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, being bound to maintain absolute secrecy regarding the car hiding door.
   
   

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, as per the unambiguous and mathematically fully correct scenario of Krauss&Wang, Henze and others.

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)

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)

In response to Gerhardvalentin, I know that this version is correct. Everyone knows that. However, what I am saying is that a section on the wrong explanation should be created, without saying that it is wrong, unless adequate sources cannot be cited for this method. You're right that a standard problem does indeed exist, but we still should display the unstandard problem.

Newellington (talk) 00:58, 5 August 2012 (UTC)

In response to Martin Hogbin, you are absolutely right. The article repeats the explanation of the 2/3-1/3 paradox, using a selection of various charts that all say the same thing. Thank you for your added complaint in this section. This page really needs to be improved.

Newellington (talk) 14:08, 5 August 2012 (UTC)

Every citable source on the Monty Hall problem mentions the naïve 50:50 answer and says that it is wrong. It doesn't even qualify as a fringe theory: it is just a mistake. The whole point of the puzzle is that the common sense first impression is wrong. ~ Ningauble (talk) 18:36, 5 August 2012 (UTC)

Newellington, just to be clear, is your complaint that the answer might not be 2/3 or that the article fails to explain well why 2/3 is the correct answer?— Preceding unsigned comment added by Martin Hogbin (talk • contribs) 21:34, 5 August 2012 (UTC)

Neither of the above. My complaint is that I believe there isn't enough coverage of the 1/2-1/2 theory, and why it is or isn't wrong. I agree that 2/3 is the correct answer, but I don't agree that the 1/2-1/2 theory should be ignored completely. It should be given coverage in the article, regardless of its accuracy.

Newellington (talk) 17:44, 7 August 2012 (UTC)

There used to be a section explaining why the answer is not 1/2 (see, for example, [3]). I haven't chased down exactly when this went away - but is this sort of what you're talking about? -- Rick Block (talk) 18:52, 7 August 2012 (UTC)

Newellington, the problem is that this article is a mess. It does not clearly distinguish between the world famous "PARADOX" where there are two still closed doors, one of them is the door originally selected by the guest, and the second one is the still closed host's door that he just has offered as an alternative, to switch on. You know that one of these two doors for sure contains a goat (the second goat!), while the other door for sure contains the prize. No one knows which door hides the prize and which door hides the goat. So most people say the chances are 50:50. But in the only correct scenario of the world famous "PARADOX" the chances for staying:switching are exactly 1/3:2/3. As said, this is the world famous "PARADOX".

The article should clearly treat this paradox, making it intelligible and then, in later sections, should show quite other scenarios, where the "PARADOX" cannot exist at all. See what I wrote above: The Monty Hall Paradox or The Monty Hall "Problem" ?.

There are some scenarios, outside the "PARADOX", where the chances are 50:50 indeed. For example in a scenario where the host does not know the location of the car. In 1/3 of cases, if he has two goats to show (chances staying:switching 1:0), this is no problem. And in another 1/3 of cases , when he disposes of a goat and the car or vice-versa (chances staying:switching 0:1), and he just by chance opens the door hiding the goat, this is no problem, too. But in the rest of 1/3 of cases, disposing of a goat and a car (chances staying:switching 0:1), he inevitably will unfortunately open the door with the car, and in this 1/3 of cases was eliminating the guest's chance to win. By eliminating 1/3 of the chances to win reduces the probability to win by switching from 1/3:2/3 to exactly 1/2:1/2. And there are other variants where the score can be 50:50. For example if the host is extremely biased to open just the door with the smaller number. He can do that if he got two goats (chances of staying:switching 1:0), and he can do that if he got the car and a goat, given the goat is behind his preferred door (chances staying:switching 0:1). So even here, whenever he opens his preferred door, the chances to win by switching are 1/2:1/2.

The article should clearly show what the clean "PARADOX" really is, with it's unambiguous mathematically fully correct symmetric scenario, and then, in later sections, the article should very clearly show what NEVER CAN BE the famous paradox at all: The paradox cannot exist in deviant scenarios. The article is a mess in not explicitly and clearly distinguishing between the only correct and clear scenario representing the paradox, and other scenarios where the paradox never existed and never can exist. Gerhardvalentin (talk) 20:30, 7 August 2012 (UTC)

In response to Rick Block, I understand now. That is what I was talking about. In response to Gerhardvalentin, I understand that the article is a mess. That's why I came here. Thanks for the info, my issue can be considered resolved. Thanks.

Newellington (talk) 23:04, 7 August 2012 (UTC)

Newellington, you are welcome to help making the article on the world famous PARADOX what it should be: no more misty and opaque dissimulation, but clear and intelligible for the readers. Gerhardvalentin (talk) 10:04, 8 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)

What exact problem would this be the fix for? -- Rick Block (talk) 00:16, 5 August 2012 (UTC)

Ten years and a million talk page words without any agreement or any indication that you will reach an agreement if we give you another ten years. Article quality declining as different editors thrash trying to shape it into conflicting visions of what they think it should be. That's a significant problem. --Guy Macon (talk) 11:58, 5 August 2012 (UTC)

Most 'good ideas' aren't and this one certainly isn't. You efforts to help are appreciated but this is not the way, do you have any other suggestions Guy? Martin Hogbin (talk) 12:47, 5 August 2012 (UTC)

I have thought and thought, and I really can not see anything else that will prevent this going on for twenty years and two million words without any resolution. Unless, of course, one of you is 97 years old, in which case nature will resolve the dispute...

