Talk:Monty Hall problem

Please note: The conclusions of this article have been confirmed by experiment

There is no need to argue the factual accuracy of the conclusions in this article. The fact that switching improves your probability of winning is mathematically sound and has been confirmed numerous times by experiment. If you find the article's arguments unconvincing, then please feel free to use the space below to discuss improvements.

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Very important topic

Template:Technical (expert)

The below suggested simple "frequency" solution should replace the current confusing conditional probability solution
See discussion further down for reasons why it should replace the current solution.

Summary and Solution

There are three doors. Two doors have a goat behind them and one door has a car behind.
(Remember goat is bad and car is good)

1) Choose one door
2) Game host opens one door containing a goat
3) Do you change door?

For example:

Door 1	Door 2	Door 3
Goat	Goat	Car

Now you have two opions: Not switch or Switch

Not switch

pick	show	outcome
1	2	lose
2	1	lose
3	2 or 1	win

If you do not switch the probability of winning is 1/3

Switch

pick	show	outcome
1	2	win
2	1	win
3	2 or 1	lose

If you switch the probability of winning is 2/3

You should therefore choose to switch doors

---

Note that there also exist an simpler and more elegant solution to the above problem.

From above we know that the probability of winning if we do not switch doors is 1/3.

We also know that we only have two options: Not switch or Switch

This means that:

P(switch) + P(not switch) = 1

which means that

P(switch) = 1 - P(not switch)= 1 - 1/3 = 2/3

Which is the same solution we had previously

---

tildes ( 82.39.51.194 (talk) 10:40, 17 August 2008 (UTC) )

Repetitive new solution section

I've undone this change twice. It renames the existing "Solution" section as "Discussion" and adds an additional "Solution" section that basically repeats what's in the existing Solution section - without references, and with a variety of style issues per WP:MOS (such as directly addressing the reader). If there's some shortcoming in the current Solution section please say what it is and we can work on addressing the concern. -- Rick Block (talk) 18:57, 14 August 2008 (UTC)

I've undone it myself too, and it's back. Can we perhaps lock the article until this user responds in the talk page?The Glopk (talk) 14:13, 15 August 2008 (UTC)

--

The problem I have with your so called solution section is that it is general confussing! You are confussed, your glossy matrix and tree are confussed. It is not about where the car is !! The solution will look the same irregardles. It is a question whether you choose to switch or not and which door the contestant choose.
The whole point of the Montey hall excercice is to answer the question wheter you switch or not and still you refuse to take that into account !

Therefor you should divide your matrix it into

Switch
Not switch

and then review

1)all the possible doors the contestant can choose

2)the response of the host

3)the outcome of such a choice

Your original matrix might be glossy and fancy but regret to inform you that it is inaccurate

It is better if you put your glossy matrix under the section "Sources of confusion"

I can also inform you that I am going to change back my solution until you either, stop removing my solution or redesign your matrix.

Ps I found it quite strange that I have to fight for such an obvious point ! It just reaffirm the notion that wikipedia is good in theory but does not work in practice. Now you have to fight a war with every moron with a computer ! Where are the credentials, expertise and self critic ? Just because you have a computer doesn mean that you are an expert !
I have attached the original post
tildes ( 82.39.51.194 (talk) 10:40, 17 August 2008 (UTC) )

---

Discussion

—Preceding unsigned comment added by 82.39.51.194 (talk) 08:35, 16 August 2008 (UTC)

The existing "solution" section is clearer than the above explanation. Cretog8 (talk) 15:04, 16 August 2008 (UTC)

The solution proposed above keeps the arrangement of goats and car the same and varies the player's initial pick rather than keeping the player's initial pick the same and varying the location of the car. Both of these approaches simplify the actual 9 cases (or 18 assuming the goats can be distinguished) down to three to better show the 2/3 probability of winning. In the rest of the article (in particular, in the Bayesian analysis section) the player's pick is kept as door 1 rather than the alternative approach of examining all picks given a single arrangement of goats and cars - and this approach is the most commonly presented in the cited references (and matches the presentation of the problem published in the vos Savant Parade column).

Another reason this approach is used is that the problem as generally presented can be considered a conditional probability problem given both a specific initial pick and a specific door the host has opened. This interpretation of the problem (see the Morgan et al reference) is discussed in the last two paragraphs of the solution section and involves examining the situation after the player's initial pick and given which door the host opens. Initially using the approach where the goat/car configuration is constant but the player's pick varies makes the connection to this conditional analysis extremely difficult to see.

