User:Martin Hogbin/Monty Hall History
History of the Monty hall problem article Archives 6 - 20
- 1 Conditionalists
- 2 Simplists
- 2.1 Registered
- 2.2 Anons
- 3 Compromisers
- 4 Not sure how to classify
- 5 Editors who were interested in getting a solution to the basic problem
- 6 Editors who were interested in finding out about the conditional solution
- 7 Interesting?
- 8 I am unsure
Better rubricate me as an "anti simple solutionist". I formulate here once more the "simple solution".
- As the car is placed randomly, the chosen door has 1/3 chance to hide the car; the opened door has chance 0, hence the remaining door must have chance 2/3.
Just explain to yourself the contradiction: As the car is placed randomly, every door has 1/3 chance to hide the car, also the opened and the remaining one, yet the opened one shows a goat. Nijdam (talk) 17:26, 30 December 2010 (UTC)
The same problem with the "combined doors solution".
- The chosen door No. 1 has chance 1/3 on the car. Hence the doors No. 2 and No. 3 together have 2/3 chance on the car. As the opened door No. 3 shows a goat, the remaining door No. 2 has chance 2/3 on the car.
Then he [Rick] argues and discusses in good faith with the person, very patiently and correctly, while enduring a verbal assault [from Glkanter]. --C S (talk) 03:23, 15 February 2009 (UTC)
The problem with this article is that the main point is cluttered by so many uninteresting generalizations, whereas the original game (where the host always opens a door other than the one picked by the player, which also does not have the car) is not analyzed rigorously. The most general stratgey of the host subject to the original rules of the game is characterized by a probability p, which is that probability with which the host opens door 2, given that the player chose 1 and the car is at 1. (There could be other probabilities for the cases of the player picking 2 and 3, but it suffices to conside one case.) To ignore the full analysis of the original problem is not wise, especially when the analysis reveals the surprising result that it is always optimal to switch. To emphasize the importance of this point, the player deos not need to know p in order to conclude that switching is optimal. This is true for every p. To assume p=1/2 is too restrictive! —Preceding unsigned comment added by 220.127.116.11 (talk) 04:04, 11 Fe
Many people think they understand why the player should switch, but then they give wrong explanations. Morgan et al. demonsrtate the latter with six such false solutions. In some of the false solutions all the statements are correct, but a complete solution is not established by the argument. The heart of the Monty Hall question is, when the player sees that a certain losing door has been opened, what are then the probabilities from the point of view of the player, that the car is behind either of the remaining doors. The simple solution to this question could be stated rightaway in the solution section without descending to lower levels. It would be a disservice to state a "solution" that lay persons "understand" whereas experts do not accept it as a correct solution. I do not think I wish to continue this fruitless discssion any more. 18.104.22.168 (talk) 01:52, 24 February 2008 (UTC)
What about this two-part approach:
- 1. Present the simple solution to the "main version" of the problem, with the least amount of fussing over the problem statement we can get away with to make it into a solution. Also present some other analysis methods of the problem, leading to the same solution. Following these presentations of the solution, briefly discuss unintended interpretations of the problem statement, and the difficulty in ruling them out unambiguously, while referring to a later section on variations.
- 2. Have (like now) a section with variations on the problem, such as those in which the player may know more than in the main version, depending on the host's behaviour.