At the very least, can we all agree that the above numbers, demonstrate a real problem? --Guy Macon (talk) 13:31, 5 August 2012 (UTC)

New guy to the scene chiming in here. Surely the article has improved from 10 years aago, hasn't it? One could make the same argument against the Wikipedia project as a whole. --RacerX¹¹ Talk to me^{Stalk me} 14:11, 5 August 2012 (UTC)

If you go to the top of this page and click on the "Show" next to "Article milestones", I think you will agree that any of the three points where it was decided that it was good enough to be a featured article are better than it is now. Especially interesting are some of the arguments made when it was demoted:

"There are major, long-term disagreements among a number of editors on how to present the topic [which] have been going on for years [...] those disagreements have negatively impacted the end product: the article itself."

and

"I agree that the article appears a bit like patchwork (as a result of the 'eternal disagreements'. Imho the best solution would be if all old editors voluntarily withdraw from the article (other than commenting) and a few new qualified editors attempt a complete overhaul."

If my proposal comes to pass, there is nothing to stop you or any other editor who has not been involved in this fiasco from looking at old versions and cutting and pasting anything you find to be an improvement into the article. Basically I am advocating that you (Racerx11) and other uninvolved editors be given a free rein to improve the article as you see fit while I and a handful of others with over 100 edits be ordered to not interfere in any way. And even though I know nothing about you, I am pretty sure that you can do a better job. --Guy Macon (talk) 17:27, 5 August 2012 (UTC)

These statistics are sadly amusing, but I am not laughing. This is the second time Guy has proposed a blanket ban (cf. "The Final Solution", 9 July 2012), and this time he is including me by name. Since he says he is serious about banning me, I must ask a couple serious questions:

1. Is there any precedent for this sort of blanket ban affecting persons not found to be culpable?
2. Conversely, is there any precedent for Arbcom providing injunctive relief from indiscriminate sanction proceedings?

It might be wise to reinforce the asbestos suit with Kevlar. ~ Ningauble (talk) 18:10, 5 August 2012 (UTC)

No precedent at all. Nothing even close has ever been tried. I would be shocked if my proposal was accepted as written. Plus, of course, it can't happen based on a proposal here; it would almost certainly have to be an Arbcom decision. BTW, I pulled "100 edits" out of a hat. You could use a cutoff of 200, 500, or 1,000, any of which would exclude you. I sure would not want to try to make a case that out of those million words and 14,463 edits your 106 edits are a significant part of the problem. --Guy Macon (talk) 18:23, 5 August 2012 (UTC)

Do counts by themselves indicate a problem? What are the similar counts for, say, Intelligent Design (a featured article about a controversial topic)? I agree the article's quality has declined, but I can't see how this particular "solution" addresses this issue. -- Rick Block (talk) 18:35, 5 August 2012 (UTC)

That's an excellent question. I will do a count at Intelligent Design later today and post the results here so we can compare. Good thinking! --Guy Macon (talk) 19:07, 5 August 2012 (UTC)

Got the statistics. Comparing with Intelligent design was very good thinking on Rick's part.

Statistics for Intelligent design and Talk:Intelligent design

2002 - 2012 -- ten years, just like MHP.

Total Edits: 37.084 edits (24,551 talk page, 12,533 article)
MHP has      14,463 edits (10,610 talk page,  3,853 article)

Top Five Individual Edit Counts:
K             = 4192 (3339 talk page,  853 article)
FeloniousMonk = 2479 (1317 talk page, 1162 article)
Dave souza    = 1558 (1236 talk page,  322 article)
Jim62sch      = 1317 (1140 talk page,  177 article)
Hrafn         = 1023  (883 talk page,  140 article)
...
(MHP has counts of 2,304, 2,058, 1,428, 790 and 672)

ID has 68 archive pages and archives every 15 days
(MHP has 28 and archives every 14 days)

I am not going to count every word, but 5 archives chosen at 
random had counts of 56,996, 25,588, 30,119, 9,743 and 28,712.
30,000 x 68 = 2,040,000, so roughly two million talk page words.
(MHP has 1.3 million talk page words)

Conclusion: by most metrics, Intelligent design has about twice the volume of MHP. --Guy Macon (talk) 12:27, 6 August 2012 (UTC)

A more meaningful statistic

Surely this is a more meaningful statistic. Anyone who doubts my summary is welcome to check the facts for themselves to ensure that I have not missed anyone out or misrepresented anyone. Martin Hogbin (talk) 21:42, 5 August 2012 (UTC)

An alternative

In the discussion between myself and Rick Block in Sunray's user space, Rick and I agreed on the following wording for an RfC.

