Talk:Monty Hall problem/Archive 8

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Very important topic

{{technical (expert)}}

{{technical}}

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.

Since these have had plenty of expert attention, I am going to disable the templates. — Carl (CBM · talk) 12:16, 16 September 2008 (UTC)

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) )

I propose a very compact, straightforward solution with no mention of probability per se. The problem is, in fact, simple so we should treat it that way. After reading most of this discussion and some of the references about how to explain it in an understandable way, I've yet to see the following type of explanation.

The problem is easy to solve - every time the game is played, one of two sequences occur:

A)Contestant picks a door that has the car behind it; host opens one of the other doors revealing a goat; behind the remaining door is the remaining goat.

B)Contestant picks a door that has a goat behind it; host opens the other door with a goat behind it; behind the remaining door is the car.

Sequence A happens 1 time in 3, so, doing the math, sequence B happens 2 times in 3.

Stickers win in sequence A and switchers win in sequence B.

I told you it was easy.

Millbast5 (talk) 22:03, 17 September 2008 (UTC)

Hang on a second.

Not switch

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

3 win? 3 is the goat, which is a known lose... Did I miss something? And this seems to presume knowing the outcome. If I pick 2, & switch, & 2 is the winning door, how, exactly, is this a win? TREKphiler ^{hit me ♠} 05:36, 22 October 2008 (UTC)

Yes you are missing something and that is your brain! :-) Door 1 & 2 are goats —Preceding unsigned comment added by 92.41.207.32 (talk) 10:42, 6 February 2009 (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)

Wait a minute. The expression P(A) + P(not A) = 1 refers to set complementation in a universal set - which has probability 1. And the not-switch strategy is not the complement of the set consisting of the switch strategy in the universe consisting of all possible strategies. For one thing, if there a two strategies there an infinite number of mixed strategies in that universe. While it is true that the probability of winning by switching plus the probability of winning by not switching is 1, it is something that you have to demonstrate by analyzing the problem. Its not hard but it sure doesn't follow from the tautology given. Just think, changing the name of the not-switch strategy to 'sticking strategy' sends the argument down the tubes.

In this particular game, the contestant always has the opportunity to switch so the switch strategy and the not-switch strategy always end up selecting two different doors and since the the other door is open and showing a goat, the car must be behind one of those two doors. Thus probability of winning by switching plus the probability of winning by not switching is 1. However in a variant of the game where the host did not always present an opportunity to switch then some of the time the switch strategy and the not switch strategy pick the same door, hence the sum of their probabilities of winning would be less than 1. Is that an example that shows P(A) + P(not A) = 1 is not true?Millbast5 (talk) 07:43, 18 September 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)

I also agree with 82.39.51.194 and 92.41.46.220. The frequency solution is most optimal.--84.92.246.30 (talk) 17:26, 30 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)

I don't know if it's necessary to repeat it, but in case someone doesn't read the section above, I'll note that the proposed replacement section is formatted poorly, lacks references, and contains less explanatory text (both quality and quantity) than the proposed replacement. TenOfAllTrades(talk) 17:48, 30 August 2008 (UTC)

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Can I make sure that I now understand what the current debate is about? Some editors want to keep the page as it is but recognise that there is a rather subtle problem with it. Others wish to replace it with a more rigorous approach which some feel has been badly presented and is not necessary. Martin Hogbin (talk) 09:39, 31 August 2008 (UTC) On a second look at things I can see that I got that wrong, I was confusing the current dispute with and earlier discussion that Rick pointed me to.Martin Hogbin (talk) 15:01, 31 August 2008 (UTC)

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Now that I have some idea of what is going on here, I can give my views on the current dispute. I do not think that the current solution should be replaced with the proposed one. However, I do think that it is more natural to have a fixed car position and vary the pick, but I also understand the reasons that the solution is the way that it is. I do have some criticisms of the article as it is and hopefully I will have some suggestions for improvements, but scrapping what has been done and replacing it with the proposed replacements not the way to improve in my opinion. Martin Hogbin (talk) 18:08, 31 August 2008 (UTC)

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I'm in favour of including both solutions. Because one method, to quote, "is formatted poorly, lacks references, and contains less explanatory text" is not grounds to remove it altogether, but to bring it up to standards. If we removed everything that was poorly written rather than improved them, Wikipedia would have about 20 articles. Besides, other topics like Zeno's paradoxes have entire articles dedicated to the various proposed solutions, so the "we already have a solution" argument doesn't seem valid here. Android 93 (talk) 06:14, 6 September 2008 (UTC)

As you will see I have made a similar suggestion here but any proposals for change get a very frosty reception. Have a look at the history of this page and you will see that an IP editor recently added an extra explanatory section only to have it immediately deleted, with no explanation and with the deletion being marked as a minor edit!

Another person tried to start a separate article only to find it set upon with a voracity I have never seen before in WP and deleted with days.

I have been trying hard to work with the existing/historical editors to find way to improve the page without compromising the integrity of the current article. So far my advice has been that I can make minor edits but if they are reverted I must immediately discuss them.

I Strongly believe that the current article fails to be convincing in respect of the basic paradox and that this must be improved. My view is supported by the fact that since the RFC there have been two people on the talk page who did not believe the article. One was eventually convinced by discussions here. As you say, there are other articles which make a better job of explaining their central paradoxes.Martin Hogbin (talk) 10:09, 6 September 2008 (UTC)

In an effort to allow this RFC to proceed unimpeded by "current editors" I have refrained from commenting, however I feel compelled to respond to this. I can see how editors new to this page might interpret responses to suggested changes as frosty, but in defense of the current editors this article is a featured article that has been through two featured article reviews [1] [2]. This certainly doesn't mean it's perfect, but it does mean there's been a significant amount of effort by multiple people spent on making the article not only mathematically correct with all major claims referenced to reliable sources but also comply with all other featured article criteria which include adherence to the WP:manual of style and a "professional standard" of writing. The recent addition by user:Technicalmayhem (a new, but not an IP, editor) was first deleted (by me) with this edit summary: revert - many WP:MOS issues with this addition. Technicalmayhem re-added it and this repeat addition was reverted by another editor also apparently watching the page. The re-addition was done with no explanation and marked as a minor edit (!), as well as the re-deletion. The separate article, "Introduction to the Monty Hall Problem", was created by user:Pello-500 (arguably the same user as 82.39.51.194 and 92.41.46.220) for the apparent purpose of including this user's preferred "solution" without gaining consensus to include it here. This is not how content disputes should be handled per Wikipedia:Dispute resolution, and since this user previously initiated this RFC he or she is clearly aware of the proper steps for handling content disputes but for whatever reason chose to ignore them. It's hard to interpret creating this article as anything other than bad faith tendentious editing.

Rick, I was not referring to your reversion but this one: 4 September 2008 Jonobennett m (48,223 bytes) (Reverted 1 edit by 68.2.55.158.)Martin Hogbin (talk) 20:24, 6 September 2008 (UTC)

From my point of view, the discussion below of Martin Hogbin's suggestions for improvement has been proceeding pretty much exactly how such discussions should go. I'm not, and I don't think anyone else is, saying no changes can be made. The caution about making a change once and if it's reverted then discussing it was not intended to inhibit changes but to encourage them (per WP:BOLD). I partially agree with the point that the article fails to be convincing (not just one, but both "doubters" have apparently been convinced) however I'm highly skeptical that ANY explanation or even combination of explanations would convince all readers. To be clear, I'm not saying it can't be made more convincing but like it says in the lead - even when given a completely unambiguous statement of the Monty Hall problem, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief. -- Rick Block (talk) 18:09, 6 September 2008 (UTC)

I agree it will not be easy but surely it is worth a go. As with similar problems, explanations from many different angles could be a good place to start. What is the best way to proceed? I do not want to start editing the current article by adding a new section, which will start off badly written, formatted, and referenced and may well get removed. Do you agree that a development version of the article might be a good place to start, with the intention of a merger when agreed, or perhaps leave it as a separate article with a link from this one? Martin Hogbin (talk) 20:24, 6 September 2008 (UTC)

Creating a new article to be merged later is definitely not the way to proceed. Proposing the text for a change, or even a new section, here (well, probably not here but in a section below) on the talk page would be perfectly reasonable. If you're talking about major structural changes, you could create a sandbox version, perhaps on a page in your user space like user:Martin Hogbin/Monty Hall problem (draft). -- Rick Block (talk) 20:54, 6 September 2008 (UTC)

I have set up a user page as described - see section below.Martin Hogbin (talk) 10:14, 7 September 2008 (UTC)

• keep original. If I understand correctly the two solutions that are being offered (no one has bothered to supply diffs), then I think the current (original) version is better. having taught this problem in the past, however, I think it needs some revisions. does anyone object if I do some editing? --Ludwigs2 03:07, 7 September 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)

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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)

Looks like it is OK to answer here. Here is a very simple way to see that the probability of getting the car if you switch is 2/3. If you switch, you always get the opposite of what you started with. You have a 2/3 probability of starting with a goat. Martin Hogbin (talk) 16:51, 30 August 2008 (UTC)

This is simple, but it relies on an assumption that when the host opens a door nothing changes (or changes the question from "what is the probability of winning after the host opens a door given the Monty Hall rules" to "what is the probability of picking a car from among 3 doors"). Under the standard rules it turns out that the host opening a door doesn't affect the player's initial chances of having selected the car (this is reflected in the Bayes expansion above with the two 1/2's that cancel each other out), but this simple explanation does not use those rules so it is fundamentally incomplete (there is much more discussion of this in the #Suggestion to replace existing solution section thread just above). Why does this explanation not apply to Deal or No Deal? -- Rick Block (talk) 17:25, 30 August 2008 (UTC)

Yes Rick, you are quite right. However if I add to my explanation the observation that, with the standard rules, no information about the location of the car is given when the host opens a door and thus the original probability of getting a goat holds after the door has been opened, then my explanation is a good one for the standard rules case. However it is then not such a simple explanation and cannot be directly applied to other cases. At least I understand what the RFC is all about now. Martin Hogbin (talk) 18:01, 30 August 2008 (UTC)