The terminology "unconditional solution" is unfortunate. The notion of a "conditional" solution has no meaning for the main problem version. It is only in variations of the main version that conditional issues arise. --Lambiam 08:28, 2 March 2008 (UTC)
[In response to Rick Block] You agree to conditionally present things in an unconditional simplified way provided that we make sure we introduce all kinds of conditional complications first. I'm going to drop out of the discussion; my contributions have had zilch effect. --Lambiam 11:34, 5 March 2008 (UTC)
The fact that every unconditional explanation present in the article keeps getting butchered into a conditional explanation (or worse still, a pseudo-conditional one) is a source of anguish to me. I'm serious, it anguishes me. It results in us having an article that is of negative value to the overwhelming majority of our readers, as it fails to give them a nice, simple, accurate, easy-to-understand solution to the paradox in its most general form, and misleads them into thinking that that solution is for some reason an invalid way to analyze the paradox -- which it is not. It's an invalid way to analyze the question-as-usually-asked -- but the question-as-usually-asked is not the subject of the article -- the paradox is. The question-as-usually-asked is merely one of the "sources of confusion", which has led our explanation of the paradox to be a confused one as well.--Father Goose (talk) 22:06, 16 March 2009 (UTC)
Drop the textbooks for a second and think about how you would write this article to be useful to an audience of 10-year-olds. Then write that article, and add to it whatever is needed to make it useful to a statistician as well.--Father Goose (talk) 16:47, 16 March 2009 (UTC)
No, we are not really told that the player opened a specific door. In the Parade quote (which is an ambiguous formulation), it is "You pick a door, say No. 1." The number is an example used to establish a way to distinguish the doors in the description, without saying which has a car or goat. The wording we are supposed to be using makes that even clearer by actually describing the process and naming the doors only at the end, in an example of the process. Neither gives a right-to-left ordering of doors, so your example is useless. And even if it was a specific door, it only matters (to the player) if the player knows how the host favors the specific doors, which most definitely is not in either problem statement. And the reason you can't use the calculation you asked for, is that it assumes the host's strategy is known to the player. [Added: But to avoid any confusion here, and when reading Morgan, just assume that the player and the host have each assigned their own sets of numbers, at random, to the three doors. The two sets are independent of each other. That is the problem intended. JeffJor (talk) 16:03, 27 October 2009 (UTC)]
There is no absolute-correct probability for the two remaining doors, in either the conditional or unconditional problem. Because all of the random elements of the problem have already been determined at the time the decision is to be made. The car 'is behind one of the two doors. Any probability that can be assigned is a only measure of the information you assume is avaliable to calculate it, not a measure of that door's worth. If the contestant has only the information available to Marilyn's Little Green Woman, that probability is 1/2 (yes, I know that is less information than the problem says the contestant has). If the contestant has only the information that the problem says she has, the probability is 2/3 EVEN IF THE CONDITIONAL PROBLEM IS CONSIDERED. And if you are going to assume the contestant has more information than the problem says - i.e., that she knows the host has a bias measured by a parameter q - you might as well assume the contestant also knows everything that the host knows. Because only the host really knows how his q was determined on this particular day. Even if he is consistent with past games, and we assume the contestant knows all that history (which the problem does not allow), there could be unseen factors that influence the actual q so that on that day, it is not what this omniscient contestant thinks it is. The only virtue in considering a possible q, is that whatever its value is, it is never detrimental to switch. JeffJor (talk) 13:16, 19 November 2009 (UTC)
Finally, and you seem to keep ignoring this, the Parade version of the problem, and the K & W one this article is based on, do not describe the so-called "conditional" problem. Morgan misinterpreted it, and I have shown you exactly how their problem statement is different. The vast majority of peopel who see teh word puzzle do not even consider the "conditional" problem. But even in Morgan, they admit the "conditional" answer only applies when the host's strategy is known to fit within certain boundaries, that include knowing p23. Several times, Gillman points out that not knowing it makes the answers to his Game I and Game II equivalent. This is why they deserve to be a sidebar, at best, in the article. It simply is not "about the mathematical intricacies of the Monty Hall problem in its usual form." Their problem is just an interesting fact about one possible variation of what everybody sees the the usual form of the puzzle to be. JeffJor (talk) 19:25, 23 November 2009 (UTC)
That version still contaminates the simple solution section with the statement "Although the reasoning above is correct it doesn't answer the precise question posed by the problem, which is whether a player should switch after being shown a particular open door." It might be better to present a problem that this is the exactly correct solution to, and then treat the other problem where's it's not separately. A simple way to do this is to introduce the condition that the host choose randomly when there are two goats left, in a way that reveals no additional information to the contestant (since, as I pointed out, if he chooses random by flipping a coin only when he has to, and the contestant can see that, then it's a completely different ball game), so that the contestant's strategy of always switching is clearly going to lead to 2/3 chance of winning, without any conditionalizing on which door the host opens. In the diagram that Father Goose linked, I think one can rigorously show that "The player has an equal chance of initially selecting the car, Goat A, or Goat B. Switching results in a win 2/3 of the time." as it says in the caption, and that no strategy of maybe not switching based on which door the host opens can beat that. Dicklyon (talk) 02:25, 17 March 2009 (UTC 