The aim of this RfC is to resolve a longstanding and ongoing conflict involving many editors concerning the relative importance and prominence within the article of the 'simple' and the more complex 'conditional' solutions to the problem. The 'simple' solutions do not consider which specific door the host opens to reveal a goat (see examples [23] [24]). The 'conditional' solutions use conditional probability to solve the problem in the case that the host has opened a specific door to reveal a goat (see examples [25]).

One group of editors considers that the 'simple' solutions are perfectly correct and easier to understand and that the, more complex, 'conditional' solutions are an unimportant academic extension to the problem.

The other group believes that the 'simple' solutions are essentially incomplete or do not answer the question as posed and that the 'conditional' solutions are necessary to solve the problem. Both sides claim sources support their views.

That argument is unlikely to ever be resolved but two proposals have been made to resolve the dispute. Both proposals aim to give equal prominence and weight to the two types of solution.

One proposal is for the initial sections including 'Solution' and 'Aids to understanding' to be based exclusively on 'simple' solutions (with no disclaimers that they do not solve the right problem or are incomplete) then to follow that, for those interested, with a section at the same heading level giving a full and scholarly exposition of the 'conditional' solutions.

The other proposal is for the article to include in the initial 'Solution' section both one or more 'simple' solutions and an approachable 'conditional' solution (showing the conditional probability the car is behind Door 2 given the player picks Door 1 and the host opens Door 3 is 2/3) with neither presented as "more correct" than the other, and to include in some later section of the article a discussion of the criticism of the 'simple' solutions.

The discussion stalled on the proposed addition of introductory wording. If there is a consensus here, how about you start an RfC using this agreed wording with/without any additional introductory wording that you consider necessary. Martin Hogbin (talk) 21:59, 5 August 2012 (UTC)

I emailed medcom (3 days ago), asking for a volunteer to finalize the question and suggest a decision rule (after individually asking Sunray, Andrevan, AGK, and WGFinley). -- Rick Block (talk) 22:20, 5 August 2012 (UTC)

Would you be happy to leave the finalising completely up to the volunteer? Would you be happy for Guy to do this, if willing? Martin Hogbin (talk) 23:25, 5 August 2012 (UTC)

Yes. -- Rick Block (talk) 03:01, 6 August 2012 (UTC)

And you? Just FYI, Guy (WGFinley) has suggested we submit a mediation request to agree on the RFC. I think if we both agree to let a third party finalize what we have there's no need to do this. BTW, if you don't agree, I will be proceeding with an RFC on my own (with or without a 3rd party finalizing anything). -- Rick Block (talk) 05:35, 6 August 2012 (UTC)

I would be happy for Guy Macon to do it if he is willing.Martin Hogbin (talk) 18:18, 6 August 2012 (UTC)

I am willing to do anything that I can to help. As an aside (not in response to the comment above), given the fact that there is another editor in this thread with "Guy" in his name, it is probably best to call me "Guy Macon", at least until I achieve my goal of having the Wikipedia community award me the title of Dalek Supreme... --Guy Macon (talk) 21:57, 6 August 2012 (UTC)

Thanks Guy. Rick and I agree the wording above. Perhaps you could start an RfC with any additional wording you think necessary. If neither proposal is accepted than we can continue looking for other solutions. I think it would be worth making an effort to get the maximum number of users to respond to the RfC.

I think we both agreed to limit the number of words used by Rick and myself to support our proposals and respond to others. It also might be a good idea to ask other regulars here to (voluntarily) limit their contributions also. Martin Hogbin (talk) 08:22, 8 August 2012 (UTC)

Rather than trying to hammer out such details, I suggest that one or both of you hand me the text you prefer, and if you know it, a brief note about what you think the other one prefers. I will put it into RfC format with what I think are a good set of ground rules (subject to what is and isn't allowed in an RfC). I will run it by everyone who follows this talk page for comments and suggestions (but not approval or veto), tweak it as needed, and ask "is there one editor who is willing to be the submitter of this RfC?" If no, in the dustbin it goes. If yes, (only one editor needs to agree) I will submit the RfC and advertise it in the appropriate places.

So what happens if some other editor objects to the RfC? Nothing. It only takes one person to submit an RfC. I will, of course try very hard to write it so that both sides are willing to put their name on it as submitter, but agreement is not required. I really have have no preference as to the outcome and I have the same generally positive feelings about everyone involved, so I should be able to do a reasonable job of avoiding obvious bias. --Guy Macon (talk) 15:08, 8 August 2012 (UTC)

We've agreed to everything in the first show/hide box at http://en.wikipedia.org/wiki/User:Sunray/Discussion_of_Monty_Hall_RfC#Outside_comments_on_including_NPOV_in_the_question except the five words Martin objects highlighted in yellow. Discussion about a consensus rule, without agreement, is at http://en.wikipedia.org/wiki/User:Sunray/Discussion_of_Monty_Hall_RfC#Where_are_we.3F. -- Rick Block (talk) 15:21, 8 August 2012 (UTC)

As Rick confirms we have agreed almost everything and are happy for you to use your discretion and judgement to decide the rest. As you say, it does not require the agreement of anyone to have an RfC. If either proposal is accepted that will be a consensus decision. If neither proposal is accepted then anyone else is quite free to make their own proposal and have an RfC on it. Martin Hogbin (talk) 23:40, 8 August 2012 (UTC)

Edit war

I am perplexed by a recent sequence of reverts:

1. On July 21-23 Gerhardvalentin rewrote portions of the article.
2. On July 24 Nijdam reverted the changes with the comment "revert to by most editors approved version".
3. On July 26 & August 5 an IP sock attempted to restore the changes, and was reverted by Nijdam and Guy Macon.
4. On August 5 Martin Hogbin reverted to the change by Gerhardvalentin with the comment "These changes have not been discussed and there is certainly no consensus for them."