The solution section was actually changed fairly recently to use the "fixed pick, varying car location" approach (rather than the "fixed configuration, varying pick" approach) to be consistent with the majority of the references and with the rest of the article. We can certainly discuss whether including the alternative solution is worthwhile, however given that they are effectively equivalent (both relying on a "without loss of generality" assumption), and that most sources use the "fixed pick, varying car location" approach, and that the approach currently used easily extends to the conditional analysis, and that the article is quite long already, I don't think you'll find much support for including this alternate solution. -- Rick Block (talk) 19:14, 16 August 2008 (UTC)

---

I dont agree with what you are saying ! It is about presenting an easy to understand and approachable article. as of now none of this is taking place. Who gives a crap about "keeping the player's initial pick the same and varying the location of the car". That is not what the Monty Hall problem is about. The sooner you will realize this the better of this article will be! You need to keep the most critical stuff and just remove everything ells ! Howevere I am starting to lose interest in all this bull crap ! This is all about you ! You have written something that you protect like hamster. This is exacly the reason why wikipedia will never gain any credential in academic circles because it is not about the optimal soluton but rather individual ego! —Preceding unsigned comment added by 82.39.51.194 (talk) 19:53, 16 August 2008 (UTC)

Please 82.39.51.194, no personal attacks. Feel free to discuss your displeasure and/or disagreement with the article but refrain from attacking the editors. -hydnjo talk 20:41, 16 August 2008 (UTC)

yes, I agree no personal attacks! It is just that I get so frustrated when the solution is staring them in the face and they still refuse to accept it because they are biased towards their own writing (which apparently is a lot). Just because someone wrote something first doesn't mean that such a person has the right to "validate" (oohh daddy please can I) or delete everything that comes after. You seam to have a lot of rules! Dont you have a rule for that? —Preceding unsigned comment added by 82.39.51.194 (talk) 21:26, 16 August 2008 (UTC)

There is indeed a rule for that: WP:OWN says that nobody should behave as if they "own" a particular article. I've put a welcome message at your talk page which you can look over to find out more stuff. All the same, you have to have some humility, too, and accept that maybe the reason your suggestions aren't being implemented is because several people don't think they're an improvement. Cretog8 (talk) 22:47, 16 August 2008 (UTC)

As a Featured (July 23, 2005) and high-traffic (392 thousand views, Jan-Jun '08) and heavily edited (500 edits to date in '08) article, there are lots of inputs for further improvement. The article has been scrutinized by the community at large in a process called Featured article review and the criticisms from that review have been successfully addressed so that the article meets today's FA standards. Given that bit of perspective, the article is absolutely not frozen and the editors that who are watching continue to to demonstrate (IMO) an open mind when it comes to addressing suggestions for improvement. It may be helpful to scan the archives of this talk page to gain further perspective about the responses to your suggestions. -hydnjo talk 23:26, 16 August 2008 (UTC)

Moreover, the current version attempts to stay as close as possible to the academic references. Have you read the references that are in the article and do you have other references that present the solution using your preferred approach? Even if you're not going to respond to the explanation above about why the current approach is used, a suggestion that you find "so and so's explanation from this paper or book" easier to understand would be received quite differently from "I like my explanation better". -- Rick Block (talk) 23:33, 16 August 2008 (UTC)

With all due respect I am not convinced that you (any of you) fully understand the rules of the Monty Hall Game.

1) First the allocation of goats and the car is taking place (remains fixed for the rest of the game).

2) Then the contestant choose one door.

Your "fixed pick, varying car location" approach totaly contradict this line of reasoning. The location of the car IS NOT changing. The allocation is stationary, it does NOT change over the period of the game. Again the first thing that happens is the allocation of the car. Again that location remaine FIXED for the ENTIRE game. But still you in insist like some stubborn child that the location is varying. It is NOT ! and it never will be ! The Monty Hall game is a SEQUENTIAL GAME which means that you CAN NOT just switch around the location of the car based upon your preferences in a later stage of the game. This means that the ONLY valid way of approaching this is the way I have put forward which means that you keep the allocation fixed and you evaluate each pick individually (the pick, the game show respons, and the outcome).

Also the argument that "majority of the references" do it one way isn't valid either. This is not an excercise in burping up some solution from some jerk off text book. Fine! you should have references to other articles but in the end it is about evaluating which approach is the most correct and not the least which approach is the easiest to understand. I hate to burst you bubble but the current approach dosent fulfill neither of these criterias irregardles if you have won a nobel price for the article.