Here is rather extreme example to show what Rick means and why my original answer was incomplete. Suppose we work to the standard rules, so you would expect my explanation to apply, but, after he as opened a door, the host tells the contestant whether he has initially chosen a car or a goat (rather spoils the game) then the probability of getting a car if the contestant swaps is clearly not 2/3 but either 1 or 0 depending on what the host tells the contestant. Martin Hogbin (talk) 18:12, 30 August 2008 (UTC)

The question still remains how we know given the standard rules that "no information about the location of the car is given when the host opens a door". This is an assertion, not reasoning, and (as it turns out) showing this with reasoning is not overly simple. Again, I think the best reference for this is the Morgan et al. paper (in the references), or the current version of this article. This topic was discussed at length several months ago, shortly before the latest featured article review - starting with Talk:Monty Hall problem/Archive 6#Rigorous solution. In my opinion the ensuing discussion (that continues in the archives into Talk:Monty Hall problem/Archive 7), somewhat contentiously spearheaded by an anonymous user (using multiple IP addresses in the range of user:70.137.168.95) who originally pointed us to the Morgan et al. paper, in conjunction with scrupulous referencing improvements to satisfy concerns raised during the last FARC have led to significant improvements in this article (as well as to my personal understanding of the subtleties involved in this problem). One of the key points this user brought to the discussion (backed up by the Morgan et al. paper) was that most solutions presented to the problem are overly simplified and include but don't justify the assertion that the host opening a door doesn't change the initial 1/3 probability. Another very good reference about this is the Falk paper (also in the references). According to Falk the answer that most people present (after the host opens a door, it's a 50/50 chance) is rooted in a deeply intuitive "equal probability" assumption; however many "correct" answers commonly presented simply replace this intuition with an appeal to another one, i.e. the "belief that exposing information that is already known does not affect probabilities". Neither of these is based on particularly sound reasoning and problems can be constructed where either one leads to an incorrect answer. This problem is contentious at least in part because it's not simple. -- Rick Block (talk) 21:21, 30 August 2008 (UTC)

There certainly is more to this than meets the eye. Might a way forward be as follows? Formulate a version of the problem where a simple analysis, such as I gave, is correct. We are at this stage picking a problem to suit our desired answer. In other words construct a problem where what has been called the most attractive false solution is actually a good solution. Now clearly state our reformulated problem as an idealised version of the Monty Hall problem and give the, now correct, simple solution. My guess is that this will be what 90% of readers will want and expect. It is just a continuation of the process started by Krauss and Wang.

The discussion about the hosts behaviour and the contestants understanding of this then can become a secondary, much more complicated, and perhaps contentious, discussion for those interested in the fine details.Martin Hogbin (talk) 22:25, 30 August 2008 (UTC)

This has been suggested before - the problem is that if we do this we drift into WP:OR unless the version of the problem we present clearly meets WP:RS. I think this boils down to whether the solution currently presented is accessible to a lay person. I think the answer is yes (but I am sufficiently "into" this problem that I realize my opinion on this is basically worthless). -- Rick Block (talk) 04:14, 31 August 2008 (UTC)

Maybe we should consider the occasional exception (WP:IGNORE) here. The dividing line between OR and verifiable is rather vague anyway; not all third party references should carry equal weight and the process of assessing how authoritative any given reference is is a form of OR. With this particular problem we could probably find more references, from normally authoritative sources, that give the completely wrong answer than we could good ones. On the other hand experience with the problem has shown that our intuition is not to be trusted either. Regarding the RFC, I have continued above.Martin Hogbin (talk) 09:35, 31 August 2008 (UTC)

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Hi Martin, thanks for the information.
I agree with much of what you say, but I disagree with the assertion that:

"no information about the location of the car is given when the host opens a door".

It seems to me that when the host opens a door the player is given information about the location of the car. In particular, the host's action informs the player that the car is not behind Door 3 and that the car must therefore be behind either Door 1 or Door 2.

Yes, you are right, I put it badly.Martin Hogbin (talk) 12:59, 31 August 2008 (UTC)

This information does not alter the fact that at the outset of the game there was a 1/3 chance that the car could have been behind any particular door.

That is right but also the chance a car is behind the door that I initially chose remains at 1/3 if the host must open a goat door and, if given a choice, must chose randomly, and I know all this. This is generally agreed to be true in the above circumstances but my mistake was to take this as being self evident. If the host acts differently, then the above statement is often not true. If you agree that, after a door is opened, the chances of the car being behind your initial door is still 1/3 then my argument still holds. Martin Hogbin (talk) 12:59, 31 August 2008 (UTC)

I can't see the logic of the probability remaining 1/3. If it were 1/3 and it is known that there was a goat behind Door 3 this would mean that the probability of the car being behind Door 2 had risen to 2/3. I know that is what the website says, but in my response to Glopk below I explain why I believe that probabilities can't be transferred as suggested on the website. Can you point me to any of the previous discussion about transferring probabilities? --Paul Gerrard at APT (talk) 12:23, 1 September 2008 (UTC)

Let me try a slightly different version of the problem. The contestant picks a door. We agree that he has a 2/3 chance of picking a goat. We also know that if he switches he will get the opposite of his original choice. Suppose, therefore, that this person decides, before a door is opened, that he will switch. Do you agree that he then must have a 2/3 chance of getting a car? Martin Hogbin (talk) 18:08, 1 September 2008 (UTC)

Hi Martin - I agree that at the outset there is a 2/3 chance of picking a goat, but I am not sure what you mean when you say that "if he switches he will get the opposite of his original choice". Also, if the player decides to switch why would the probability of picking a car increase to 2/3. If you extend the logic to switching a second time would you say that the probability of picking the car was 1 (i.e. a certainty). What would be the reason for suggesting that deciding to switch would increase the probability of picking the car? —Preceding unsigned comment added by Paul Gerrard at APT (talk • contribs) 00:11, 2 September 2008 (UTC)

By "if he switches he will get the opposite of his original choice" I refer to the simple fact that , if the player initially chooses a goat and the switches he must get a car(because his original choice is one of the goats and the host has removed the other one from play leaving the car as the only other door) and if the player initially chooses a car and switches he must get a goat (because there is only one car and he has it). The above always applies, without exception, regardless of the strategy of the player or the host (we are here assuming that the game rules require the host to always open a goat door and offer a swap).

Regarding a second swap, no of course, the probability does not rise to 1. The whole point of this problem is that after a door has been opened, the player has more information as to the whereabouts of the car, making the unchosen door more likely to have it. I have to say, and indeed am saying, that the current explanation is not as convincing as it should be. I am currently trying to improve it. Martin Hogbin (talk) 08:46, 2 September 2008 (UTC)

However, while the original probabilities remain unchanged the information that the car was not placed behind Door 3 effectively changes the game from a "3 door game" to a "2 door game".
The effect of knowing that the car is not behind Door 3 on the probability that the car is behind Door 1 can be demonstrated as I have done above using Bayes' Theorem.
Bayes Theorem can also be used to calculate the probability that the car is behind Door 2 given that it is not behind Door 3. (The answer is 1/2.)
Our different perspectives are as intriguing as the puzzle itself.
What am I missing? (I have tried to make sense of the references with the exception of Morgan et al which costs $14)--Paul Gerrard at APT (talk) 11:29, 31 August 2008 (UTC)

No, because the opening of a door does give you extra information about the chances of the car being behind the other door. Martin Hogbin (talk) 12:59, 31 August 2008 (UTC)

Paul, please re-read the "Bayesian Analysis" section. As others have already pointed out, the proposition of interest for the MHP is not $\qquad C_{i}|\neg C_{3}\,=\,$ "The car is behind door i, given that it is not behind door 3", for $i=1,2\,$. Rather, the player is interested in the truth of $\qquad C_{i}|H_{j3},I\,=\,$ "The car is behind door i, given that the host opens door 3 after the the player initially chooses door j", for $i,j=1,2\,$, and where $I\,$ represents everything the player knows prior to making the first choice, including the host's behavior. The two propositions are obviously different, and therefore it is to be expected that they evaluate to different numbers. But only the latter one matters. Further, the evaluation of $\qquad P(C_{i}|H_{j3},I)\,$ using the Bayes theorem shows the need to make explicit the assumptions concerning the host's behavior through the value of $\qquad P(H_{j3}|C_{k},I)\,$, the probability of the proposition "The host opens door 3 after the player chooses door j, given that the car is behind door k". All possible host's behaviors are specified by assigning values to this function of j and k. The Bayesian Analysis section in the article shows only one particular such function, the one consistent with the "standard" interpretation of the MHP, i.e. "The host shows a goat behind one of the two doors not chosen by the player. If two such doors are available, they are equally likely to be opened".The Glopk (talk) 16:46, 31 August 2008 (UTC)

Thanks The Glopk, it is good to be reminded that $\qquad P(C_{i}|H_{j3},I)\,$ is not the same as $\qquad P(C_{i}|\neg C_{3})\,$. This a very important distinction and I agree that these expressions are very different and that there is no reason to believe they should be equal.
It is therefore important to reconsider the problem to make sure we are trying to answer the same question. My understanding of the problem is that we are trying to determine whether it is to the advantage of the player to switch after the host opens a door.
We also need to review how the expressions relate to the issue of whether it is to the advantage of the player to switch after the host opens a door.
The differences between the expressions are subtle.
My understanding is that $\qquad P(C_{1}|H_{13},I)\,$ = 1/3 shows that the rules the host follows to select a door do not alter the probabilities - they are still 1/3.
However, $\qquad P(C_{1}|H_{13},I)\,$ does not take into the fact that the host has revealed that there is a goat behind Door 3 even though the rules of the game require the host to choose a goat door.
It is really hard to find words which explain why this is a problem, but please consider $\qquad P(C_{2}|H_{13},I)\,$. If you evaluate this expression in the same way that website evaluates $\qquad P(C_{1}|H_{13},I)\,$ you will find that:

$\qquad P(C_{2}|H_{13},I)\,$ = $\qquad P(C_{1}|H_{13},I)\,$ = 1/3

Similarly, you will find $\qquad P(C_{3}|H_{13},I)\,$ = 1/3 unless you change the probabilities to reflect the fact that it is known that a goat is behind Door 3.