Note that the final revert (4) actually restored the recent changes (1). I agree with the two editors (2, 3) who attempted to revert these changes, and there does not appear to be a consensus for keeping them. I think the changes can be criticized for injecting awkward or inappropriate language, personal opinion, and polemical tone into the article.

Do we need to discuss the changes in detail, or can we get a quick consensus on keeping or removing them? ~ Ningauble (talk) 15:59, 5 August 2012 (UTC)

My revert had nothing to do with content. If a change needs to be made it needs to be made by a non-banned editor -- if a sockpuppet of a banned editor makes an edit, I don't even look at it before reverting it, at which point the real editors can continue editing the article as if the sockpuppet had never been there. --Guy Macon (talk) 17:03, 5 August 2012 (UTC)

To wit: My edit of July 21-23 was fully based on the sources that Kmhkmh had just mentioned. Nijdam reverted with an inapplicable comment. As to awkward or inappropriate language, please help to clean it. Tone should not be polemical (?). Please help the article to separate the clean effectual paradox, existing in a well defined unambiguous and mathematical correct scenario, from other scenarios that also should be discussed. But please help to get rid of the mingle-mangle state of the article. Stop to fob the readers. Gerhardvalentin (talk) 17:19, 5 August 2012 (UTC)

And as to the "polemical tone": It's what reliable sources say: "Beispielsweise könnte der Moderator für den Fall, dass er eine Wahlmöglichkeit zwischen zwei Ziegentüren besitzt, mit einer bestimmten Wahrscheinlichkeit q die Tür mit der kleineren Nummer wählen (s. Übungsaufgabe 15.3). Beispiele wie 15.3 finden sich häufig in Lehrbüchern zur Stochastik. Ihr einziger Zweck besteht darin, mit bedingten Wahrscheinlichkeit schematisch rechnen zu üben." It is not enough to dislike what modern reliable sources say, to keep them out of the article. Gerhardvalentin (talk) 18:30, 5 August 2012 (UTC)

I agree with Ningauble that these changes are mostly not for the better. -- Rick Block (talk) 17:56, 5 August 2012 (UTC)

Please regard WP:NPOV and WP:Technical. Once more: First of all, your POV of disliking what modern reliable academic sources explicitly and almost bluntly say and are commenting on this very matter, is not enough to permanently and intolerably trying to blanking them out from the article. And observe WP:Weight Gerhardvalentin (talk) 19:56, 5 August 2012 (UTC)

I reverted only for the reasons stated in my edit summary. Maybe I misread the changes. It looked to me as though Nijdam had removed the, long standing, 'Combining doors' solution from the article.

I also agree with Gerhardvalentin's comments above but accept that his changes could do with some rewriting by a native English speaker and some reorganisation. With the current impass however, collaborative editing is impossible. Martin Hogbin (talk) 20:35, 5 August 2012 (UTC)

Thanks. Please help to get the article what it is expected to be by actual reliable academic sources and their weight and, first of all, by Wikipedia policy and by the readers. Please help. Everyone is invited to forget about any unnecessary conflict, in favor of the article. Gerhardvalentin (talk) 21:02, 5 August 2012 (UTC)

Nijdam has reverted Gerhard's changes again, so it seems like we may want to discuss them. I think there are basically three.
1) Change all instances of "standard problem" to "symmetric problem", including changing the heading "Other simple solutions" to "Other simple solutions based on symmetry".
2) Add a paragraph in the "simple solution" section comparing a strategy of staying with a strategy of switching (sourced to Henze).
3) Move the paragraph about Falk's "equal probability" assumption from the "Sources of confusion" section to the "Variant" section, renaming this section "Variants of asymmetry, assuming a non-confidential host".
If we're going to discuss these, I think it may be helpful for someone not strongly vested in the topic (Guy Macon?) to help structure the discussion. Per the thread above, we may be on the verge of a related RFC, so another possibility is that this discussion is premature. In any event, I strongly suggest all edit warring stop immediately. -- Rick Block (talk) 05:58, 6 August 2012 (UTC)

To make clear what the text of the article actually is talking about, we should explicitly say that

1. the "standard version of the problem" is a "symmetric standard version of the problem", and that all "simple solutions", without exception, are clearly based on this "symmetric standard version", so on the standard version of symmetry.
2. Henze is explicitly basing on the symmetric standard version, and he shows it by Bayes, and then he shows it by "strategy" what the actual chances" staying : switching" are "1/3 : 2/3" in any actual case.
3. Falk is clearly excluding any difference of "before : after" if we do not have "knowledge" of some host's bias (otherwise we "would have learned nothing to revise the odds on the door first selected by the guest").
   As she clearly addresses "known host's bias", and not to confuse the readers, these words of Falk belong to the section "Variants of asymmetry by assuming a non-confidential host", yes.