Further, it is not about Bayesian statistics ! To allow Bayesian statistics to dictate how the original problem and its solution is presented is WRONG. Especially when such a solution contradict(is incorrect)the original game and the sequential steps such a game is based upon. If you want to have a section of Bayesian statistics that is fine! But you need to very carefully point out the different assumption such an analysis is based on! This is not taking place at the moment! The way I see it is that you have all been so blinded by the complexity of the Bayesian analysis that you have completely surrendered to its preachings! --82.39.51.194 (talk) 10:37, 17 August 2008 (UTC)

Perhaps it is not clear to you what is the meaning of a clause like "Assume, without loss of generality, X.". To put it simply, it means: "I could repeat the same argument for cases W,Y,Z,etc., But since these are obviously equivalent to X, as they differ only by a change of names, I'll just write down the case for X and be done with it". For example, consider the proof of the Pythagorean theorem: the logic of the proof would be unchanged if you rotated the names of the triangle's vertexes so that A becomes B, B becomes C, and C becomes A. That is, the particular labeling of the vertexes is irrelevant to the proof. Similarly, in the analysis of the Monty Hall problem, the naming of the doors is unimportant, so long as it is kept consistent within the reasoning. Because of this, for example, the Bayesian analysis of the problem is written concisely using the "Assume, without loss of generality" clause.The Glopk (talk) 16:48, 17 August 2008 (UTC)

(to 82.39.51.194) First, please read Wikipedia:No original research. Taken to its extreme what this policy says is that even if the preponderance of references are provably incorrect (which, just to be clear, is not at all the case here), the Wikipedia article must summarize what they say rather than presenting something we make up ourselves. Second, although you're perfectly correct about the sequencing of the problem we're talking about the probability of winning by switching, either over all iterations of the game or (in the Morgan et al interpretation) given a specific player's initial pick and the host's specific response. Also note that the Parade description of the problem specifically includes the words:

You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?"

If the goal is to determine the overall chance of winning by switching, over all possible scenarios, in a strict sense we should enumerate all possible car locations, player picks, and host responses (per the fully expanded decision tree as presented in Grinstead and Snell). On the other hand, from the player's point of view, the player (who doesn't know where the car is) picks a door and then the host opens a door, so following through what happens when the player picks a specific door (say #1) over all possible car locations (the approach currently in the article and in most references) is entirely sufficient. This is not varying the location of the car after the player has picked, but examining all possibilities of where the car might be to determine the probability of winning. Indeed, given the Parade version of the problem your preferred approach makes almost no sense at all (why are we considering what happens if the player picks door 2 when it's stated the player has picked door 1?). Morgan et al take this one step further, and consider only the scenario where the player has picked door 1 and the host has opened door 3 (eliminating the possibility that the car is behind door 3! - and, even in only this one case, the probability of winning by switching is still 2/3 [if the host chooses which of two goat doors to open with equal probability]).

I suspect you're thinking the problem is asking about the probability of winning over all possible car locations, player picks, and host responses. What would your analysis be if the problem is about the chances of winning for a player who's literally initially picked door #1 followed by the host opening door #3? -- Rick Block (talk) 17:38, 17 August 2008 (UTC)

You dont seem to understand! The most critical question we should answer is NOT what happens if the contestant chooses one specific door and the the host opens another door! The problem is more general than that ! The whole point with the Monty Hall exercise is to answer the question whether or not the contestant should switch doors in GENERAL. This is also indicated in your quote

"You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?"

The word "say" that appear before the door numbers indicate to me that this is just an example. The individual case of door 1 and door 3 is not that important! It could have been any configuration! Again the important question is whether or not the contestant should switch doors in GENERAL. Also note the sequential nature of your quote. 1)first the allocation has taken place (which remain fixed for the rest of the game 2)Then the contestant choose one door. When you start to shuffle around the location of the car it CONTRADICT the sequential nature of the game. It also obscure the whole purpose of the Monty Hall exercise which is to prove that the contestant will benefit from switching doors in ANY situation not just "for a player who's literally initially picked door #1 followed by the host opening door #3? "--82.39.51.194 (talk) 08:39, 18 August 2008 (UTC)

My points are:

1. If the goal is to answer the general question then all goat/car configurations, all initial picks, and all host responses should be considered. Your approach is considering only one configuration using an assumption that the same logic pertains to all other configurations. The approach currently in the article considers only one initial pick and uses an assumption that the same logic applies to all other picks. Your assertion seems to be that the latter approach is wrong. Either one of these is valid (they are like looking at two sides of the same coin), although most references use the approach currently in the article. If you're not seeing that these approaches are fundamentally equivalent, I think it is you who is not understanding the problem. In addition the current approach is (IMO) more consistent with the form of the Parade problem statement which uses the "say No. 1" terminology (which encourages thinking through scenarios involving a given initial pick).
2. Some references (e.g. Morgan et al) explicitly consider the situation at the point the player has picked a door and after the host has opened a door. Your approach makes this analysis difficult, however the current approach makes this a relatively simple extension of the analysis.

The approach currently in the article is used because most references use it and this approach easily extends to cover the "conditional" (Morgan et al) interpretation of the problem. Even though you think the critical question is the general question, the conditional interpretation exists in the literature (the Morgan et al and Gillman references) so it is considered in the article. I'll ask again - how would you address this interpretation? -- Rick Block (talk) 19:15, 18 August 2008 (UTC)

---

First of all the current solution is NOT more consistent with the form of the Parade problem statement. Actually my proposed solution is MORE consistent with the Parade problem! The reason for that is that it is more logically consistent (sequential game nature which means that the location of the car remain fixed), more general (consider a larger amount of cases) and easier to understand (iteration nerver suns out of styl).

Secondly, How I would deal with the conditional interpretation? For me that is an interpretation that is not critical for the Monty Hall problem nor its solution. You can and should understand the Monty Hall problem and its solution only with the basic frequency statistics approach (without Bayesian statistics). Note that my proposed solution uses frequency which is the most consistent with the original formulation of the Mony Hall game since again

1) It DOES NOT contradict the logic and sequential nature of the original Monty Hall game. The location of the car (and the goats for that matter) remain fixed for the entire game.
The logic is intact
2) Is not based upon any prior assumtions about the door selection. what you see it what you get.
3) Does not start in the middle of the game (door selecton), again sequential nature
3) The formulation is much more general (consider a larger amount of cases)
4) Much easier to extend to alternative configurations for example D1=C,D2=G,D3=G or D1=G,D2=C,D3=C or D1=G,D2=G,D3=C

As I said previously if you want to include a section on conditional probabilities as an extra bonus you have to very carefully point out the different assumption such an analysis is based on. If you dont correctly point out such differences the Bayesian section will actualy contradict the original sequential reasong of the Monty Hall game.

1) Firstly you you need to explain that in order for a conditional probability approach to work we have to take the steep from frequency statistics to Bayesian statistics. This means that we have to depart form the sequential nature of the problem where we first have the allocation of the car and goats and then we we have the door selection. Now instead we are going to assume that the player already has done his door selection. In order to evaluate such a decision in the middle of the game we have to vary the location of the car. So basically we are inverting the origial order of the game. 1) door selection 2) allocation

2)Secondly you need to explain the reason for that). You need point out that Bayesian statistics in the form of conditional probabilities are based upon the assumtion of serial correlation (normal distribution with fat tails) which means that the observations are not independent. For example

WHOA THERE!!! The above paragraph is pure nonsense. The Bayesian formulation of probability theory, based on Cox's axioms, is completely general and has been shown to be equivalent to any other standard formulation (e.g. Kolmogorov's). See the references ited in that section, e.g., E.T. Jaynes's "Probability Theory as Logic". Please do not make absurd statements to (try to) make a point. Also, please stop talking about "Bayesian statistics". There is not statistics involved in the Monty Hall problem: rather, it is purely a problem of probability theory.The Glopk (talk) 01:25, 21 August 2008 (UTC)

P(A∩B) is the probability of A and B happening
P(B) is the probability of B happening
P(A|B) is the probability of A happening given that event B has happened

Bayesian expression: P(A∩B)=P(B)*P(A|B)

When we have serial independence ( A and B are independent) then the P(A|B) expression is reduced to P(A)

Which means that our Bayesian expression is reduced to the frequency expression

P(A∩B)=P(B)*P(A)

These two explanations are important because it helps the reader to understand the some what contradicting and confussing set up! Here I am open for suggestions! Any solid and easy to understand explanations that can simplify the transition is most welcomed !