However, if you change the probability of a goat being behind Door 3 to zero then you must also change the probabilities of the car being behind the other 2 doors such that their sum is 1. This raises the question of what would the new probabilities become. I see no alternative other than 50% chance for both Door 1 and Door 2.

Also, I know the existing website suggests that the host's showing of the goat behind Door 3 has the effect of "transferring" the 1/3 of probability to Door 2. The following example helps demonstrate why I believe such transfers do not occur. Imagine a similar game with 3 doors, two goats, and a car. The probabilities for each door are 1/3. Also the probability of the car being behind any pair of doors is 2/3 (e.g. the probability that the car is behind either Door 1 or Door 2 is 2/3). Now imagine that any goat door is opened - what then happens to the probability that the car is behind either of the other doors? If we assume that probabilities are transfered as per the website then for each of the remaining two doors the probability that the car is behind that door becomes 2/3 and the sum of these 2 probabilities is 4/3 whereas the sum of the probabilities must be 1. Thus, there is a problem in transferring probabilities.

Now let's consider $\qquad P(C_{1}|\neg C_{3})\,$. It reflects everything that is known by the player. The process by which the host chose Door 3 has given no more information to the player than that a goat was behind Door 3. Thus, $\qquad P(C_{1}|\neg C_{3})\,$ = 1/2 seems to me to indicate that it is the true probability of the car being behind Door 1 and there is no advantage in switching.
Finally, lets consider what happens to $\qquad P(C_{1}|H_{13},I)\,$ if we adjust the probabilities to reflect the fact that we know the car is not behind door 3 as suggested above.
That is $\qquad P(C_{1})=P(C_{2})=1/2\,$ and $\qquad P(C_{3})=0\,$. If you do this you will find that:

$\qquad P(C_{1}|H_{13},I)\ =P(C_{1}|\neg C_{3})\ =1/2,$

That is, the Bayesian Theorem actually gives the same result as $\qquad P(C_{1}|\neg C_{3})\,$ providing the probabilities used in the formula are adjusted to take account of the fact that a goat is behind Door 3.
Thus, it seems to me that there is no advantage in switching. --Paul Gerrard at APT (talk) 11:47, 1 September 2008 (UTC)

$\qquad P(C_{i}|H_{j3},I)\,$ is the probability that the car is behind door i given the host opens door 3 after the player initially selected door j (under the rules of the game). This is exactly the question we're interested in. The rules of the game ensure the host opens a goat door. You say this "does not take into the fact that the host has revealed that there is a goat behind Door 3" and admit that it "is really hard to find words which explain why this is a problem". That's good, because it is not a problem.

$\qquad P(C_{2}|H_{13},I)\,$ evaluates to 2/3, not 1/3, since

• $\qquad P(H_{13}|C_{2},I)=1\,$ (if the car is behind Door 2 and the player picks Door 1 the host is forced to open Door 3)

Similarly, $\qquad P(C_{3}|H_{13},I)\,$ is 0, not 1/3, since

• $\qquad P(H_{13}|C_{3},I)=0\,$ (if the car is behind Door 3 and the player picks Door 1 the host never opens Door 3)

No "changing of probabilities" is required to get these to sum to 1. Before the host opens Door 3 the probabilities are 1/3, 1/3, 1/3. After the host opens Door 3 (under the rules of the game) the probabilities are 1/3, 2/3, 0. The effect (in this case, under these rules) is that Door 3's probability has been "transferred" to Door 2. This doesn't say anything like you can group any two arbitrary doors and probabilities transfer between them.

Our differences of opinion seem to hinge around what probabilities we feed into the Bayesian formula.

If it is correct to say that all probabilities are 1/3 then I agree with your calculations.

If it is correct to adjust probabilities to 1/2, 1/2, 0 for Doors 1, 2, 3 respectively then I believe my calculations are correct.

I believe you are right if we are talking about probabilities prior to the door being opened and I am right if we are talking about probabilies after the host opens a door. As we are trying to take account the information available to the player at the time the player must decide to switch or not not to switch, we must use the information available to the player and at this time the player knows only that there is a goat behind Door 3. The rules of the game provide no more information than this. Thus I believe the probabilities must be 1/2, 1/2, 0 for Doors 1, 2, 3 respectively rather than 1/3 for each door.

With respect to the grouping of doors I agree that one cannot group doors and probabilities transfer between them. My example was to prove that one cannot do it. However, it seems to me that your solution does exactly this. Isn't your solution efectively saying that:

The probability of the car being behind Door 1, 2, or, 3 is 1/3 for each door.

You then group the 2 doors that are not chosen by the player and say there is a 2/3 chance the care is behind one or other of those doors.

The host then selects a goat door from the two doors that were not chosen by the player.

You have then concluded that the entire 2/3 probablity that the car was behind either of these 2 doors must therefore be the probability of the car being behind the door that was not selected by either the player or the host.

I disagree with the logic for two reasons.

Firstly, in the same way as you have grouped the unselected doors you could have said that the probability of the car being behind either Door 1 or Door 3 is 2/3. Then when the host opened Door 3 I think your logic would suggest that the probability of the goat being behind Door 1 would become 2/3.

Secondly, if we go back to Bayes and consider the probability of there being a goat behind Door 2 if it is revealed that a goat is behind door 3 assuming that the probability of a goat being behind either door is 1/3. If we calculate:

$\qquad P(C_{2}|\neg C_{3})=P(\neg C_{3}|C_{2}).P(C_{2})/P(\neg C_{3})\,$

we get $\qquad P(C_{2}|\neg C_{3})=\,$ 1 x 1/3 / 2/3 = 1/2
I don't see how the rules of the game can push this to 2/3 as the constraints on the host do not reveal any information other than the car is not behind Door 3.
--Paul Gerrard at APT (talk) 02:36, 2 September 2008 (UTC)

---

$\qquad P(C_{1}|\neg C_{3})\,$ is indeed 1/2. The difference is this ignores the constraints on the host (in particular, that the host MUST open Door 3 if the player picks Door 1 and the car is behind Door 2). This would be the probability of winning if the host happens to open Door 3 choosing randomly between Door 2 and Door 3 (this is the "forgetful Monty" variant discussed in the article). -- Rick Block (talk) 16:30, 1 September 2008 (UTC)

Let's consider if the constraints of the game affect the probability of the host selecting Door 3. If we have no information about the location of the car the probability of the host opening Door 3 is 50% given that he is not going to open Door 1. This probability is true because:

There is a 1/3 chance the car is behind Door 1 and if it is there is a 50% chance the host picks Door 2 and a 50% chance the host will pick Door 3.

There is a 1/3 chance the car is behind Door 2 and if it is there is a 100% chance the host picks Door 3

There is a 1/3 chance the car is behind Door 3 and if it is there is a 100% chance the host picks Door 2

That is the probability of the host selecting Door 3 is the same as the probability of the host selecting Door 2. (This probability is 1/6 + 1/3 = 1/2)

I cannot see how the constaints on the host influence the probability that the car is behind Door 1.--Paul Gerrard at APT (talk) 03:04, 2 September 2008 (UTC)

Let's try this again. You're ignoring the effect of the game rules on the host's actions which is why saying we're given Door 3 has a goat is different from saying the host (under the rules of the game) has opened this door after the player initially picked Door 1. These are different givens, yielding (in this case) different probabilities. I'm not "adjusting" any probabilities or "grouping" any doors, only using what the probabilities are and plugging them into Bayes formula. I AGREE

$\qquad P(C_{2}|\neg C_{3})=1/2\,$

but disagree that this is the question we're answering.

Now it's time for you to say where you disagree with the following:

$\qquad P(H_{13}|C_{2})=1\,$ (probability of host opening Door 3 after the player initially picks Door 1 given the car is behind Door 2 is 1 - the game rules enforce this)

$\qquad P(C_{2})=1/3\,$ (probability of the car being behind Door 2 is 1/3)

$\qquad P(H_{13})=1/2\,$ (probability of the host opening Door 3 after the player initially picks Door 1 is 1/2 - the game rules enforce this as well and you correctly derived it, just above)

So, by Bayes

$\qquad P(C_{2}|H_{13},I)=(P(H_{13}|C_{2})\times P(C_{2}))/P(H_{13})\,$

$\qquad P(C_{2}|H_{13},I)=(1\times 1/3)/(1/2)\,$

$\qquad P(C_{2}|H_{13},I)=2/3\,$

Please note the big bold banner at the top of this discussion page which says the article is mathematically sound and the result has been experimentally verified numerous times. If you disagree, it's FAR more likely (I'd say certain) that you're incorrect rather than that the article is wrong. -- Rick Block (talk) 03:31, 2 September 2008 (UTC)

The player's initial choice is relevant to the extent that it restricts the choice of the host to the other remaining doors. However, is it relevant to whether there is any advantage in switching?

It is true that the player has a 2/3 chance of picking a goat and given the host must choose a goat it is very tempting to conclude that there is a 2/3 chance that the car will be behind the unselected door. That is, 2/3 of the time the rules of the game will cause the host to reveal with absolute certainty the location of the car. But, is it really true that the probability that the car is behind the unselected door is 2/3? Do the rules of the game really make this conclusion valid? It is tempting to say yes.

However, hmmmm ... BOTHER

I think you are right and I am wrong.

At the outset the player has a:

1/3 chance of selecting a door with the car behind it and if he switches he certainly loses.

2/3 chance of selecting a door with a goat behind it and if he switches he certainly wins.

That is, a switching strategy gives a 2/3 chance of winning.

This is a really good problem. Thanks for your patience in helping to me to work through this.