The simple solution needs to be presented as a solution; if there are sources that comment on how "rigorous" it is, that can follow, but we can't say it's somehow less without saying who says so. Even saying it solves a different problem is the opinion of its critics, not of its adherents, typically, so needs to be presented as such. Dicklyon (talk) 15:24, 8 April 2009 (UTC)
So, lacking such other sources, we should obviously present the "usual" unconditional solution, with sources, and also the "conditional" solution and the fact of who says it's necessary and that the other is inadequate. Or does someone argue for more than this, or less than this? Dicklyon (talk) 23:28, 15 March 2009 (UTC)
The beauty of the simple unconditional pure MHP is that it is a true good mathematical paradox in the sense that everyone is first fooled but afterwards easily agrees what is the good answer (unless you are lawyers: in a survey at Nijmegen university, it was not possible to convince a lawyer that their initial answer was wrong). A paradox for talking about in the pub or at a party is not good if it requires tables of calculations and mathematical notation. The unconditional version is the version which you can solve by simulation, repeating over and over again (a splendid exercise, which forces one to fix each of the stages of the game). Gill110951 (talk) 04:43, 7 December 2009 (UTC)
The excerpted quote above is a perfect example of the muddlement of this article. I refer to the quote: "The overall probability of winning by switching is determined by the location of the car." Huhh??? What does this mean? Again, let me emphasize, I have no idea who it is that was being quoted, which quote I just repeated. I don't mean to attack anyone in particular. But I have to ask ... First, what is the difference between the "probability" and the "overall probability"? Second, what does the probability have to do with the location of the car? I don't get it. The location of the car has nothing to do with the probability. If the writer has a point to be made, then he/she should try to state the point non-cryptically. Sorry for this rant, but this kind of thing is very frustrating ... End of rant. Shalom. Worldrimroamer (talk) 04:26, 10 December 2009 (UTC)
Well, in-spite of my preference for the popular solution to be the main focus of the article I still feel that the conditional solution should be mentioned in the Variants section. The FARC version linked above seems to have nothing in it about the conditional solution at all. Also I like the current lead much better. Personally I think that simply moving the conditional section to the Variants heading would be the best compromise as that way no information will be lost from the article but the flow will still proceed from the most common understanding of the MHP towards more in-depth analysis. I would be more than willing to make the change myself as an editor who has only been part of this discussion for a short time. How does that sound to you guys? Colincbn (talk) 07:28, 10 December 2009 (UTC)
Yes, yes I have [read the sources]. And I am not suggesting we remove any content from the article. I am simply pointing out that the majority of academic sources, and this of course includes non-maths sources, use the simple/unconditional version of the problem and never bring up conditional probability. Therefore this simple version should be more prominent in the article. Conditional probability can come later, not to hide or bury it, but to put it in the most logical place considering what the most accepted form of the problem is. Besides there is no page on 1+1=2, or 1+2=3, the only reason there is an MHP is because of the psychology of why people get it wrong, therefore if anything the psychology should take the lead in defining the problem. Colincbn (talk) 04:09, 17 May 2010 (UTC)
Just from reading the present Wikipedia article, I agree with Martin Hogbin's suggestion, because I don't see why allowing the host to prefer one goat over the other is a more relevant generalization than allowing the host other behaviors. Melchoir (talk) 06:47, 3 December 2009 (UTC)
Clear and precise formulation! Thank you. – Makes it "to the point", finally showing the real problem: (delete this if you like) In exactly 2/3 of millions the player will win by switching, and only in 1/3 of millions she will win by staying. For most people it is hard to believe, let alone to understand. But a fact. No doubt. The famous 50:50 paradoxon. But no one can change this fact, neither Morgan. Some smart, but totally fabricated, say perverse interpretation occured: If the host gives information about the actual status (although he never can change the actual, current status!), e.g. having a preference for opening one particular non selected door, then he can meet his preference in 2/3 of millions only:
In 1/3 (when he has got two goats to choose from), i.e. when staying will win! And in another half of 2/3 (when he got the car and just one goat, and when the goat by chance is behind his preferred door), i.e. when switching will win!
In these 2/3 of millions, both switching and staying in average will have a chance of 1/2 each. But in 1/3 of millions the car will be behind his preferred door, and by exceptionally opening the "avoided and unusual" door he shows that switching will win anyway with a chance of 1. But the host never can change the current status, though. He just may indicate information about the actual constellation. Why not opening all three doors at once? But all of that is a dishonorable game, when secret facts about the actual constellation should be revealed. Just suitable for students in probability. Not really affecting the famous 50:50 paradoxon in any way.