I now feel that I have put forward a solid argument why the current solution is not optimal. If you still fail to take my points into consideration I suggest that we seek outside help on the matter since otherwise we will continue this discussion for all eternity. To tell you the truth I have more productive things to do that arguing with you about these things!
--82.39.51.194 (talk) 10:50, 19 August 2008 (UTC)

I will not go into the details of what you wrote here; others can do that much better than I can.

First, I want to applaud your change of attitude instead of changing the article time and time again, you are now discussing the matter.

However, you should note that you seem to be the only one arguing this point. That doesn't in itself mean you are wrong, of course; but you should appreciate the fact that many editors, some of them very well versed in this theory, have already been over this article in great detail. That in turn means you should consider the possibility that you are in fact mistaken. In your first posts here you seemed to take the stance that you were inquestionably correct, but as I said before, you have changed your attitude a bit, instead putting forward arguments to support your opinion. Kudos.

One point that you haven't responded to yet is the tenet of no original research. The references cited in the article support the problem statement we currently use, and so far you have proposed no references to support your version. Oliphaunt (talk) 11:57, 19 August 2008 (UTC)

You are right I have to considered the possibility that I might be wrong! I have done that and the answer was that I am not! It is not that much to wrong about! Further, the best references is the original Monty Hall game and its corresponding rules! I can probably dig up a reference on the frequency statistics approach (which I assume is my suggested approach) but I am not sure it is necessary. It is like asking for a reference for why 2+2=4. It is a simple exercise on calculating probabilities! If you find that difficult then you should probably not take on the conditional probability approach which dominates the current article ! --82.39.51.194 (talk) 13:31, 19 August 2008 (UTC)

What you're not right about is your assertion that the current approach contradicts the sequence of the game. It absolutely follows the sequence, from the player's perspective. First two goats and a car are placed behind 3 closed doors, but the player doesn't know what is behind each door. In your solution, the configuration is given. Why doesn't the player just pick door 3 and stick with their initial choice? The reason of course is that the player does not know the configuration. In the version currently in the article, the configuration is left unknown (matching the situation from the player's point of view). The configuration is certainly fixed before the player picks a door, but the player doesn't know the configuration. What the player does know is the door he or she initially picks. The current solution, most sources, and the Parade problem description implicitly say (and the Bayesian analysis section explicitly says) "let's call the door the player picks door 1 (renumbering the doors if necessary)". The analysis proceeds given the unknown, but fixed, configuration of goats and car with the (now fixed) initial player choice. We examine all possible configurations of the goats and car not because we're moving them around after the player has picked (which would indeed violate the sequence of the game) but to enumerate all possible scenarios using the same frequency based approach your solution uses. This approach

1) exactly matches the player's view of the game. The location of the car is fixed, but unknown (in your solution, the car location is known but the player's pick is treated as variable - this is distinctly not the player's view and, literally shows only that switching wins with 2/3 probability if the car is behind door 3)

2) is not based on a specific arrangement of goats and car, which (again) matches the player's view of the game.

3) does not start in the middle of the game (if the current section is at all confusing in this regard, we could certainly work on clarifying it)

4) easily extends to the conditional analysis, without needing to switch to Bayesian logic (indeed, the current solution section includes the conditional analysis, without mentioning anything about Bayesian analysis)

5) considers all initial configurations (and, through renumbering of the doors, all initial picks and host responses)

6) is consistent with the view of the problem throughout the article (all the images and all the text consistently call the player's initial choice "door 1" and the door the host opens "door 3")

You're quite welcome to seek outside help, although I would suggest the best help would be sources. The existing references are (as far as I know) the best, most authoritative sources on the topic of the Monty Hall problem. The existing approach follows the solution typically presented in these sources. I've said this about 3 times already, but your solution and the solution currently in the article are essentially equivalent (both rely on a "without loss of generality" assumption). Given two essentially equivalent approaches, using the one that more closely matches the references, more closely matches the player's view of the game, and more easily extends to cover the conditional analysis seems like the obvious choice. -- Rick Block (talk) 16:20, 19 August 2008 (UTC)

yeahh yeahh what ever !--82.39.51.194 (talk) 17:29, 19 August 2008 (UTC)

I've clarified the sequencing in the Solution section. I hope this helps to address your concerns. -- Rick Block (talk) 02:54, 21 August 2008 (UTC)

I actually agree with 82.39.51.194 I also find the current article difficult to understand. I think a frequency statistic approach is easier to understand and more consistent with the original game.--92.41.172.75 (talk) 08:50, 21 August 2008 (UTC)