We should go back to my proposal and then discuss these items one by one, making the article clearer for the reader, step by step. Gerhardvalentin (talk) 11:09, 6 August 2012 (UTC)

I'm lost. Nijdam (talk) 11:25, 6 August 2012 (UTC)

Rick, will you help to make the article on the world famous PARADOX clear and intelligible, by distinguishing the distinct PARADOX that only can exist in its world famous symmetric scenario, separating it from quite other scenarios where the PARADOX never existed nor ever can nor will exist (biased host has NOT to keep secrecy regarding the door hiding the car, etc. etc.)

The article - after Nijdam's revert - again is a mingle-mangle of PARADOX and NON-PARADOXICALS. Please let us discuss how the PARADOX can be separated and clearly distinguished from quite other scenarios, scenarios that the paradox has never been based on? I would like to go back to my last version, and we could discuss item per item. Gerhardvalentin (talk) 21:04, 7 August 2012 (UTC)

Gerhard - 3 editors (Nijdam, Ningauble, and I) have indicated a preference against at least the totality of your changes. Rather than "go back" to your last version I think it would be more appropriate to discuss them individually item per item leaving the article as is for now. Is my list above accurate? If so, how about taking them one by one (one at a time) creating a new section to discuss each. Note that if the RfC being discussed just above actually happens, it might be better to wait until it is resolved. -- Rick Block (talk) 15:30, 8 August 2012 (UTC)

Thank you Rick for your proposal. Your interpretation of my concern is correct, let me repeat my answer from above:

Separating the valid clean PARADOX from asymmetric scenarios

To make clear what the article actually is talking about, the reader should be informed that

1. the "standard version of the problem" is a fully symmetric standard version of the paradox, and that all "simple solutions", without exception, are clearly based on this standard version of full symmetry.
2. Henze, not being a mulish "frequentist", and others are "of course" explicitly basing the clean paradox on the symmetric standard version of a host who is bound to maintain absolute secrecy regarding the car hiding door. Show Henze's "strategy" that explains why the actual chances of "staying : switching" are exactly "1/3 : 2/3" in any actual case.
3. Falk is clearly excluding any difference of "before : after" if we do not have "knowledge" of some host's bias (she says "and you know about that bias"). In absence of a "known bias", "we've learned nothing to allow us to revise the odds" (on the door first selected by the guest).
   As she clearly addresses some "known host's bias", and not to confuse the readers, these words of Falk belong to the section "Variants of asymmetry by assuming a non-confidential host".

Which item do you prefer to begin the discussion? And as to your proposed RfC: I'm going to ask Guy Macon myself to help me with my own proposal to clearly distinguish from now on the clean valid paradox from any other asymmetric scenario, and to stop indistinctly mingle-mangle different sources of quite different vantage point in indistinctly hiding such discrepancy. I am asking you to help me, also. Regards, Gerhardvalentin (talk) 17:56, 8 August 2012 (UTC)

Gerhard this is exactly what my proposal on the impending RfC will do. Both sides will gain. The simple, notable, unintuitive probability puzzle will be first covered as exactly what it is, a simple, notable, unintuitive probability puzzle with no distractions. When that is done, we will be free to discuss the finer points of probability, problem variations, the 'right' and 'wrong' ways of solving the problem in a calm and scholarly manner based on what the sources have to say on the subject. Martin Hogbin (talk) 23:47, 8 August 2012 (UTC)

Yes, Martin. The world famous PARADOX only "exists" in – and consequently "is" based on its fully unambiguous and mathematically correct defined scenario. (Uni of San Diego calls it "senario"). This "paradox" never existed and never can nor will "exist" anywhere else.

And – after the paradox has completely and exhaustively been presented – the "core reason" for the deficient and faulty 50:50 intuition should be presented in the first section of showing "Other scenarios outside the paradox", of leaving the correct scenario. It is the "Monty does not know"-version, where in a full  "1 out of 3"  the chance to win by switching will be wasted and destructed and so the chance to win by switching can only be 1/2.