This fact being emphasized in every university, and you can find it in multiple internet pages of universities all over the world. This should clearly be stated in the article, and Morgan et al. may not be allowed to confuse. Facts should be presented clearly in WP. Sources? Wherever you have a look. Regards, --Gerhardvalentin (talk) 22:58, 7 May 2010 (UTC)
If you find that difficult then you should probably not take on the conditional probability approach which dominates the current article ! --22.214.171.124
Chuck, my hat is off for you! I like your way of thinking. "P(not switching) = 1/3 and P(switching) = 1 - P(not switching) = 2/3. It uses nothing more than P(A) + P(not A) = 1" So simple but still so powerful! Only that well crafted sentence says more than the complete article in my opinion. I think that reasoning in combination with the new suggested "frequency" solution would greatly improve the article--126.96.36.199 (talk) 20:53, 28 August 2008 (UTC)
so the "we already have a solution" argument doesn't seem valid here. Android 93 (talk) 06:14, 6 September 2008 (UTC)
When Monty opens a door, he doesn't tell us anything we didn't already need to know. He ALWAYS shows a goat. It makes no difference to this puzzle WHICH remaining door he shows. 188.8.131.52
Glkanter I totalt agree with you the current article is Bull crap! 99% useless and 1% useful. It does not do a good job communicating a simple and straightforward unconditional solution! I totaly agree with you and I want you to change it!! --184.108.40.206
Exactly! Thank you! There is no reason why the Monty Hall problem should be conditional! A simple unconditional solution is the best solution! --220.127.116.11 (talk) 12:37, 14 February 2009 (UTC)
When you have realize that the best answer is a simple one then you are welcomed back.......--18.104.22.168 (talk) 20:56, 10 February 2009 (UTC)
Conditional probabilities are for morons! We know from optimal control that the closes way between two points is a straight line. The unconditional solution is located on that line. The conditional solution is not on that line. Occam's razor forever!! !--22.214.171.124 (talk) 22:09, 15 February 2009 (UTC)
I might have missed something, but surely it's not a difficult problem. Two times out of three, the host indicates to the contestant what door the car is behind by having no choice but to open the other door, therefore it's always worthwhile to swap, because two times out of three he will have chosen the correct door for you. It's more of a psychological trick or an illusion than any great mathematical conundrum. —Preceding unsigned comment added by 126.96.36.199 (talk) 23:57, 2 April 2009 (UTC)
I don't understand why the solution in the article has to be so long? It's quite simple really.
According to the traditional description of the game the host cannot show you the car if you picked a goat. Therefore, if you picked a goat the host will always eliminate the other goat. Thus when you switch you are basically betting that you picked a goat. The only time switching loses is when you picked the car in the first place. Thus when you don't switch you are basically betting that you picked the car. It's trivial to say that there's a 2/3 probability you pick a goat and only a 1/3 chance you pick the car. Thus, betting you picked a goat, and therefore switching, will win more often since you will more often, randomly choose a goat. 188.8.131.52 (talk) 19:57, 10 July 2009 (UTC)
Crikey, i didn't realise I had stumbled into a hotly debated topic. I am not a mathematician and wikipedia is not a site specifically for mathematicians so the feedback I would offer is to also give an explanation of the problem that does not require understanding of mathematical jargon. A mathematical explanation using correct scientific terms and equations should certainly be included, however, (imo) this tends to switch most people off - hence this suggestion for the dual forms of explanation. In my experience the general public tend to think in verbal-type concepts with everyday words rather than equations and official mathematical terminology.
Best of luck 184.108.40.206 (talk) 07:10, 25 March 2010 (UTC)
My advice is to start as if you're teaching your 5-year old the ABCs and then progress to more elegant, more mathematically technical explanations. 220.127.116.11 (talk) 10:00, 29 March 2010 (UTC)
My general impression of the article at the moment is that it makes the problem, in its (as I believe) generally understood form seem more complicated and elaborate than it actually is. The whole actual problem, as I understand it, really can be dealt with in a handful of paragraphs. Then there could maybe be a stronger indication to the reader that "the rest of the article is nit-picking for hard-core pedants". 18.104.22.168 (talk) 13:11, 27 May 2011 (UTC)
John Tierney in the New York times has an explanation - not sure if it's fancy enough to qualfy as a 'solution'.
When you stick with Door 1, you’ll win only if your original choice was correct, which happens only 1 in 3 times on average. If you switch, you’ll win whenever your original choice was wrong, which happens 2 out of 3 times.