I WAS WRONG! THERE IS AN ADVANTAGE IN SWITCHING --Paul Gerrard at APT (talk) 13:14, 2 September 2008 (UTC)

And, to your original point, the Bayes expansion supports this as well. Please also note that you are in very good company - folks who initially got this wrong include Paul Erdős. -- Rick Block (talk) 13:54, 2 September 2008 (UTC)

Player's initial choice is irrelevant

The first choice of the player is irrelevant. The host will ALLWAYS open the door with a goat behind it leaving the player with two doors of which either one has the car behind it. The palyer then picks one of these two doors and so has a 50-50 chance of winning the car. Mathematically you should not count the hosts action because the propability that he opens the door with the car behind it is 0. So the number of actual choices are four of which two leads to the car and two to the goat. —Preceding unsigned comment added by 81.197.22.58 (talk) 21:12, 1 September 2008 (UTC)

The player's choice absolutely matters. The host can't open this door, and must open a door with goat. Yes, the host always opens a door with a goat behind it but because the host can't open the player's door and must open a door with a goat, if the player initially picks a goat (there's a 2/3 chance of this happening) the host is showing the player which of the other two doors (the unselected ones) has the car. Please read the Solution section and look at the diagram. If you're still confused we can talk about it. -- Rick Block (talk) 01:02, 2 September 2008 (UTC)

Allright, let's then ask what is the propability for the player to win the car AFTER (for the rest of the game) the host has opened the door? That must most certainly be 50%. Then we only have left the discussion about the relevance of the player's first choice. —Preceding unsigned comment added by 85.156.86.237 (talk) 09:30, 2 September 2008 (UTC)

Just because there are two doors left after the host opens one does not mean they're equally likely to have the car. Under the rules of the MH problem, they're not. Do you agree the playing card simulation is equivalent, i.e. shuffle two red twos (losers) and the ace of spades (the winner), deal one to the "player", and you (playing the part of the host) look at your two cards and turn up a red two? If so, try this, with playing cards, maybe 20 or 30 times and keep a record of how often you (the host) end up with the ace (in which case, switching would win). I suspect if you actually do this you'll discover not only that switching wins more often than not, but that you can take a shortcut which is that when you (the host) look at your hand you don't even have to turn up a two to decide if the player will win by switching. If the ace is one of the two cards (2/3 chance), switching wins. The same thing is true with the MH problem. After the player picks, if the car is behind either of the two unpicked doors then ... switching will win (and this is known before the host opens a door, based only on whether the player initially picked the car!). -- Rick Block (talk) 10:03, 2 September 2008 (UTC)

You seem to be right. Now when I get the idea, it leads me thinking that this problem would be easier to understand seen from the host's point of view. When the player doesn't switch his choice, the host can lose the car only if the player has picked the door with the car behind it in the first time. This is 1/3 chance so the host has 2/3 chance for keeping the car. By switcing his choice the player lets the host keep the car if it is behind the door he picked first (with 1/3 chance) and the player gets the 2/3 odds which the host had before switching. I think this pretty much complies with Chuck's (?) P(not switching) = 1/3 and thus we have P(switching) = 1 - P(not switching) = 2/3. I thank You very much for your time and comments. —Preceding unsigned comment added by 81.197.28.197 (talk) 22:09, 3 September 2008 (UTC)

JUST ONE MORE THING. To make it even clearer, you could present an analogyous case where there are 100 doors with 99 goats and one car behind them. First you let the player choose one door and then the host opens one door which he knows there is a goat behind it. Then the host would ask if the player wants to switch the door he picket first to all the 98 doors that are left. The most favourable choice must be obvious! The cunning thing in the original problem, is the number three, which reduces the choice to two doors in the second phase. It would though be interesting if someone could present a similar (simple and without too much mathematics) example in the three-prisoner case! —Preceding unsigned comment added by 80.186.57.76 (talk) 08:53, 4 September 2008 (UTC)

The way results are recorded in the playing card simulation makes it equivalent to a game in which the player chooses door #1, and then Monty says, “In a minute I’ll open one of the doors. Would you like to stick with #1 or switch to whichever of the two doors remains closed?”

A better simulation for the game, as stated, would be to keep track of four sets of numbers: wins by switching to #2, wins by switching to #3, wins by sticking with #1 when #2 is the remaining choice, and wins by sticking with #1 when #3 is the remaining choice. Out of all runs, the player will tend to win 1/3 of the time by switching to #2 and 1/6 of the time by sticking to #1 when #2 is the remaining choice.

Keeping track of these numbers might help the player understand not just that switching doubles ones chances, but why switching doubles one chances even when the puzzle uses a scenario that has only a 50% chance of occurring once door #1 is selected. Since the odds of winning by switching have been split into two, with one half being ignored, the odds of winning by sticking should be split as well: compare the 1/3 chance of winning by switching to #2 to the 1/6 chance of winning by sticking with #1 when #2 is the door that remains closed. Simple314 (talk) 02:10, 26 January 2009 (UTC)

Confusing text

In the Sources of Confusion section, the last paragraph is confusing.

Another source of confusion is that the usual wording of the problem statement asks about the conditional probability of winning given which door is opened by the host, as opposed to the overall or unconditional probability. These are mathematically different questions and can have different answers depending on how the host chooses which door to open if the player's initial choice is the car (Morgan et al., 1991; Gillman 1992). For example, if the host opens Door 3 whenever possible then the probability of winning by switching for players initially choosing Door 1 is 2/3 overall, but only 1/2 if the host opens Door 3.

The example of the difference between conditional and unconditional is not illuminating. Specifically, why the probability is 1/2 if the host opens Door 3. What probability are we referring to here? --dhawk (talk) 17:06, 31 August 2008 (UTC)

The probability of winning by switching is reduced in the above circumstances because the fact that the host has opened door 3 indicates a greater likelihood (1/2 rather than 1/3) that you have chosen the car. This is because, if you had chosen a goat there is the possibility that he might have had to open door 2 because the car was in 3. If, on the other hand, he opens door 2, you know it is your lucky day because the car must be in 3 as otherwise he would have had to open it. Martin Hogbin (talk) 17:55, 31 August 2008 (UTC)

Ahh. That's interesting enough that I think it should be expanded, perhaps to include the fact that you would be certain that switching would get you the car if the host opened door 2. Maybe it's just me, but that paragraph by itself doesn't make those details clear. dhawk (talk) 05:43, 1 September 2008 (UTC)

Yes, perhaps further explanation would be in order but I believe that the issue of different host strategies is a sideline that greatly and unnecessarily complicates the the main problem, which is that nearly everybody does not believe that you get a 2/3 chance of the car if you swap (conditions apply). Martin Hogbin (talk) 18:47, 1 September 2008 (UTC)

Suggested improvements

Although I agree that the solution given should not be replaced with the suggested one I do think that the proposer has a point in that the page does not do a very good job of addressing the fundamental issue that the chances of getting a car if you swap is 2/3 and not 1/2 (Morgan et al. conditions plus the constraint that the host picks randomly when he has a choice, and the contestant knows this).

I do understand that there are issues regarding verifiability in deviating from the current solution but I believe that we should make every effort to make the current solution more convincing. I am sure that there are ways in which issues of verifiability can be addressed if we try.

I think the issue of different host actions is a red herring that complicates and already exceedingly difficult problem and that this should be omitted from the main part of the solution and from 'Sources of confusion' - the last thing that section needs is another source of confusion.

Two starting suggestions are:

1 Describe the problem at the start of the solution in sufficient detail that only one answer exists. This would include describing the host's exact policy.

2 Describe the problem in a way that eliminates the issue of host startegy, for example by including the quote from Morgan et al,'You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat. What is the probability that you win if your strategy is to switch?'. Martin Hogbin (talk) 18:59, 1 September 2008 (UTC)

The various comments on this page since I have been here indicate to me that the solution needs to be better explained. Martin Hogbin (talk) 22:09, 1 September 2008 (UTC)

Certainly no article is ever perfect. The existing "Problem" section provides both the well-known (ambiguous) Parade version and a fully qualified version that permits only one possible (numerical) answer. Given this, I'm not sure what your first suggestion is.

Your second suggestion (to use the Morgan et al. phrasing) would be odd, since they use this as a description of what they consider to be a different problem. Their issue is not so much "host strategy", but how to interpret the question. Nearly all phrasings of the Monty Hall problem put the player at the point the host has already opened a door and ask whether switching to the other door is better. This phrasing is consistent with Morgan et al.'s interpretation of the problem as a conditional probability question, which leads to a 2/3 chance of winning by switching only if the host is constrained to randomly (with equal probability) select from among two goat doors in the case the player initially happens to select the car. The unambiguous problem statement in the Problem section includes this constraint. Without this constraint what Morgan et al. call the "unconditional" question (corresponding to the "overall" chance of winning across all hypothetical players, i.e. if 1000 players play this game and all of them choose to switch, how many would we expect to win the car) still has the same 2/3 win by switching solution - but (as they show in their paper) permits other host behaviors where knowing which specific door the host opens yields a probability of winning by switching anywhere from 1/2 to 1. This is in some sense a technicality, but it's a significant enough technicality that it needs to be mentioned here. Nearly all people have a VERY hard time understanding that these are two different questions and that they require slightly different reasoning. We had an extended discussion about this some time ago (it started with the thread I referred you to - and continued for weeks if not months) and ultimately ended up with what's in the article now. The goal is to be pedantically (mathematically) correct but to avoid introducing the concepts of conditional vs. unconditional probability until after the (usual) unconditional solution is presented.

As an encyclopedia article, I think we're certainly in the ballpark of a reasonable approach. The goal is NOT to convince everyone that the 2/3 answer is correct but to provide the "normal" answer (in a way that most people can understand) and then to expand on this to cover the more technically accurate Morgan et al. approach (hopefully without losing too many people). That it's possible to have different "conditional" and "unconditional" answers for the same problem (with the same host constraints) is (I believe) extraordinarily unusual (I've tried, but have been unable to come up with other examples). This isn't exactly the crux of the Monty Hall problem, but I think putting the player in front of two closed doors and one open door is. The "normal" solutions certainly depend on an assumption that the host does nothing that changes the probability of the player's first pick, i.e. the door the host opens is a goat so the other door has a (1 - 1/3) chance of being the car or if you pick a goat first with a 2/3 chance and then switch you now have a 2/3 chance of winning the car. These are very common, but (per Morgan et al.) mathematically insufficient ways to "solve" the problem. I don't think the slightly more rigorous approach presented in the Solution section is tremendously more difficult to comprehend.