Only thereafter quite other unfounded and inapplicable scenarios of some assumed host who is NOT bound to keep secrecy regarding the car hiding door, outside the PARADOX, should be shown as examples used in classrooms of schematically training probability & statistics, based on relevant textbooks. Such clear structure of the article is indispensable to finally stop fobbing the readers. Will you sign my RfC, also? Gerhardvalentin (talk) 10:18, 11 August 2012 (UTC)

What is 'a fully symmetric standard version of the paradox'? Nijdam (talk) 07:48, 9 August 2012 (UTC)

And what is there meaning of 'the clean valid paradox'?Nijdam (talk) 07:55, 9 August 2012 (UTC)

The fully symmetric scenario is is what MvS said to have been the basics of her answer: the objects are supposed to have been placed uniformly at random before the show (as we have no better evidence), the guest is supposed to have selected her initial door uniformly at random (as we have no better evidence), and the host is obliged to maintain absolute secrecy regarding the car hiding door (as we have no evidence of defraud). Thats the fully symmetric standard version of th paradox, as per Krauss and Wang and others. And the paradox is depicted as follows: Just only two still closed doors, 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 still closed host's door that he just offered as an alternative, is not 1/2, but exactly 2/3. Henze refers to http://math.ucsd.edu/~crypto/Monty/monty.html, e.g. Gerhardvalentin (talk) 08:12, 9 August 2012 (UTC)

Well, you haven't got a lucky hand in explaining things. Let me ask it this way: What is the difference between 'the standard version of the paradox" and the 'fully symmetric standard version of the paradox'? And the question: What is the meaning of 'the clean valid paradox', still stands. Nijdam (talk) 15:20, 9 August 2012 (UTC)

I don't think so. Do you know, for example, what the department of mathematics of the University of California, San Diego says? Did you read Richard Gill? He once invited you on a cup of coffee, but you denied. Just read the actual relevant sources. Gerhardvalentin (talk) 16:17, 9 August 2012 (UTC)

Is this an answer to the above questions?? (BTW I never denied to drink coffee with Richard, don't write things you do not know about.)Nijdam (talk) 09:15, 10 August 2012 (UTC)

Yes. Gerhardvalentin (talk) 12:48, 10 August 2012 (UTC)

Comment

In don't really want to participate in the discussion, but since Gerhard Valentin refers to "my source", I'd like to point out, that his quote from the source may be somewhat misleading and out of context, as he ommits important context and actually combines lines from different sections without indicating it. The correct full quote can he found be found here: Henze p. 106. Note that the section from which the first part of the quote was taken comes with some introductory commentary on modeling and assumptions, which was omitted in the quote. Furthermore the final bold part actually stems from the next chapter.--Kmhkmh (talk) 07:21, 6 August 2012 (UTC)

Out of context? Full text: In chapter "7 Laplace Modelle", middle of page 50, Henze says:

7.5 Zwei Ziegen und ein Auto
In der Spielshow Let’s make a deal! befindet sich hinter einer von drei (rein zufällig
ausgewählten) Türen ein Auto, hinter den beiden anderen jeweils eine Ziege. Der
Kandidat wählt eine der Türen, etwa Tür 1, aus; diese bleibt aber vorerst verschlossen.
Der Spielleiter, der weiß, hinter welcher Tür das Auto steht, öffnet daraufhin eine der
beiden anderen Türen, z.B. Tür 3, und es zeigt sich eine Ziege. Der Kandidat kann nun
bei seiner ursprünglichen Wahl bleiben oder die andere verschlossene Tür (in unserem
Fall Nr. 2) wählen. Er erhält dann den Preis hinter der von ihm zuletzt gewählten Tür.
In der Kolumne Ask Marilyn des amerikanischen Wochenmagazins Parade erklärte
die Journalistin Marilyn vos Savant, dass ein Wechsel des Kandidaten zu Tür 2
dessen Chancen im Vergleich zum Festhalten an Tür 1 verdoppeln würde. Das war
die Geburtsstunde des sog. Ziegenproblems im Jahre 1991, denn die Antwort von Frau
Marilyn löste eine Flut von Leserbriefen mit gegenteiligen Meinungen aus. Es gab
Standhafte, die ihre ursprüngliche Wahl beibehalten wollten, Randomisierer, die sich
nur mittels eines Münzwurfes zwischen den verbleibenden Türen entscheiden mochten,
und Wechsler, die ihre ursprüngliche Wahl verwerfen wollten.
Etwa 90% der Zuschriften an Frau Marilyn spiegelten die Meinung wider, die Chancen
auf den Hauptgewinn hinter Tür 1 hätten sich durch den Hinweis des Moderators von 1
zu 2 auf 1 zu 1 erhöht, da jetzt zwei gleichwahrscheinliche Türen übrig geblieben seien.
Frau Marilyn blieb jedoch bei ihrer Empfehlung und führte die folgende Argumentation
ins Feld: Das Auto ist mit Wahrscheinlichkeit 1/3 hinter Tür 1 und folglich mit
Wahrscheinlichkeit 2/3 hinter einer der anderen Türen. Öffnet der Moderator eine dieser
beiden Türen, so steht die Tür fest, hinter welcher das Auto mit Wahrscheinlichkeit 2/3
verborgen ist. Folglich verdoppelt Wechseln die Chancen auf den Hauptgewinn.
Dieses Argument tritt noch klarer hervor, wenn man die Erfolgsaussichten der Strategien
eines Wechslers und eines Standhaften gegenüberstellt. Der Standhafte gewinnt dann
den Hauptgewinn, wenn sich dieser hinter der ursprünglich gewählten Tür befindet,

and he continues on page 51:

und die Wahrscheinlichkeit hierfür ist 1/3. Ein Wechsler hingegen gewinnt das Auto
dann, 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 Erfahrungen sammeln möchte, lasse sich unter der Internet-
Adresse http://math.ucsd.edu/~crypto/Monty/monty.html überraschen.