Suggestion to replace existing solution section

It appears 92.41.17.172 is suggesting that the current solution section be replaced with the contents from #Summary and Solution (above). My understanding was that per #Discussion (also above) we basically came to an agreement, albeit not very enthusiastic on the part of 82.39.51.194, about this. I won't repeat the discussion from above, but does anyone have anything more to say about this? -- Rick Block (talk) 18:30, 28 August 2008 (UTC)

The proposed change (diff) replaces the existing solution with text which is less detailed, has no supporting references, and which includes less appealing figures. The existing section should remain. TenOfAllTrades(talk) 18:56, 28 August 2008 (UTC)

I tend to agree with Rick, too, that the existing solution should remain. In my 35 years of teaching mathematics at a major university, including many, many probability courses, I've dealt with the Monty Hall problem many times in classes. What intrigues me most, though, is not so much the fact that the direct solution given seems to bother so many (perhaps because of the supposed veridical nature of the statement of the problem), but the fact that if one merely steps back and looks at the entire problem, there is a remarkably simple solution. The key to this solution is not to focus on switching but instead to focus on not switching. Choosing the strategy of not switching is tantamount to ignoring any and all extra information with the problem, and it is thus manifestly clear that:

P(not switching) = 1/3

and thus we have

P(switching) = 1 - P(not switching) = 2/3.

It uses nothing more than P(A) + P(not A) = 1. This is no more (and perhaps a lot less) counterintuitive than all the rest of the considerations which focus directly on choosing the switching option. I have always given perfect marks for this elegant solution and wonder why it is not mentioned in the article. Focussing on the switching strategy directly involves something like a probability tree or other construct (entailing conditional probability), fraught with potholes that can trap or fool the not-so-careful reader.

Some of the most elegant solutions to problems in mathematics arise by looking at such problems from a different perspective. This is just one such example. -- Chuck (talk) 19:29, 28 August 2008 (UTC)

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--92.41.17.172 (talk) 20:53, 28 August 2008 (UTC)

I don't mean to be rude, but have you read the article? The solution section says Players who choose to switch win if the car is behind either of the two unchosen doors rather than the one that was originally picked. In two cases with 1/3 probability switching wins, so the overall probability of winning by switching is 2/3 as shown in the diagram below. The diagram then shows (visually) what happens if you switch (Chuck's point is simply the inverse of this). The killer argument against this simplistic analysis (and Chuck's) is that it completely ignores the fact that the host opens a door. Why should this same argument apply both before and after the host opens a door? The initial pick is 1/3 (surely), but then the host opens a door. Why is this any different from Howie opening a case on Deal or No Deal (or would you say never take the deal because the chances of your case being the grand prize go up every time a case is opened)? I don't teach math at a major university, but this solution simply fails to address the problem (per Morgan et al.) and doesn't seem worth more than a B- (correct answer, completely inadequate reasoning). Consider three different hosts - host 1 who opens a randomly selected "goat door" if the player initially picks the car, host 2 who always opens the rightmost "goat door" if the player initially picks the car and host 3 who opens one of the unpicked doors randomly. With host 1 the correct answer is 1/3 chance of winning if you don't switch. With host 2, assuming the player stays with the initial pick (and it's door 1), the correct answer is 0 chance of winning if the host opens door 2 but 50% chance of winning if the host opens door 3. With host 3 the correct answer is 1/2 chance of winning (assuming the host didn't reveal the car). What this means is that the argument that the initial pick results in a 1/3 chance of winning is, um, not quite correct. If you don't also consider how the host selects what door to open the answer may (coincidentally) end up with the right answer (if it's host 1) but this is fundamentally a coincidence. Given any possible host behavior the average chance of winning if you don't switch is 1/3, but if you pay attention to which door the host opens (and how the host chooses to open doors) the chances of winning by staying are anywhere from 0 to 50% depending on which door the host opens. The point is the real solution has to include a consideration of how the host chooses which door to open. If it doesn't, then the answer applies regardless of which host we're talking about (host 1, host 2, or host 3) - i.e. you arrived at the right numerical answer but your reasoning is faulty. -- Rick Block (talk) 04:45, 29 August 2008 (UTC)

Sorry, Rick, but you are being rude (which is out of place here) and it seems you have been very much caught up in the veridical nature of the posed problem. Yes, I have indeed read the article, and it beats around the bush with various possible host actions (which is another matter that I'm not addressing) ... but of course I am observing the generally agreed on rule that says the host always opens a goat door different from the door chosen by the contestant. What you don't see is that there are two and only two options which under the host assumption exhaust all possibilities: (A) be obstinate (not switching no matter what is shown) or (not A) always switch when a door (different from the one selected by the contestant) showing a goat is opened. While (not A) has all the folderal about choosing to switch), option (A) does not (if you just think about it for a moment). Moreover, (A) union (not A) is the entire event space, so P(A) + P(not A) = 1. And since P(A) is manifestly eequal to 1/3, P(not A) = 2/3.

