeds given the normal assumptions (car is randomly placed, player picks a door, host opens a different door choosing randomly if the player's initial pick hides the car), the probability the car is behind the door the player picks is 1/3 while the probability the car is behind the other door is 2/3. By "probability" I mean the limit of the relative frequencies of these outcomes if you repeat this same set up numerous times (see "frequentist probability"). This is also the probability you would deduce in a Bayesian analysis of this situation (which allows an analysis of events you might not repeat using information available to an observer). You might have noticed some of the folks on this page don't seem to get along - the difference between "frequentist" and "Bayesian" is possibly the root of much of the conflict.

Back to your question. The frequentist and Bayesian probability the car is behind the door the player initially picked is 1/3, while both of these are 2/3 it is behind the other door. If you arrive late, we can now see the difference between these two interpretations of probability. The frequentist probability is and will forever remain the same. The Bayesian probability however is a function of the information you know. Rather than measure the probability of "what is" (in the sense of what we'll see if we repeat the same thing over and over again) it really measures the probability that someone knowing a certain amount of information can make a correct choice. The difference is subtle but very real. If you don't know the setup, to pick the car you must effectively pick randomly between the two remaining doors. A random choice between two alternatives, regardless of any "real" (frequentist) probability has a 50/50 chance of being correct. So, in a Bayesian sense, the "probability" the car is behind each door for you (not knowing the set up) is 1/2 - while at the same time the frequentist (and, for those in the audience, Bayesian) probability the car is behind the player's initial pick is 1/3 (and 2/3 for the other door).

A little algebra may (or may not) make this more clear. Let's say there are two alternatives and the frequentist probability between them is p and 1-p (where p is anything between 0 and 1 - i.e. we hide a car between two doors using any algorithm ranging from "always hide it behind door 1", to "flip a coin to pick which door to hide it behind", to "always hide it behind door 2", to "hide it randomly behind one of three doors, let a player pick one door and have the host open another door picking randomly if the player's door hides the car"). You arrive on the scene not knowing the algorithm we used. The frequentist probability that the car is behind door 1 is p, and that it is behind door 2 is (1-p) - meaning if we do this over and over again these will be the limits of the relative frequencies of where the car is. You don't know this. If we do this 100 times (where, whatever the algorithm, we've ended up with door 1 and door 2), then roughly 100*p times the car will be behind door 1 and 100*(1-p) times the car will be behind door 2. If you pick randomly, of the 100*p times the car is behind door 1 you'll pick door 1 half of these, so you'll be right about 50*p times. Similarly, of the 100*(1-p) times the car is behind door 2 you'll be right about 50*(1-p) times. Altogether, you're right 50*p + 50*(1-p) times, which is 50 times, which is 1/2 of the time.

The point here is that to someone who doesn't know the "actual" (frequentist) probability, picking randomly between two choices makes the result a 50/50 choice. Another way to say this is if you don't know the history and there are only two choices left, you have a 50/50 chance of making a correct choice. On the other hand, this definitely doesn't mean that if you're choosing between, say door 1 and door 2, if you repeat whatever the set up was it will converge to 50% of the time the correct choice is door 1 while 50% of the time the correct choice is door 2. -- Rick Block (talk) 05:16, 2 August 2012 (UTC)

Oh dear! Martin Hogbin (talk) 12:39, 2 August 2012 (UTC)

Or simply put: walking in after the initial phase, you do not know which one of the remaining closed doors is the one originally picked by the player. Repeating will show you also the cases where door 2 is initially picked and door 3 opened by the host. The relative frequencies for both doors to hide the car will both tend to 50%. Nijdam (talk) 11:10, 2 August 2012 (UTC)

Oh dear, yes! Gerhardvalentin (talk) 00:45, 3 August 2012 (UTC)

A little clearer. If only the article concentrated on issues that people are actually interested in, like this one. Martin Hogbin (talk) 12:39, 2 August 2012 (UTC)

The fact that many people wade through the whole article and are left asking the same questions does seem to indicate that the article is wanting something. On the other hand, it may be a certain something that defies any possible appeal to common sense. ~ Ningauble (talk) 17:51, 2 August 2012 (UTC)

My answer to JeramieHicks's original question is this: What the person arriving in the middle does not realize is that the visible goat does not just reveal information about one of three doors; but specifically reveals information about one of two doors not initially chosen. The narrower scope makes it much more informative: 1/2 is more than 1/3. (This is a Bayesian perspective, which does not correspond to what frequentists define as probability analysis.) The gambler's fallacy does not apply here: it pertains to independent events (e.g. separate rolls of the dice), not to incremental information about one thing (i.e. the location of the car). ~ Ningauble (talk) 17:51, 2 August 2012 (UTC)

@JeramieHicks: First, as I wrote above: I myself would formulate the central condition of the task set as follows: The contestant now determines two doors, of which the host has to open one with a goat. For every understanding of the problem which is not equivalent to this has no 2/3-solution.