And then some examples for training purpose are following: "Trainingsaufgaben".

Page 99 begins with the headline "15 Bedingte Wahrscheinlichkeiten" - chapter 15 Conditional probabilities, and page 105 shows "15.9 Das Ziegenproblem". He says that the price has been placed uniformly at random, and the candidate chooses one door uniformly at random. He says that the host has no choice in case the guest should have chosen a goat. But if the guest should have chosen the prize, then we assume that the host choses one of his two goats uniformly at random. He shows by Bayes that the guest doubles his chance by switching. And then on page 106 he adds:

Die oben erfolgte Modellierung soll die Situation des Kandidaten vor dessen Wahlmöglichkeit
so objektiv wie möglich wiedergeben. Man mache sich klar, dass ohne konkrete
Annahmen wie z.B. die rein zufällige Auswahl der zu öffnenden Ziegentür im Falle
einer Übereinstimmung von Autotür und Wahl des Kandidaten eine Anwendung der
Bayes-Formel nicht möglich ist. Natürlich sind Verfeinerungen des Modells denkbar.
Beispielsweise könnte der Moderator für den Fall, dass er eine Wahlmöglichkeit zwischen
zwei Ziegentüren besitzt, mit einer bestimmten Wahrscheinlichkeit q die Tür mit der
kleineren Nummer wählen (s. Übungsaufgabe 15.3).
15.10 Beispiel (Fortsetzung von Beispiel 15.3)
Beispiele wie 15.3 finden sich häufig in Lehrbüchern zur Stochastik. Ihr einziger Zweck
besteht darin, mit bedingten Wahrscheinlichkeit schematisch rechnen zu üben.

and he adds some decision trees, evidently for training purpose.

English: "Examples like 15.3 are often found in educational textbooks on stochastic. "Their only purpose is to schematically training to calculate with conditional probability"

I repeat:

Bei allen diesen Betrachtungen ist natürlich entscheidend, dass der Moderator die Autotür geheimhalten muss, aber auch verpflichtet ist, eine Ziegentür zu öffnen (the confidential host has to keep secrecy regarding the door hiding the car) and

Ihr einziger Zweck besteht darin, mit bedingten Wahrscheinlichkeit schematisch rechnen zu üben.

Their only purpose is to schematically train calculations of conditional probability.

Not a matter of the paradox, but a matter of maths lessons.

So, not out of context. Read it. Henze means what he says regarding the paradox, and he says it bluntly clear. The article needs to factor this view of modern academic sources of Henze and many others. Rick, please help to shape the article accordingly. Gerhardvalentin (talk) 09:42, 6 August 2012 (UTC)

I only point to the fact that the referred example 15.3 is about two dice thrown out of sight. So, Gerhard, what is your point? Nijdam (talk) 10:55, 6 August 2012 (UTC)

Furthermore Henze quotes Vos Savant with her erroneous way of arguing. It seems he follows here there in her reasoning (the combined doors "solution"), making him a bad mathematician. Nijdam (talk) 11:06, 6 August 2012 (UTC)

TRUTH? What is your reputable source for this disqualification of Norbert Henze? You clearly highlight the dilemma of this messy article and of the never ending conflict. No respect to WP, no respect to the academic sources, no respect to the readers. Gerhardvalentin (talk) 11:30, 6 August 2012 (UTC)

@nijdam: Calling Henze "bad mathematician" because you don't like his approach is frankly nonsense. For all we know, he's most likely a better mathematician than you are.

Norbert Henze

@Gerhardvalentin: First I'm not sure why you are posting here long quotes (again in a slightly different way) which can been seen much better with the original sectioning in the link I've provided. And yes your original quote was somewhat out of context as it lacked the most important part being: "Die oben erfolgte Modellierung soll die Situation des Kandidaten vor dessen Wahlmöglichkeit so objektiv wie möglich wiedergeben. Man mache sich klar, dass ohne konkrete Annahmen wie z.B. die rein zufällige Auswahl der zu öffnenden Ziegentür im Falle einer Übereinstimmung von Autotür und Wahl des Kandidaten eine Anwendung der Bayes-Formel nicht möglich ist. Natürlich sind Verfeinerungen des Modells denkbar." (The modeling above should reflect the candidate's situation before his choice as objectively as possible. One needs to beware however, that without [additional] concrete assumptions as for instance the random choice of goat door in case of the candidate's [first] choice and the car door coinciding the application of the Bayes' formula won't be possible. Of course one could consider refinements of the model.) This came before your originally quoted text but without the bold print part. In essence he says, that additional assumptions are required and that introducing the parameter q is valid refinement. Now you've added the bold printed part from the next section, which talks about a complete different problem, that is example 15.3 (a dice problem). And the bold printed line that you've (falsely) added to your quote (Beispiele wie 15.3 finden sich häufig in Lehrbüchern zur Stochastik. Ihr einziger Zweck besteht darin, mit bedingten Wahrscheinlichkeit schematisch rechnen zu üben.) refers to the dice problem of example 15.3 (Beispiel 15.3). Note in the earlier section about the MHP in your quote Henze talks about exercise problem 15.3 (Übungaufgabe 15.3) and not example 15.3. So the bold line you've added to Henze's comment on the MHP is from another section and refers to another problem (the dice problem). I assume that aside from quoting somewhat sloppily you missed the difference between example 15.3 and exercise problem 15.3. This would not have happened if you had read whole section 15.10 rather than just its first few lines (context !) as then it becomes apparent to which problem he was referring in the first lines of that section. In any case you clearly misquoted Henze and by that creating claim, that Henze never stated. Henze never stated the MHP refinement with q (or the conditional treatment as such) have as their sole purpose the schematical training of calculating conditional probabilities, instead he stated that the dice problem (and alike) have as their sole purpose to schematically train the calculation of conditional probabilities. --Kmhkmh (talk) 12:35, 6 August 2012 (UTC)