Works for me ! Anyone else find that helpful ? --22.214.171.124 (talk) 17:31, 6 January 2012 (UTC)
There's no need for all these mathematical proofs. If you stick, you have a 1/3 chance of picking the prize. If you switch, however, you are trying NOT to pick the prize. By switching when you do not win the prize, you will win every time. This means a 2/3 chance of winning. — Preceding unsigned comment added by 126.96.36.199 (talk) 01:23, 21 February 2012 (UTC)
On point 5, unfortunately, there is no way to appease probability purists, at least not all of them, because there is more than one school of thought. Indeed, there is an unresolved (unresolvable?) problem in trying to define the meaning of probability and randomness in a fundamental and rigorous way: they tend to be circular. But that is a story for another topic. Seriously: it is a whole other article; it does not belong here. The only way to achieve point 2 is to agree on point 3, i.e. to adopt a notably conventional meaning for what is notably agreed to be unfortunate wording of the original problem, and give poor Marilyn a rest. 188.8.131.52 (talk) 23:41, 29 February 2008 (UTC)
This proposal, or something like it, meets the needs of both the casual reader who has heard of the problem and wants to know what the deal was, and the reader who is interested in a greater understanding of the issues.Simple314 (talk) 01:10, 9 February 2009 (UTC)
Not sure how to classify
The question is: should the requested chance be regarded as a conditional probability, and if so, why? Morgan et al. say it should, but their only argument or hint is that the information in the number of the door is otherwise not used. What information is in the number? Does it matter if the (blind) player has knowledge of that? What other information might be relevant? Should it be limited to events which restrict the possible outcomes? Even if it doesn't affect the outcome?
The article Conditional probability states that P(B|A) = P(B) if A and B are statistically independent. In other words: conditional and unconditional chances are the same if A does not change the probability of B. And it doesn't. Heptalogos (talk) 15:34, 10 February 2009 (UTC)
This is the simple solution in it's simplest form. Nothing new. Vos Savant did the same. Some people raised objections because they assumed that the problem is a conditional one. They forgot to explain why. They only said something about a numbered door being opened, maybe assuming that this number was already there before it was opened. This would reveal new information. But even then it would surely not affect the 1/3 chance of the picked door, so you are still right. But some people here insist that if part of the problem is conditional, the whole problem is. I am still waiting for the mathematical rule by which a problem is defined as conditional. I think I found it at Conditional probability where is stated that there should be statistical dependency, which means that the asked probability is affected, which is surely not the case here. But I don't get any answers to that. They just stopped explaning there. Heptalogos (talk) 14:00, 15 February 2009 (UTC)
[Referring to the claim that the simple solutions only give the correct answer by chance]
One probabilistic point that I observe is, (ir)relevance of the (un)conditional probability. I'd say that in this case they are equal not just by a numeric coincidence. Rather, the conditional probability (treated as another random variable) is constant (a degenerate random variable) in this case, due to an obvious symmetry. Taking into account the total probability formula we conclude that the conditional probability must be equal to the unconditional probability in this case. Thus I feel indifferent. Both are relevant in one sense or another. Do you agree? Boris Tsirelson (talk) 21:42, 10 February 2009 (UTC)
Yes. And here is an analogy. The (elementary) geometry of a circle is a special case of the (differential) geometry of an arbitrary curve. However it would be terrible to treat the circle (in an encyclopedia) only inside an article on curves in general. The symmetric case is more notable, more important, this is the point. And let me repeat: we surely have our POV about importance (rather than content). Boris Tsirelson (talk) 12:19, 4 December 2009 (UTC)
I don't think your understanding of the statement you quote assumes good faith. By its plain language, the statement you object to is a prediction that consensus for your change is not going to arise. Right or wrong as it may be, Rick is offering this prediction as a reasoned guess based on his experience with the editors of this article, and which kinds of edit ordinarily seem to achieve consensus. Do you have any particular reason to believe that this prediction actually masks an intent to forbid such a consensus from forming (assuming, arguendo, that this was within Rick's powers)? –Henning Makholm 05:41, 15 February 2009 (UTC)
OK. I have tries to edit the description accordingly. It is not particularly elegant, though -- feel free to simplify the language if you can. An interesting observation here is that (a) it is never a disadvantage to switch no matter which door and what p; (b) the "always switch" still wins with probability 2/3, independently of p; (c) therefore the assumption that Monty flips a fair coin if he has a choice is not necessary for the standard answer to be correct (though it does simplify the analysis). –Henning Makholm 06:13, 15 February 2009 (UTC)
Editors who were interested in getting a solution to the basic problem
Thanks guys, got it now. The falacy in my proposition is in the assertion that because there are two closed doors and the car is behind one of them, the second contestant has a 50:50 chance of finding it. This is only true from the second contestant's limited perspective. The conditions of the scenario is posited have in reality already shifted the odds to 2:1. Thanks again -- Timberframe (talk) 19:29, 26 October 2008 (UTC)
As I understand it, if a contestant chooses a door at random, the remaining doors will both have goats 1/3 of the time and a goat/car mix 2/3 of the time. 1/3 of the time the remaining doors both have goats, so by switching you would lose on this 1/3 of occasions. But on the other 2/3 of occasions the host will pick the door with a goat to avoid the door with the car. Since this is the case 2/3 of the time, by switching to the other door not chosen by the host you will succeed 2/3 of the time.