Again, no article is ever perfect. If you have specific suggestions for wording improvements please offer them up (or boldly make them - but only once without discussing them here if they're reverted). -- Rick Block (talk) 02:32, 2 September 2008 (UTC)

I think you are missing the point of what I am trying to do. I know much thought and hard work has gone into the page and those editors responsible would not be happy to see their work replaced but there has been a criticism of this page and a suggestion that the solution is replaced with another. I do not support this, but I do think that some of the criticisms are valid and should be addressed. As I originally said in response to the RFC, my basic suggestion is to add rather than replace.

Wikipedia should reliably inform, but it cannot do this if the reader does not believe what they are being told. The current attitude seems to be, 'We have done a good and rigorous job of explaining the paradox, that is all we can do. If readers do not believe the answer that is just too bad'. In this respect some of the articles on the related problems do a better job even though the articles are of poorer quality by Wikipedia standards.

What about a separate section called, 'Still not convinced? or even a separate article. We could then try different approaches to help people understand the paradox. We could state at the top that this section ignores some of the subtleties of the problem in favour of clarity. We could also consider my suggestion 2 above. It is not uncommon in mathematics to first answer a different question from the one originally posed and then to consider how the two questions differ. This approach might lead to a better discussion of the conditional/unconditional issue.

I would like to do some serious addition to this subject in cooperation with others, but in such a way that the integrity of the current article is not compromised. Perhaps a development article called 'Explanations of the Monty Hall problem', with consideration to a possible future merger - it might fall flat on its face but that way the current article is unharmed. Martin Hogbin (talk) 09:24, 3 September 2008 (UTC)

I feel WP:NOR should be pretty much sacrosanct; what we write in Wikipedia must be verifiable. What we could do, however, is to add a page in the Talk namespace with some more explanations, or even a Wikibooks article (are there any guidelines on referring to Wikibooks?)

Also, your suggestion of "Still not convinced?" immediately made me think of Talk:0.999.../Arguments. We could follow a similar approach here for those who dispute the theoretical result. Oliphaunt (talk) 17:33, 3 September 2008 (UTC)

We may be able to achieve some of what I am hoping for without OR. For example, my suggestion to start with a different question was based on the statement of a similar question by Morgan. Regarding WP:NOR there are reasons why we should also consider WP:IGNORE and WP:COMMON SENSE. Because of the concentration on NOR, the article does not do as well as it might in terms of being convincing. Also, in this particular case there is in fact a degree of OR in the selection of references that editors have chosen to cite. A perverse editor could rewrite the article giving the wrong answer but with it still being fully verifiable, citing letters and statements from many eminent organisations.

I do not think "Still not convinced?" should be just relegated to the talk pages although I agree that a dedicated talk page is a good idea. Can we not also try to improve the article's explanatory power by adding something? Martin Hogbin (talk) 20:16, 3 September 2008 (UTC)

Article deletion discussion

A newly-created article related to this topic – Introduction to the Monty Hall Problem – has been nominated for deletion. Interested individuals may participate in the discussion at Wikipedia:Articles for deletion/Introduction to the Monty Hall Problem. TenOfAllTrades(talk) 11:59, 4 September 2008 (UTC)

It's gone now. Wow, that was fast... —Preceding unsigned comment added by 69.36.227.135 (talk) 23:45, 4 September 2008 (UTC)

User space for development of a new section

I have set up a user page user:Martin Hogbin/Monty Hall problem (draft) for the purpose of developing clearer and more convincing solutions and explanations of the problem. All editors please feel free to add your pet solutions there for discussion and futher editing.

All editors (current, historical, and prospective) please note that I hope this page will eventually be edited to a standard where all, or parts of it, can be added here under a separate section.Martin Hogbin (talk) 10:21, 7 September 2008 (UTC)

Section Aids to understanding in the article exists for this purpose, and I think it covers the best of the notable common sense explanations well. If you can find a notable explanation that is easier to grasp then more power to you, but the suggestion that we ignore rules in favor of common sense is ironical, for as a veridical paradox the very point of the problem is that common sense is wrong. The fact that some people remain unconvinced by any explanation is a testament to its excellence as a veridical paradox: common sense rebels against it. ~ Ningauble (talk) 20:23, 9 September 2008 (UTC)

I do not understand what you mean by 'notable common sense explanations'. I do not believe that many people remain unconvinced by any explanation, it may be that there is one that will do it for them. Martin Hogbin (talk) 20:54, 10 September 2008 (UTC)

By "common sense explanations" I was referring to those which sacrifice rigor in favor of clarity and ability to convince. Perhaps "informal explanations" would be a better term, but I was straining to emphasize a point about common sense being at issue. By "notable" I mean to emphasize that, since there are so many erroneous treatments in circulation, even an ordinarily WP:RS needs some vetting akin to WP:GNG. Perhaps that is an overstatement, but the situation calls for careful scrutiny.

I didn't say many people remain unconvinced, but that some do. I know several who are normal, functioning adults. Don't underestimate the cognitive dissonance that must be overcome when common sense is challenged. ~ Ningauble (talk) 20:51, 11 September 2008 (UTC)

Not ambiguous

The article says "Some of the controversy was because the Parade version of the problem is technically ambiguous since it leaves certain aspects of the host's behavior unstated, for example whether the host must open a door and must make the offer to switch." I don't think this is true. I don't think that the original version is ambiguous. It doesn't say, if host behaves every time the same, but that is not important, because the question is about a single decision. That single decision has probabilities for every choice you take. The host could behave differently at a different point in time, and the resulting probabilities might change. But that is not the question.

You can use a computer to simulate this, in which case you would have to put the car behind a different door every time, let the host open a different door all the time, and then let you decide to stick with the door you chose first or not. Then you can use the results of the simulation as an estimation of then probability of the original setup. When you simluate that with a computer, you of course have know if the host always behaves the same. But that was not the question. In the original question, the host *does* open a door, behind it is *not* the car, and he *does* let you choose again. --195.227.10.66 (talk) 17:49, 10 September 2008 (UTC)

If one assumes Monty opened a door at random then the odds are different. His intention, or the rule he follows, does matter. Vos Savant acknowledged the ambiguity by admitting in a follow up article that the constraint was implicit in her solution but had not been explicit in her statement of the problem:[3] "... remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. ... Anything else is a different question." ~ Ningauble (talk) 18:23, 10 September 2008 (UTC)

Of course it is "ambiguous", or, more accurately, "incorrect". Lacking knowledge of the rules of the host's behavior introduces an independent variable into the equation. I don't see how "technically incorrect" differs from "incorrect" in mathematics. As far as I can tell, the words "ambiguous" and "technically incorrect" are media spin, trying to hide the fact that the problem and solution given by vos Santos were wrong.Apollo (talk) 16:23, 11 September 2008 (UTC)

Further ambiguity if host behaviour not clear from start: If you are in a quix programme where you suspect the producers want to avoid having to pay a car for you, you may suspect the extra choice is given because you already picked the car. Or if you suspect they WANT you to win the car - to boost the popularity of the programme, say - the extra choice may be an indication you DIDN'T pick the car to begin with. So, without info on host behaviour, this is not a pure math problem; it's one about psychology, TV company economy, viewer behaviour, and other issues way outside math. What is the information content of the fact that you are given the second choice? Only if you knew in advance that you WOULD be given this second choice is the content nil, and the math of the standard solution to the problem correct.--Noe (talk) 16:36, 11 September 2008 (UTC)

"Ambiguous" doesn't mean a little bit wrong (as in a little bit pregnant). Rather, it conveys more information about the nature of the error than "incorrect." Her answer to the problem stated was incorrect because she failed to state unambiguously what she meant the problem to be. I find it amusing how she tried to spin it as a misinterpretation by the reader rather than a misstatement by herself, for ambiguity may always be construed against the author. ...but we digress. ~ Ningauble (talk) 18:39, 11 September 2008 (UTC)

Essentially none of the objections to her answer were based on grounds that the problem statement was ambiguous (there's a quote in one of the references about this - maybe we should add this). The problem statement IS in a technical sense ambiguous, but given the context of "math puzzles" the missing assumptions are both what most people intuitively assume and highly reasonable. -- Rick Block (talk) 19:15, 11 September 2008 (UTC)

I agree. Almost any statement of any problem leaves some questions unanswered. These issues were used as a means of avoiding embarrassment by people who got the answer wrong, based on reasonable assumptions as to what the question was.Martin Hogbin (talk) 20:06, 11 September 2008 (UTC)

I agree, most of the confusion does not arise from this ambiguity. Although the objection of Morgan et al. could (arguably) be dismissed in similar manner as "a different question" than that intended, it is an analysis too important to leave out. ~ Ningauble (talk) 21:28, 11 September 2008 (UTC)

The problem as stated is not mathematically ambiguous nor incorrect - because it describes a situation which could occur and asks a question which has an answer. Yes, the hosts behavior is not specified to the degree you might wish but as a math problem, you just have to solve it without knowing more. You have to deal with the possibility that the host opened the door at random, and it happened to have a goat behind it; you have to deal with the possibility that the host knows the contestant picked the car and is merely trying to convince him or her to switch; you have to deal with every behavior that is consistent with the information given.

For all the furor, once the hosts behavior is fully specified it is a straightforward math problem but it becomes much more interesting if you assume you're supposed to solve the problem exactly as it stands. (Almost certainly, that was not the intent of the person who asked the question originally, but that is another matter.) Yes, it does have an answer and the answer is that it is better to switch - how much better doesn't have a simple numerical answer but the form it takes shows that switching is superior. And that is all thats needed to answer the question as posed.