You surely are right. Just to see what Henze says: Page 99 starts with chapter "15 Conditional probabilities", and at the beginning he shows some examples. In section

15.1 Example (5 lines) he mentiones an urn with 2 red, 2 black and 2 blue balls. After some balls have been taken out, he asks what is the probability that the two balls that had first been taken out were red?, continuing straight with section 15.2 (three lines). He mentions again the "Ziegenproblem" (goat problem) that was subject of chapter 7.5, and he says:

15.2 Example

In the situation of the "goat problem" (see section 7.5), door #1 had been pointed at, and the host has disclosed door #3 to show a goat. What is the probability that the guest wins the car by switching to door #2? and he straight continues

15.3 Example

In the neighboring room two dices are rolled ... and he again asks What is the probability that at least one dice shows a six?

And after those three examples of coloured balls, of three doors with two goats and a car, and of two dices, and after his three perspective questions "What is the probability that ...", he immediately continues, still in section 15.3, to treat these three examples a little closer. He outlines (still in the beginning of section 15.3 !) that all three examples have something in common. They have in common, that there exists a result (though unknown to us) of some stochastic event.

The whole section 15.3 is on these three examples: coloured balls, doors and dices. After the later section 15.9, treating again the MHP, in section 15.10 he refers again to section 15.3 (please remember what section 15.3 is about: It is the summary on the coloured balls, the three doors and on the dices, and Henze says that "such examples as in 15.3" (imo he evidently means conditional probability examples) can be found in textbooks for training conditional probability calulus. "Their sole purpose is to schematically train to calculate conditional probability." So I did not cite "out of context", but I said exactly what Henze says. Thank you for your efforts, your objections are helpful to clearing what the matter is. The clearer, the better. Kind regards, Gerhardvalentin (talk) 19:21, 6 August 2012 (UTC)

Yes, you did cite out of context for the reason explained above and by that (accidentally) constructed a completely fictious statement of Henze. When he is talking about the about the common feature of all 3 examples, he is talking about a typical feature of conditional probabilities, which is present is all 3 examples 15.1-15.3. But he is not talking about all 3 examples having just the purpose of schematically training conditional probabilities. That statement is just made about example 15.3, when it is detailed and extended in example 15.10. Note that Henze treats all 3 introductory examples 15.1-15.3 in greater detail in the later sections, but only for the treatment of example 15.3 he states that its sole purpose is to schematically train calculating conditional probabilities. Literally he states: "Beispiele wie 15.3 finden sich häufig in Lehrbüchern zur Stochastik. Ihr einziger Zweck besteht darin mit bedingenten Wahrscheinlichkeiten schematisch rechnen zu üben." (Examples likes 15.3 are often found in probability & statistics textbooks. Their sole purpose is to train the schematically computation of conditional probabilities.). Note he states examples like 15.3, he does not state examples like 15.1-15.3. Furthermore if you carefully go through detailed treatment of example example 15.3 in example 15.10, then you see, that he explains in detail, why he sees example 15.3 as (problematic) schematical training. This is specific to example 15.3 (and alike) and has nothing to do with example 15.1 or 15.2. There is absolutely nothing in Henze's text that suggests, that "examples like 15.3" is to be understood to include examples 15.1 and 15.2, on the contrary of the 3 given examples he considers 15.1 and 15.2 as valid and sees only 15.3 as problematic and constructed for the (sole) purpose of schematical training. So to be clear here, rather than "doing exactly what Henze says", you did not understand at all what Henze was saying. But be that as it may I'm not on a mission to ensure your proper understanding of Henze's book and I'm not interested in turning this into yet another neverending subthread on MHP. If you insist on reading something into it, which is not there, that's your business. Alright before I'm finally going to shut up for good one more piece of information. The German noun "Stochastik" can't really be translated with "stochastic(s)", the latter is more or less identical to the German adjective "stochastisch", which has a somewhat different meaning. The former is correctly translated as "probability [theory] & statistics" (see Henze p. 1)--Kmhkmh (talk) 20:47, 6 August 2012 (UTC)

Thank you, Kmhkmh. Gerhardvalentin (talk) 10:04, 8 August 2012 (UTC)