- This guy got it right. He says, my capitalization: IF A CONTESTANT CHOOSES A DOOR AT RANDOM!
- You HAVE to choose YOUR INITIAL door at random YOURSELF, in order to get the guarantee!!!! This guy is a born winner (ie instinctive game theorist who knows you have to randomize to optimize) Gill110951 (talk) 13:31, 15 August 2010 (UTC)
I'm sure someone will tell me if i've misunderstood the problem 184.108.40.206 (talk) 08:10, 22 March 2010 (UTC)
I was similarly confused when I first read this article, until I wrote a perl script to simulate the problem 1000 times, and was surprised to discover that switching really DOES win 2/3 of the time. Then I made the image I put up under "Not Switching" to help me wrap my head around it. Scenario 1 and scenario 2 ARE different scenarios, but they're the same choice, and the problem concerns the possible choices, not the possible scenarios; it's not important whether the host reveals goat A or goat B, what's important is that the host reveals "an option that is both a goat and not the player's choice" (as I hope is illustrated well in the image I added), thus scenarios 1 and 2 represent only one of the three possible choices. Luvcraft 01:18, 24 August 2007 (UTC)
What actually confuses people
I am just wondering if it is a flaw in one of the Variations à Other host behaviors à “The host does not know what lies behind the doors, and the player loses if the host reveals the car.”
The answer is allegedly “The player loses when the car is revealed a third of the time. If the prize is still hidden, switching wins the car half of the time.” Shouldn’t switching here also result in a 2/3 probability of victory (since it is purportedly the same result (so far) as in the original Monty Hall problem)? --220.127.116.11 14:15, 9 July 2007 (UTC)
I like the idea of having a diagram in the solution section of the article. However, with the current diagram it looks like if you replace the words "Host must reveal" with "Host luckily reveals" in parts 2 and 3, you would have an 'explanation' that switching wins 2/3 of the time, even if the host forgets what is behind the three doors but happens to luckily open a door with a goat.
Last year I brought up the version of the problem where Monty luckily reveals the goat when I noticed that the probability of winning upon switching does not increase in this version. This observation confused the heck out of me until I saw the diagrams in the "To Switch or Not To Switch" section on the following webpage.
The diagrams there are a little confusing at first but they definitely capture the difference in these two versions of the problem. Can we use the two diagrams in the article? I think they are useful for visualizing why the host's knowledge of what is behind each door matters. This alternate version is probably why many people don't feel sure about the solution to the standard version after they are shown it. Synesthetic
It's not that the problem is smart, it's that the presentation is misleading and easy to miss the obvious that the host is picking explicitly a goat door, purposely not opening a car door. If this is stressed, "scientists and nobelists" would not have such a problem. It's pretty certain this 'problem' is used by idiots that want to say "see, those smart people are idiots" when the only thing they do is not present the premises of the problem clearly or at least without any doubt. --18.104.22.168 (talk) 23:26, 10 May 2011 (UTC)
The article gives the impression that it's important that the host knew along where the car and goats are. Aren't the results the same whether he knew or not, since you've already eliminated the cases where he opens the 'car' door? Robin Johnson (talk) 14:07, 1 February 2008 (UTC)
Editors who were interested in finding out about the conditional solution
Leaving out the door numbers in the wording of the problem, essentially changes the problem and takes away not only its charm, but also changes it to a form it never was intended to. I.e. it no longer would be equivalent to the three prisoners problem, from which it seemes to be derived, but especially it no longer reflects what happens in the Monty Hall show. Nijdam (talk) 00:31, 17 February 2009 (UTC)
On second thoughts: why should the popular wrong "solution" be presented first? Beter to give an easy to understand right solution first! Nijdam (talk) 16:24, 29 March 2009 (UTC)
The fuss is that mathematicians say the problem clearly asks about a conditional probability, and that the answer "2/3" is correct only if the host is constrained to pick randomly between two goats (in the case the player picks the car). See the second question in this FAQ. -- Rick Block (talk) 13:47, 3 April 2009 (UTC)
I suggest we:
- 1. Archive this talk page.