You complain that my solution ignores the fact that the host opens a door showing a goat. Note that it makes no difference in the outcome if the host opens either of the doors not initially selected. In the obstinate case (never switch no matter what is opened) the probability of winning the car is 1/3. If the contestant has initially chosen a goat door and her strategy is always to switch, then no matter which of the other two doors is opened, she will always with the car by switching to the opened door if it reveals a car or to the other closed door if it reveals a goat.

In either of these two scenarios, the problem you and many have is in realizaing that (always switching) and (never switching) exhaust all possibilities. That is precisely where the problem seems like a (veridical) paradox ... and that is why the solution I give is so elegant and, at first glance, seems that it must be wrong - even though it is completely correct.

I am reminded how, in teaching how to find areas of regions to students, we tell them that one way is to decompose the region into simpler regions and add up the areas of the pieces. Then we watch the solutions come in for a problem such as: find the area of the shaded region shown - which is nothing more than a large rectangle with a smaller rectangle removed - and students slavishly decompose the region into four rectangles, compute the area of each, and add them up, correctly of course - but completely missing the obvious solution to subtract the area of the smaller rectangle from that of the enclosing rectangle! Too bad people can't think outside the box or (in this case) inside the box.

People's dogged insistence that the result for never switching has something to do with opening a different door from the one selected is a bit like this, the only point to the problem is to see that since the host opens a different door from that chosen by the contestant, then the obstinate and always switch options are discjoint and do indeed exhaust all possibliities.

Oh well, some people look at a sphere and see perfection - others look at one and see a snowball and duck for cover. -- Chuck (talk) 14:11, 29 August 2008 (UTC)

Sorry for the rudeness. However, see below as well. -- Rick Block (talk) 15:26, 29 August 2008 (UTC)

ha, ha. I and many with me prefer Chuck's simple solution plus 82.39.51.194 "frequency" solution anyday over that lengthy and confused outburst by Rick Block!
There are no such thing as host-1,host-2 and host-3. The only host that exist is host-1 = Random selection of "goat door". Why would I consider an example that does not apply?
With all due respect there might be a reason for why you are not teaching math at a major university. --92.41.46.220 (talk) 09:06, 29 August 2008 (UTC)

Respect acknowledged, however your reasoning is still inadequate. Yes host 1 is our host, but there's nothing in the simple solution that makes use of this fact. Assume for the moment we're talking about host 3 (makes it more like Deal or No Deal), not host 1. The same simple argument can be used, but in this case this argument results in an incorrect solution. The flaw is the implied claim that because P(not switching) = 1/3 when the player initially selects a door it remains so after the host opens a door, i.e. that how the host chooses to open a door is "extra information" that can be ignored. Not so.

You have to at least explicitly say that P(not switching) doesn't change when the host (our host, host 1) opens a door because of the constraints on the host's behavior, but then the question becomes how do you know this? Asserting it to be true with no reasoning doesn't seem sufficient.

If you have access to it, please read the Morgan et al. paper. This paper distinguishes host 1 from host 2 which is a much more subtle distinction than the difference between host 1 and host 3. With host 2, the average chance of winning by not switching (ignoring which door the host opens) is 1/3 - just like host 1. However, with host 2 there are two distinct probabilities depending on whether the host opens the rightmost or leftmost door. If the host opens the rightmost door (which he does unless the car is behind it) the chance is 1/2. If the host opens the leftmost door (which he does only when the car is behind the rightmost door) the chance is 0 (of winning by not switching). You can in some sense say the probability of winning by not switching in this case is 1/3, however the probability for any given player (with knowledge of which door the host opens) is either 0 or 1/2.

So, back to the article. I believe the solution as stated is as simple as possible without being too simple. -- Rick Block (talk) 15:26, 29 August 2008 (UTC)

I also agree with 82.39.51.194 and 92.41.46.220. I think we have consensus now to change the solution section!
--Pello-500 (talk) 13:51, 29 August 2008 (UTC)

1. It appears that Chuck actually doesn't agree with your change;
2. It's incredibly obvious that Pello-500 is just a new account created by 92.41.