Then, if you "choose" say door 1 - what means that the host has to open door 2 or door 3 with a goat - you know before the host opens a door that you will win the game by switching if the car is behind door 2 or door 3. For if the host opens door 2 you will take door 3, and if he opens door 3, you will take door 2. So you have a 2/3 chance by switching before the host has opened a door. The question is now whether the chance to win by switching is still 2/3 if the host has opened door 2 or 3 with a goat. Suppose the chance is now 1/2. Then there would be a strange situation: You have invented a game which you always start with a 2/3 chance which will always change to a 1/2 chance during the game.

Surely the "history" matters. You know many examples of this. The essential mathematical consideration of the MHP is, that the probability for the host to open his door is twice as big if the contestant has chosen a goat (p=1) as if he has chosen the car (p=1/2). (This simple fact is hidden in the section Bayes' theorem.)

You may also consider the following two cases, where the contestant chooses door c, the host opens door h, and the other remaining door is s:

1. Car behind c and host opens h: p1 = 1/3 * 1/2 = 1/6

2. Car behind s and host opens h: p2 = 1/3 * 1 = 1/3

So the probability to win by switching is p = (1/3) / (1/3 + 1/6) = 2/3

The person who does not know the rules of the game or the history simply doesn't know the chances.--Albtal (talk) 19:43, 2 August 2012 (UTC)

JeramieHicks, that's the shortfall, the still lasting deficit of that article. Read above what I wrote at The Monty Hall Paradox or The Monty Hall "Problem" ?.

The striking point is the well defined scenario, the scenario that this famous paradox is based on. The symmetry of this scenario secures that the door first selected by the guest (one out of three) has and retains exactly a 1/3 chance to win the car, while the set of those two unchosen doors (with a chance of 1/3 each), as "an entity", has and retains a unite 2/3 chance to win. In spite you know for sure that – because there is only one car – that those two doors on average hide 1 1/3 goats, so are to hide "one goat at least", their unite chance to win the car is and retains 2/3. Given by the symmetry of the scenario (if two goats are behind the host's doors, then he will choose one "uniformly at random"!), these chances (1/3 resp. 2/3) cannot change, even after the host has opened one of his two doors showing a goat. Whatever door the host opens showing a goat,
be it #2 or #3, the door first selected by the guest retains its 1/3 chance, and so the host's still closed second door retains unite 2/3 chance.

Now imagine that someone, who comes in after the host has opened one door, not knowing about the scenario (rules) nor which door has been chosen by the guest. He has a 1/2 chance to pick the guest's door [say #1] and a 1/2 chance to pick the still closed host's door [be it say #2 or #3]. He does not know that [say] door #1 has a 1/3 chance only, but the still closed host's door #2 resp. #3 has a 2/3 chance. If you are a frequentist, you could argue that he should observe that only in 1/3 of cases he will win by picking #1, while in 2/3 of cases he will win by picking the still closed host's door #2 resp. #3. But as he is unable to distinguish between those doors (not knowing that [say] door #1 is the door first selected by the guest), that does not help him in any way. His chance to win by randomly picking one of the two still closed doors will exactly be only 1/2 therefore,
as (1/3 x 1/2) + (2/3 x 1/2) = 1/2.

And, as Marilyn vS said: Imagine there is a million doors with only one car and else just goats. The guest selects one door with a 1/millionths chance. Then the host, disposing of 999'999 doors, shows 999'998 goats, leaving only one of his doors closed. What would you think of the chance to win by switching to the host's only one still closed door?

But it is the still lasting deficit of the article to fob the readers, leaving them alone. Gerhardvalentin (talk) 00:35, 3 August 2012 (UTC)

Slight bias

I currently believe that there is a bias on the page that the 2/3-1/3 probability solution is the right one. However, let it be noted that Wikipedia is an UNBIASED encyclopedia. Therefore, a new section, with the reasoning for the other argument, should be created, and any bias hints found in the first should be edited out.