Millbast5 (talk) 10:42, 13 September 2008 (UTC)

I agree with you that it is possible to give a solution based on reasonable assumptions. However, I think that in your statement, 'once the hosts behavior is fully specified it is a straightforward math problem', you completely miss the point. There are interesting and relevant issues concerning the host's behaviour but what makes this problem special is that even when fully specified, nearly everyone gets it wrong, including excellent mathematicians and statisticians. Some of them have tried to use the issue of the host's behaviour to cover up the fact that they have made a mistake.Martin Hogbin (talk) 12:07, 13 September 2008 (UTC)

The problem may be tricky as a word problem but as a probability problem it is one that most students in an elementary course in probability would get right. Stated as a problem in terms of the classic balls and urn it is: an urn with three balls in it, known by everyone concerned to be two white balls and one red. One person reaches into the urn and withdraws a ball at random which he keeps concealed in his hand. A second person announces that she will withdraw a white ball then looks into the urn and lifts out a white ball, as promised. What is the probability that the ball left in the urn is white? What is the probability the ball left in the urn is red?

Stated this way it is transparent is it not? The ball left in the urn is white is if and only if the first person withdrew the red ball - which happens with probability 1/3. The ball left in the urn is red with probability 1 - 1/3, or 2/3.

I can't say why the statement in terms of doors, etc., hides the fact that the initial action determines everything that follows. But here its quite clear - the guy draws a red ball, the second person withdraws one of the two white balls, white ball left - automatic. The guy draws a white ball, the second person withdraws the other white ball, red ball left - automatic.

I stand by my claim the that as a math problem it is straightforward. (Actually, a piece of cake relative to some other ball and urn problems students routinely solve.)

Now to the interesting form of the problem - where nothing about the hosts behavior is known beyond what was explicitly stated in the given problem. I did not say that the problem could be solved given reasonable assumptions - I said it could be solved as it stood. It does require methods that are accepted as good probabilistic techniques - the use of Bayes Theorem for example - but no assumptions.

How does one deal with the possibility that the host might have opened a door at random and, by chance, a goat was revealed? That part is relatively easy. Opening a door at random leaves the switch and non-switch strategies equal, whether a goat or a car is revealed, hence if the other possibilities for the hosts behavior put the switch strategy ahead it is better overall.

I should amend my statement a bit - I should not have said the problem had no numeric solution for the amount of the difference between the switch and non-switch strategies. I should have said that I can determine that the switch strategy is better but the method I used does not quantify the difference. The way I dealt with the possibility of random door opening is what stands in the way of quantification - happily the problem doesn't require it.

The rest of the solution is fairly complicated but I will glad to post the details if someone wants to see them. The overall method is to parameterize all the possibilities for the hosts behavior with two continuous random variables, p and q, where p is the probability the host will open a door he knows has a goat behind it when the contestants initial choice is the winning one and q is the probability he'll do that when the contestants initial choice is a loser. To arrive at a distribution for these random variables, Bayes Theorem is used with an equal-likelihood apriori distribution and the data that in the one known trial, a door was opened revealing a goat. Using that distribution one averages over all possibilities to arrive at a number for the probability of getting the car using the switch strategy.Millbast5 (talk) 08:01, 14 September 2008 (UTC)

You should try asking your urn version of the problem to people that you know. You will be amazed how many get it wrong. There is nothing magic about the door-and-goat statement that makes it especially difficult - the difficulty is in understanding how the probability works. Of course, once you know the answer it is easy!

It makes no difference whether the host knows where the goat is, all that matters is that he does, in fact, reveal a goat. Let me restate this in terms of your urn problem: an urn with three balls in it, known by everyone concerned to be two white balls and one red. One person reaches into the urn and withdraws a ball at random which he keeps concealed in his hand. A second person withdraws a bell at random which proves to be white. What is the probability that the ball left in the urn is white? What is the probability the ball left in the urn is red? What is your answer to the above question?Martin Hogbin (talk) 09:18, 14 September 2008 (UTC)

It is of course possible to "solve" the problem without making any assumptions about the host's behavior as User:Millbast5 suggests by assigning probabilities to all possible variants. Morgan et al. do this with the host's preference for choosing between two goat doors while keeping the other constraints intact (host must open a goat door and must make the offer to switch). This analysis says that as long as the host must open a door to show a goat and must make the offer to switch, the probability of winning by switching varies from 1/2 to 1 (depending on how the host picks between two goats), i.e. switching is always at least as good as staying but unless we further constrain the host (for example, adding that the host picks between two goats randomly with equal probability) we can't know a numerical answer.

There's a paper by Chun presenting a formal way to describe the fully generalized problem, "On the Information Economics Approach to the Generalized Game Show Problem" published in The American Statistician in 1999 Vol 53(1). Without digging out the paper I forget whether Chun uses the description to net out the bottom line for the probability of winning by switching, but in a similar fashion to Morgan et al. one could clearly assign probabilities to the variants and end up with a formula involving these probabilities. The "Other host behaviors" section hints at this, listing some of the published variants and the resultant probabilities of winning, but doesn't "sum it up". Note that given the variants listed in this section, the probability of winning by switching is between 0 and 1, and an analysis that concludes you're better off switching is going to rely on how probabilities are assigned to the variants. If there's a paper that shows an assumption-less analysis (Chun's or some other paper), we could try to present it here, but without a source to work from I think we're definitely in WP:OR territory. Another question is whether such an analysis would be appropriate to include in this article, which is often criticized for being too long already. I'd suggest that it might be appropriate to mention such a analysis (if one has been published) but not attempt to include it in the article. -- Rick Block (talk) 16:42, 14 September 2008 (UTC)

Are you saying that certain host actions (subject to host must always show a goat and offer the switch) can reduce the odds of winning if you switch to less than 2/3?Martin Hogbin (talk) 19:40, 14 September 2008 (UTC)

Depending on what you mean by "the odds", yes. This is the whole point of the Morgan et al. paper. Must show a goat and must make the offer makes the overall (unconditional) probability, across all players, 2/3. But, if the host opens the rightmost door whenever possible (or has any other identifiable preference for one door or another) there are two subsets of players. Those who switch when the host opens the rightmost (absolutely preferred) door win 1/2 the time. Those who switch when the host opens the leftmost (absolutely unpreferred) door always win. The usual presentation of the problem ("say you've picked Door 1 and the host opens Door 3") implies we're asking about the probability given the knowledge of which door the host has opened, i.e. the conditional probability as opposed to the unconditional probability. If we don't constrain the host to pick between two goat doors with equal probability, the best we can say about the chances of winning by switching if we already know which door the host has opened is that it's somewhere between 1/2 and 1, with an "average" probability (across all players) of 2/3. Morgan et al. assign the host's preference for the door he opens a probability p - switching wins with probability 1/(1+p). -- Rick Block (talk) 00:16, 15 September 2008 (UTC)

Thanks for that explanation.Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)

To answer the question Martin asked: the probability that the first person draws a white ball is 2/3 and nothing can change that, though information about whether he has in fact drawn the red or a white ball in specific instances can be obtained by drawing another ball at random from the urn. When the second random ball is red, which happens with probability 1/3, the conditional probability is 1 that the first ball drawn was white; when the second random ball is white, which happens with probability 2/3, we let x be the conditional probability the first ball drawn was white. Combining the two cases we get 1/3 * 1 + 2/3 * x as the overall probability of the first ball drawn being white. Since we know the overall probability is 2/3, we can solve for x: 1/3 * 1 + 2/3 * x = 2/3 <=> x = 1/2. Happily, that agrees with our intuition.

Yes, after some thought, I agree with you. There is a difference between does draw a white ball and must draw a white ball. Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)

We can get the conditional probability for the case when a white ball is intentionally extracted from the urn (my statement of the problem) as follows: when the second drawn ball is red, which happens with probability 0, the first drawn ball was white with conditional probability 1; when the second drawn ball is white, which happens with probability 1, let the conditional probability be x as before. Combining the cases: 0 * 1 + 1 * x = 2/3 <=> x = 2/3.

Did you really say "It makes no difference... he does, in fact, reveal a goat."? I keep reading it and it keeps coming up the same. You must have seen the argument above in some form and rationalized that away. How about changing the rule for the second person to be that she withdraws the red ball if its in the urn and only when there are two white balls in it does she withdraw a white. Thus whenever she withdraws a white the first person must have drawn the red ball, so the conditional probability of the first person having drawn white is 0. Hopefully this is sufficiently clear that you can see, regardless of what your intuition tells you, that the method that leads to the event of a white being drawn by the second person does make a difference in the conditional probability of the first person having drawn a white ball.

Yes you are right.Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)

I think we might have stumbled onto something here explaining the population's difficulty with this problem - they somehow twist it around in their minds until it depends on grasping how two events that are exactly alike in appearance can result in different effects on a conditional probability. It seems so obvious to them that they cannot have different effects that simple, straightforward derivations of the solution are rejected, even though they can't point to a flaw.

Could be.Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)

Yes, I think you are right. My mistake was to assume, without thinking, that a white ball being removed by chance has exactly the same effect as its being removed intentionally. I therefore think it is important in our explanation to cover this point in some way. In the current solution the probability of losing by switching, if the player initially chooses a car, is split into two lots of 1/6, with no explanation of why this is so and in what circumstances it would not be.Martin Hogbin (talk) 10:54, 17 September 2008 (UTC)

I don't believe we should go into this level of detail in giving a solution for the general reader - you have to be into 'philosophy' to even worry about the information aspect of various actions. I reworded my description of the game (in terms of urns and balls), into doors, goats and cars. It came out just as clear to me - clearer in fact because I avoid probability altogether. I read in one of the references that people understand frequency better than odds, so I stated it in frequency terms. Think the guy was right. I inserted it into the solutions sections above. Seven lines, no pictures, no probability and only two symbols - 'A' and 'B'. Almost anyone can read it over several times in five minute and get the whole thing in his head at one time.Millbast5 (talk) 23:49, 17 September 2008 (UTC)

I suppose it might seem a bit childish, but one can also settle this by playing the game numerous times, applying different methods for the extraction of the second white ball and keeping track of the results for the different ones.Millbast5 (talk) 01:40, 15 September 2008 (UTC)

Oh rats, I didn't answer exactly the question Martin asked - he asked about the ball in the urn and I answered as though he asked about the ball in the first persons hand. It should make no difference because I did state what I was calculating. —Preceding unsigned comment added by Millbast5 (talk • contribs) 02:05, 15 September 2008 (UTC)

I assume the statement of the problem discussed in this thread is the one from Parade, as quoted in the lead:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. 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?" Is it to your advantage to switch your choice?