If you're not interested in engaging in serious discussion here, please stop editing the article. TenOfAllTrades(talk) 14:00, 29 August 2008 (UTC)

In addition to edit warring over the article, Pello/92.41/82.39 also erased my comment calling him on it. I've restored it above. Note that he had originally (and erroneously) claimed that Chuck had agreed with his position. TenOfAllTrades(talk) 21:43, 29 August 2008 (UTC)

RfC: New or old solution section?

The new suggested "frequency" + Chuck's solution is easier to understand. Current editors refuse to accept this due to ownership. They won't let anyone edit the article !

Initial thought - if I understand correctly the question is not about what the actual solution is, that is agreed, but it is about the best way to explain it. From my experience different people have different way of looking at things. When I am trying to get to grips with a probability problem I usually find that one particular way of looking at it helps me to get it. The may not be the way that helps other people. My suggestion, therefore, is to have both solutions. Maybe the second could be added under the heading 'Another way to understand the solution'. Martin Hogbin (talk) 17:56, 29 August 2008 (UTC)

Disagree, please leave the old solution. The new proposed "frequency" solution is poorly written and referenced. Chuck's solution is just the "naive Bayes" one (uniform prior on the host behavior). As remarked by Rich Block, the "old" solution section, i.e. the one that passed the F.A. review, offers a more comprehensive treatment that is supported by the refrences and does cover more sophisticated host behaviors.The Glopk (talk) 00:29, 30 August 2008 (UTC)

My suggestion as to what should be done depends on what people believe that the purpose of this article is. For example, is the purpose purely to consider the problem academically and with some degree of rigor or is there a secondary function of explaining to as many readers as possible what the solution is and why in a way that makes sense to them. Martin Hogbin (talk) 08:53, 30 August 2008 (UTC)

---

Bayes was right, but I think the logic of the existing solution is flawed

The Bayesian analysis is correct to the extent that it demonstrates that the probability of the car being behind Door 1 remains 1/3 after the decisions are made by the player and the host. That is the "decisions" by the player and the host do not affect the probability of the car being behind Door 1.

However, I believe the interpretation of the Bayesian analysis is flawed.

In particular, although the Bayesian analysis correctly accounts for the fact that the host knows the location of the car it does not reflect knowledge gained by the player when it is revealed that the car is not behind Door 3.

Bayes's Theorem can be used to take account of the fact that the player learns the car is not behind door 3 by calculating the probability of the car being behind Door 1 given that it is not behind Door 3.

Let's do it in English.

The probability that the car is not behind Door 1 given that the car is not behind Door 3

= [(The probability that the car is not behind Door 3 given it is behind Door 1) x (The probability that the car is behind Door 1)] / (The probability that the car is not behind Door 3)

= [1 x 1/3] / 2/3

= 1/2

This result is consistent with many people's intuition that if there are two doors and a car is behind one of them then there is a 50:50 chance of the car being behind either door.

That is, I think conclusions shown on the website are wrong.

Given the debate that has already taken place I guess there will be some opposition to my view. --Paul Gerrard at APT (talk) 11:51, 30 August 2008 (UTC)

---

Paul, I would be happy to try and persuade you that you are wrong but this is not the place. You can email me at martin003@hogbin.org. Martin Hogbin (talk) 14:02, 30 August 2008 (UTC)

The basic issue is that "the probability that the car is not behind Door 1 given that it is not behind Door 3" (which is indeed 1/2) is NOT the question posed by the Monty Hall problem. In English, what we're looking for (flipping "is not behind Door 1" to "is behind Door 1" - seems a little clearer this way) is "the probability that the car is behind Door 1 given that the host opens Door 3". With the standard rules, this probability ("host opens Door 3") is 1/2 given the car is behind Door 1 (from the problem statement: If both remaining doors have goats behind them, he chooses one randomly). This probability is also 1/2 given the car is not behind Door 1, since if the car is not behind Door 1 it's equally likely to be behind Door 2 or Door 3. Combined, what this means is the probability the host opens Door 3 is 1/2 whether or not we're also given the car is behind Door 1. This makes the Bayes expansion be (1/2) x (1/3) / (1/2), i.e. 1/3. -- Rick Block (talk) 16:28, 30 August 2008 (UTC)