Thanks,

Newellington (talk) 14:06, 4 August 2012 (UTC)

Under the rules of the fully unambiguous, mathematically explicit version of the standard problem (Krauss and Wang description from this page as well as Henze, 1997 and many others), the 1/3 - 2/3 probability solution is the correct one, indeed. Sources that show probability solutions of differing values were addressing quite different scenarios (see above The Monty Hall Paradox or The Monty Hall "Problem" ?). You are right, those differing scenarios, used in lessons on probability theory, using also a host who is not bound to maintain secrecy regarding the car hiding door are used to schematically practice conditional probability theory (Henze, 1997). Yes, those quite other conceivable but diametrically opposed versions on some unknown asymmetry, completely outside the scenario of the standard version, have to be treated in a quite separate section, but never in the main section that is based on the standard problem. Gerhardvalentin (talk) 17:14, 4 August 2012 (UTC)

WP is based on what is said in reliable sources and they all agree that, given the usual assumptions, the probability of winning by switching is 2/3.

The problem with this article is that, rather than concentrating on explaining that fact, it is drawn into academic sideshows. Martin Hogbin (talk) 17:18, 4 August 2012 (UTC)

Ten Years And A Million Words

Here is what we have "accomplished" so far:

STATISTICS:
1,269,228 talk page words

Total Edits: 14,463 edits (10,610 talk page, 3,853 article)

Average: 120 words per talk page edit.

Individual Edit Counts:
Rick Block         = 2,304 edits (1,799 talk, 505  article) 
Martin Hogbin      = 2,058 edits (1,981 talk,  77  article)
Blocked User G.    = 1,428 edits (1,363 talk,  65  article)
Gill110951         =   790 edits   (602 talk, 188  article)
Nijdam             =   672 edits   (602 talk,  70  article)
Gerhardvalentin    =   498 edits   (447 talk,  51  article)
Heptalogos         =   398 edits   (398 talk,   0* article)
Glopk              =   366 edits   (234 talk, 132  article)
Kmhkmh             =   352 edits   (352 talk,   0* article)
Guy Macon          =   304 edits   (304 talk,   0* article)
Blocked IP Sock R. =   182 edits   (182 talk,   0* article)
Dicklyon           =   164 edits   (123 talk,  41  article)
Father Goose       =   143 edits   (107 talk,  36  article)
Glrx               =   135 edits    (70 talk,  65  article)
Ningauble          =   106 edits   (106 talk,   0* article)

Notes:

Article created in Feb 2002.

"Talk page" is defined as Talk:Monty Hall problem plus Talk:Monty Hall problem/Arguments. It does not include any discussions at noticeboards or on user talk pages.

"0*" means "16 or lower" (limitation of counting tool).

Word count was done by cutting and pasting the text from all the archives into one text document and using UltraEdit to get a word count. If there are duplicates in the archives, someone wanting to read them all would still have to slog through well over a million words. Other statistics came from WikiChecker. I welcome anyone who wants to do their own count.

By comparison:

Harry Potter and the Philosopher's Stone  =  76,944 words
Harry Potter and the Chamber of Secrets   =  85,141 words
Harry Potter and the Prisoner of Azkaban  = 107,253 words
Harry Potter and the Goblet of Fire       = 190,637 words
Harry Potter and the Order of the Phoenix = 257,045 words
Harry Potter and the Half-Blood Prince    = 168,923 words
Harry Potter and the Deathly Hallows      = 198,227 words
All Harry Potter books                  = 1,084,170 words

One thing that struck me while skimming through the archives was that again and again someone would claim that all the issues are settled or that they soon would be. You can respond to this with Yet Another Claim that this will be solved Real Soon Now, but I won't believe you.

I propose the following solution. Yes, this is a serious proposal, and yes, I do realize that it cannot be decided here.

[A]: Apply a one year topic ban on everyone who has made over 100 edits, (which includes me) with it made clear that we are banning everyone and that this does not imply any wrongdoing.

[B]: Reduce the article to a stub.

[C]: Let a new set of editors expand it.

If anyone tries to invoke a policy that would not allow this, invoke WP:IAR. Nothing else has worked. I don't believe that anything else is going to work. Our best mediators, arbcom members etc. have utterly failed to solve this problem. My proposal will solve it.

(Puts on asbestos suit) Comments? --Guy Macon (talk) 21:51, 4 August 2012 (UTC)