As I stated earlier in this thread, it is ambiguous because we do not know whether there is any information content in the fact that the host is opening another door - did he choose to do so BECAUSE the guest either hit or didn't hit the car with his first choice? If the cost of the prize won by the guest is deducted from the host's salary, say, he may simply be trying to direct the guest away from the car, so THE FACT that he was given a second choice is an indication that he DID in fact choose the car from the start, i.e. DON'T SWITCH! Or, if the advertising income for the quiz show is boosted whenever a guest wins a car, the producers may want to direct the guest to the car, so that the fact he was given a second choice is an indication he did NOT choose the car to begin with, i.e. DO SWITCH! We are not given enough information to decide whether one of these scenarios may be taking place, so in order to attack the problem as a math problem, we need to make certain assumptions to begin with, e.g. that the host ALWAYS gives the guest a second choice after opening a door with a goat behind it. This assumption may be a natural one to make, and the fact that many people defend "wrong" solutions to the problem is NOT simply caused by this ambiguity, but still, it IS ambiguous.--Noe (talk) 07:41, 15 September 2008 (UTC)

The point user:Millbast5 is making is that the problem is not ambiguous (meaning it has two different, conflicting interpretations) but rather it simply has more uncertainties than we'd like. We can solve the problem even not knowing any of these things by assigning them each a probability and developing an equation for the overall probability of winning based on these probabilities. For example, we could start with three probabilities p1, p2, and p3 representing the probabilities that the host made the offer to switch because the player initially picked the car, because the player didn't initially pick the car, and because it's simply the rules of the game (respectively). Clearly,

p1+p2+p3 = 1

and a player who switches always loses in the p1 case, always wins in the p2 case, and may or may not win in the p3 case. So, we have (so far)

p(winning by switching) = p2 + x * p3

where x is the conditional probability of winning given that we're in the p3 case (i.e. neither the p1 or p2 cases). This sort of analysis can continue and we'll end up with the probability of winning being a function of 5 to 10 probabilities expressing the likelihood of a set of various scenarios. The range of this function will be [0,1]. If we assume each of the mutually exclusive probabilities are equally likely (for example, p1, p2, p3 above are all 1/3) we can compute a number - but this number would depend on how we break down the possibilities. Instead of the 3 possibilities for "host intent" above, we could say there are two possibilities - host "plays fair" (p3) or not (p1 or p2). Breaking it down this way and assuming equal likelihood we'll end up with p1=p2=1/4, and p3=1/2 (so our final numerical answer will be different).

The point is a conditional probability analysis is sufficient to analyze a problem with as many uncertainties as you'd like, however I'm not sure anyone other than a mathematician would think such a solution would be satisfying. -- Rick Block (talk) 14:11, 15 September 2008 (UTC)

Thanks for the assist Rick, but I meant something simpler - I meant the question of whether the switch or not-switch strategy is better does have a definite answer. (The answer is that the switch strategy is better.) Assigning probabilities in the way you're describing comes under the heading of making assumptions. Statisticians, mathematicians and others have developed methods to deal with this type of shortage of information which have been borne out as accurate and those methods are what I used - I would expect another mathematician solving this problem to do the same.

As Noe mentions one might be in a situation where the non-switch strategy is better even though the switch strategy is better over the full gamut of possibilities. (In game theory, a strategy which is better in every case than another strategy is said to dominate the other one - which isn't true here.) Of course, it takes information that is not given in this problem to know you are in a situation where you should use the non-switch strategy - but if you do have some other information, you use it. Or if you have a hunch about that evil looking host that you just can't ignore. If you got a peek behind one of the doors and saw a goat, use that too. (If you're using the switch strategy, of course your original choice will be that very door so you're certain to win if you get a chance to switch.)

Can I get this straight. There are two types of issue that can affect the conditional probability. The first is really down to the rules of the game (rather than host behaviour but there is no clear distinction) and is that the host must open a door to reveal a goat, in other words by the rules of the game the host can never open a door to reveal the car.Martin Hogbin (talk) 10:54, 17 September 2008 (UTC)

The second type of issue is the way in which the host chooses which goat to reveal when he has a choice.Martin Hogbin (talk) 10:54, 17 September 2008 (UTC)

I agree with Martin that this latter type of issue is far different than the rules of the game.

I'm of the opinion that bringing up the possibility of the host having a known preference for which door he opens to reveal a goat is deleterious to the article. Of course, its possible but nothing in the problem statement hinted at the possibility that contestants watched the show long enough to spot his preference - the original question applies as well to the very first time the game show was held as any other. It is not a question that can even come up if the doors are not identified and the doors were identified as for expository reasons - making things more concrete. The problem can be stated without such identification and statements of equivalent problems don't have anything corresponding to doors.

Taking an accidental feature of the problem statement plus adding an assumption that the process has been going on long enough for a preference to become known is basicly a different problem - at best it should be in a footnote. You could bring up the possiblity that host used his left hand to open the door more often than his right in some circumstances or blinked his eyes at a differing rates - distracting trivia is the category this sort of thing falls into.Millbast5 (talk) 06:33, 18 September 2008 (UTC)

I think I will clean up my computation of the probability of winning with the switch strategy and put it on my talk page so that anyone that wants can see exactly what I'm talking about. Everyone thats agreed with me up until now has misinterpreted what I was saying. Should be there by tomorrow night at the latest.

Maybe I've made a little progress because Noe did not mention the possibility of the host randomly opening doors as something making the problem ambiguous.

Millbast5 (talk) 01:44, 16 September 2008 (UTC)

My talk page contains the computation for the switch strategy.Millbast5 (talk) 09:17, 16 September 2008 (UTC)

@

Lead change

The two sentences added to the lead (see this diff) assume the host opening a door has no effect on the probability of initially selecting the car (which is true only because of the constraints on the host), and essentially explain the fairly trivial point that 1 - 1/3 = 2/3. How about if we state the major reason that the initially picked door remains a 1/3 chance? For example:

Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In the usual interpretation of the problem the door the host opens is not randomly selected and opening this door does not affect the player's initial 1/3 chance of having picked the car. The car is behind one of the two unopened doors; if there is a 1/3 chance it is behind one there must a 2/3 chance it is behind the other. The player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.

The critical point is that the host opening the door does not affect the player's initial 1/3 chance of picking the car - mostly because the host does not open a random door. I think assuming this to be true, without no explanation whatsoever, does not accurately summarize the solution. -- Rick Block (talk) 02:27, 13 September 2008 (UTC)

I prefer the version previous to the diff. It gives the lead section more punch, and gives the reader pause to think. To elaborate: (1) Stating the naive perspective and the correct result, with emphasis on the magnitude of the error, is an excellent way to explain why the subject is interesting. (2) While teasers in the lead are generally discouraged, it is appropriate and arguably necessary to make an exception for articles about puzzles or paradoxes. Explanation of the solution, even in summary, ought to be deferred to the body of the article in order to accommodate readers who are inclined to try reasoning it out for themselves. ~ Ningauble (talk) 14:14, 13 September 2008 (UTC)

Another objection - moved from main page

I moved this from the main article to here. It does not really belong here but it has been suggested that we set up a space for discussion of the subject so that it is kept separate from the article itself and discussion on how to improve it. I am not sure how to do that but I think it is a good idea. Can any one help?Martin Hogbin (talk) 08:50, 13 September 2008 (UTC)

I found this answer to this question on another website and am in no way taking credit for this answer...

"The solution given is that the selection should be switched because the door you chose has a 1/3 chance of success, while the other unopened door now has a 2/3 chance of success.

This solution is surprising, counter-intuitive... and wrong.

Marilyn vos Savant was wrong. The solution given by Mark Evanier was wrong. And the solution given by these fictional math geniuses of "21" was spectacularly wrong.

What am I talking about?! How can all these real and fictional geniuses be wrong?

Switching doors can double your odds of winning, but only if Monty Hall not only knows which door has the car, but is REQUIRED under the rules of the game to open a door that (i) is not the door you chose, and (ii) is not the door that has the car behind it. This must be stated as part of the puzzle in order for the correct solution to be that there is an advantage to switching doors.

This is not merely a semantic distinction -- it is critical to the problem. If Monty Hall is operating under those rigid rules, by opening a goat door after you've chosen, he is giving you valuable information. If you guessed correctly (a 1/3 chance), he can open either of the other two doors. But if you guessed incorrectly (a 2/3 chance), he is FORCED to open up the one door that is neither your choice nor the door with the car.

Now, the puzzle, as presented by Kevin Spacey, does not state that Monty Hall is required to open up any door. He simply says that after you choose your door, Monty "decides" to open up another one that reveals a goat.

If the rules of the game are that he opens a randomly selected door after you make your choice, then the logic of switching doors break down -- there is no advantage to switching whatsoever.

If he simply opens up a door without regard to any pre-determined rules of the game, there is also no advantage to switching. You don't know why he opened a door, so you haven't gained any information.

And in fact, in the clip from the movie "21," the solution given is spectacularly wrong because Kevin Spacey's genius character explicitly posits that you don't know why Monty Hall is opening up a door -- he says that the game show host may be trying to "playing a trick on you, trying to use reverse psychology to get you to pick a goat." Despite what these fictional math geniuses say, there surely is no advantage to switching in this case, because, again, you've gained no useful information. It may be to your disadvantage to switch under these circumstances.

Let's say you play this game 100 times, and Monty reveals a goat door only when you've initially picked the car door (and does nothing when you initially pick the goat door -- just opens your door and tells you that you lost). 67 times you'll pick a goat door and automatically lose. If you adopt a switching strategy, the 33 times you pick a car door, you'll then be shown a goat door by Monty, get "tricked" into switching, and lose. The switching strategy will yield you exactly zero wins out of 100. (Refusing to switch will give you 33 wins.)