- 2. Stop squabbling.
- 3. All admit the problem, however it's actually phrased, is not meant to have different conditional and unconditional answers.
- 4. All admit that the current statement of the problem indeed satisfies the previous point.
- 5. Tweak the wording of the solution section so that it appeases probability purists while avoiding explicitly rubbing the nose of the reader in conditional vs. unconditional probability considerations, i.e. change it so it has the form of a conditional solution but without making a big deal out of it.
- 6. Add a discussion of the conditional and unconditional consideration, referencing Morgan et al, to a later section of the article.
A conditional approach to the solution is certainly not wrong, and if the solution shows the probability of winning by switching is 2/3 for each player regardless of which door they pick and which door the host opens then the unconditional probability surely must be 2/3 as well.
Note the solution section needs to be accessible to a reasonably intelligent non-specialist, e.g. your grandmother (I realize there's a small chance your grandmother is a mathematician) or your 15 year old kid (no, not the goat kind).
Is this an acceptable approach? -- Rick Block (talk) 19:49, 29 February 2008 (UTC)
I am unsure
How to clarify the text is a good question. When you say that the intuition that "no change in information means no change in probabilities" causes problems, then you're right after a fashion. It causes problems because the intuition about what constitutes relevant information is not necessarily reliable, and you wouldn't be counting on it and thereby led astray if you didn't have the first (correct) intuition. Let's look at the notion of relevant information a bit more closely; you don't even need probabilities to see the problem here.
Clearly, any sensible notion of relevant information must at least include anything that would, under the rules of classical logic, allow you to deduce everything. For instance, if you picked the door number 1 (I'll assume this throughout for simplicity), and you were shown what is behind doors 2 and 3, then you would also know what is behind the door you have chosen. It wouldn't make any sense to insist that "no information" has been provided about what's behind door 1 in this case, simply because the the door remains closed, and you haven't actually seen what is behind it.
The same thing applies if you know that the host has joined the Cult of Even Numbers, and will choose to open door 2 whenever he can, because he would be committing dreadful sin if he were to slight an even number by preferring to open an odd-numbered door. Suppose that door 3 is opened. You know that there is no goat behind door 2, because if there were, your host would now be a very sinful man. You therefore know that there is a car behind door 2, and goats behind doors 1 and 3. In this situation you were provided with information about what's behind all the doors, even though only one was opened. The door the host chose was relevant information because of the extreme religious bias of the host.
In a probabilistic setting, the idea that "opening the door D = 2 or D = 3 to show a goat does not provide any information about whether the thing X behind door 1 is a car or a goat" makes sense, if you take it as the assumption that the mutual information I(X;D) = 0, or equivalently that P(X | D) = P(X), i.e. "knowing D does not tell you anything about X" (or "D and X are independent"). The assumption holds if and only if the host chooses uniformly at random among the possible choices. In any other case, the mutual information is nonzero, and the probabilities change in a way not consistent with the standard solution (like in the extreme case of the previous paragraph, where P(D = 2 | X = Car) = 1, which is as far from 1/2 as you can get, and D = 2 gets preferential treatment). -- Coffee2theorems (talk) 11:47, 26 May 2010 (UTC)
I think I've already made clear how I interpret Whitaker, so I'll just comment on the Seymann statement. I do not know if I agree or disagree, but it is one POV that problems that have been posed with a set wording may have a single "correct" answer despite apparent ambiguity. See e.g. Jaynes's solution to the Bertrand paradox: he insisted that the problem has a single correct solution, even though the conventional wisdom is that it is ambiguous and has at least three potential interpretations. -- Coffee2theorems (talk) 18:05, 29 May 2010 (UTC)