The same holds true if you play only once. You pick a door, and then Monty reveals a goat door. Unless his actions are explicitly part of the rules of the game, you have no idea why he showed you the goat door. He may have only showed it to you because your first pick was the car door, and he is in fact trying to trick you into switching. You're now in a mind game with Monty Hall that has nothing to do with the "statistics" and "variable change," that Kevin Spacey's super-student refers to.

This distinction is part of the reason the commonly-given solution is so counter-intuitive. The way it's presented, you're playing this game, and Monty Hall suddenly and without explanation reveals a goat door and gives you a chance to switch -- it doesn't seem as though it would be to your advantage to do so, and in fact it isn't. But once you say that Monty Hall is required under the rules of the game to reveal a goat door and give you a chance to switch, it makes a bit more intuitive sense that there may be an advantage in switching (and in fact there would be)."

source -->(http://gocomics.typepad.com/tomthedancingbugblog/)

The objection above results from the way that probability problems are often stated.

In the Monty Hall problem, if Monty picks a door at random, there is the possibility that he will pick the car. But, he is always seen to pick a goat (and this is said in the statement of the problem). Now it could be that Monty picked randomly but the TV company only showed the times when he picked a goat, however, it is far more likely that he knew where the goat was. In both cases the results are the same, of the screened shows the player who swapped won more often than the one who did not. The problem states that Monty reveals a goat thus we are expected to consider only the cases where this is so.Martin Hogbin (talk) 10:08, 13 September 2008 (UTC)

This objection is already addressed in the article (2nd to last paragraph in the lead, in the Problem section, in the Sources of Confusion section, and in the Other host behaviors section). The topic is already covered (with references), and this material is copyrighted (so can't be used directly, and isn't what I'd call a reliable source anyway). Is there a suggestion here for some change to the article? -- Rick Block (talk) 20:07, 13 September 2008 (UTC)

Not by me. I moved the text above from the article (where it hand been added by an IP editor) to here.Martin Hogbin (talk) 20:40, 13 September 2008 (UTC)

Language clean up

In the introdudtion, it currently states:

" Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3. With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty."

Two things need changing in this. The first is the reduction of words. Not advantageous to disadvantageous to begin with. The second thing is that the sentence is misleading imho. The point of the Monty Hall problem is that it is a counter-intuitive probability based problem. The point is not at what point is it disadvantageous but that probability shows that in the long run it is advantageous to switch. The addition of the statement I quoted above seems to distract from the point the article is trying to make and should not be in the introduction. --Candy (talk) 18:57, 22 September 2008 (UTC)

Agreed. Rick had suggested a clearer explanation two sections previous in this discussion (Lead change) but, for reasons given there, I favor simply removing these two sentences from the lead section. Does anyone object to this proposed revert? ~ Ningauble (talk) 19:03, 24 September 2008 (UTC)

Agreed. Anything that makes things clearer.Martin Hogbin (talk) 17:49, 25 September 2008 (UTC)

I agree with the suggestion of using the word 'disadvantageous' but not with removing the two sentences - to me they are the best explanation in the whole article of why the switch strategy wins with probability 2/3 - I believe that compactness is a big plus and that pictures are a distraction. I would favor emphasizing the idea at this stage that the host is intentionally opening a door with a goat behind it. So something like 'when the host opens the other door he knows conceals a goat'. Grammatically it is incorrect as it is because it isn't the 'losing door' that is revealed but the goat behind it.Millbast5 (talk) 00:45, 27 September 2008 (UTC)

Random host variant

Regarding the variants on host behaviors, I don't find the example of:

"The host does not know what lies behind the doors, and opens one at random without revealing the car (Granberg and Brown, 1995:712)."

to be very useful. This conditional ignores the outcome that occurs 1/2 the time when the car is behind door 2 or 3 and the host must choose a random door to open - he accidentally chooses the car. Since this example of host behavior ignores these potential outcomes of the game and the idea that host chooses randomly is covered well in the previous "aids to understanding" section, I would advocate deleting this variant from the table. However if it is to be included, shouldn't the odds of getting the car by switching still be 2/3? Of course the overall odds of the game if the host does not know the location of the prize has been shown to be 1/2, with the car being prematurely revealed in 2 of the 6 overall outcomes and the win/lose status of the game to be unknown. But in this specific case, it is restricted to the condition where the host "opens one at random without revealing the car." Isn't it correct that once you are to this point in the game, the probability is exactly the same as if the host had known the location of the car and deliberately exposed a goat? This variant eliminates the option of the host exposing the car, so he must act as if he has knowledge of the car's location even though it is stated that he does not. In other words, once you have gotten to this point, you have one of three scenarios (sticking with the idea the player has chosen door number1)

1. car behind door1 and host has opened door 2 or 3
2. car behind door2 and host has opened door 3
3. car behind door3 and host has opened door 2

Each of these scenarios has a 1/3 chance, in 1 you loose by switching and in 2 or 3 you win by switching thus the odds are the same as with a knowledgeable host switching gives a 2/3 chance of winning. Clarence3456 (talk) 20:03, 25 September 2008 (UTC)

This is a notable variant (many references discuss the difference between this variant and the standard one), so I think it should remain in the table. The odds are 1/2 because there are 6 equally likely events:

1. car behind door1 and host has opened door 2
2. car behind door1 and host has opened door 3
3. car behind door2 and host has opened door 1
4. car behind door2 and host has opened door 3
5. car behind door3 and host has opened door 1
6. car behind door3 and host has opened door 2

but because we're given that the host happened to open a door that didn't show the car we know we must be in one of (still equally likely) cases 1,2,4, or 6. We don't know what would happen in cases 3 or 5 so we simply ignore them, but that doesn't increase the probabilities of cases 4 and 6. If we're given the host didn't show the car, we're only looking at 4/6 of the total cases and among those there's a 50/50 chance of winning by switching. -- Rick Block (talk) 13:39, 26 September 2008 (UTC)

I'm not disagreeing with your explanation Rick, but I think the best method of showing the probability is 1/2 is to compute the number from an indisputable probability - that being: the probability is 0 that the winning door was chosen if the car is revealed. Randomly opening a door entails two results - revealing a goat and revealing a car. Neither of these can possibly alter the fact that the winning door was chosen with overall probability of 1/3 but each result does alter the probability in a particular case. Since the overall probability cannot change these different results must cancel each other out. One third of the time the randomly opened door reveals a car and when that happens the probability that the chosen door conceals the car is 0. Two thirds of the time the randomly opened door reveals a goat and when that happens the probability the chosen door conceals the car is x. 1/3 * 0 + 2/3 * x = 1/3 is the equation that describes the state of affairs. Solving for x one gets 1/2.

It is generally treacherous to consider one outcome of a random event in isolation because its easy to lose sight of the importance of other outcomes. Even if you avoid the pitfalls you can often use the others to verify your computations.Millbast5 (talk) 01:41, 27 September 2008 (UTC)

You can use this same argument to show that the host intentionally opening a door he knows conceals a goat cannot affect the probability in a particular case. As before let x be that probability. The probability is 1 that he will reveal a goat so the equation is: 1 * x = 1/3, i.e. x = 1/3. Or you could say that since there is no alternate outcome to cancel whatever change you might hypothesize, the change must be zero.Millbast5 (talk) 01:58, 27 September 2008 (UTC)

Milsbast5 I see you and Rick are correct. An alternate to Milsbast's equation that helps me understand is that the total probability must of course be 1, and as you state, if the host reveals a goat, the chance that door contains the car is 0. The chance the each other two contained the car was equal at the start of the game and no actions have changed that, so call the odds that either door contains the car x. So overall: 0 + 2*x = 1. So x must be 1/2. If you guys want, you can delete this section of the discussion to keep it clean or leave for others.Clarence3456 (talk) 19:02, 27 September 2008 (UTC)

Clarence your argument is simpler and simplicity is always good. The reason for my more circuitous argument is that it ties together two things which should be tied together - the different results that come from one random event. It also explains why opening a door at random and revealing a goat by chance is not the same thing as opening a door known to conceal a goat, even though to the casual observer they are identical. —Preceding unsigned comment added by Millbast5 (talk • contribs) 01:14, 28 September 2008 (UTC)

I keep trying to think of new ways to explain this problem so that even someone who firmly believes that the two strategies are equally good can finally get it. Here is another go at it.

Imagine there are two players involved, named N and S - initially they have to guess the same door but after the host has opened a door, S has to switch to the remaining door. Also imagine, to begin with, that the host opens a door at random. That is extremely easy to analyze - each of the three participants has a 1/3 chance of having or opening the door that conceals the car. You will note that in the cases where the host opens a door concealing a goat, one of the other two wins and they win equally often, of course, since they each get the winning door 1/3 of the time.

Those who believe that it matters not a whit whether the host opens a door he knows hides a goat or he opens a door at random and reveals a goat by chance, can point out that none of the trials where the host did the latter would have come out any different if he picked the same door on purpose. Couldn't possibly get a different result in those cases - N has same door, Host opens same door so S is forced to take the same door. Case closed - it doesn't matter.

Now we consider the cases where the host opens a door at random to reveal a car, which happens 1/3 of the time. Can't forget those cases if the host is going around opening doors at random. So N has a goat, host shows car, S stuck with the other goat. Now pretend nothing changes except that the host is not allowed to show the car. He's also not allowed to take N/S's initial choice so he has to take the door S was eventually forced to take in the random play - there is no other door to consider. Well, now S is forced to take the door that hides the car. Clearly not fair - S gets every car the host showed in the random play in addition to the ones he got when the host randomly revealed a goat. N gets exactly what he got before.

So if one focuses on the cases where the host opens a door randomly and reveals a goat then the result is the same if he opened the same door because it hid a goat. On the other hand, if you consider all the cases where the host could open a door he knew concealed a goat - which is every case - then there is huge difference. S wins twice as many cars as N does.

What should you do if you don't know why the host opened the door to show a goat (as you don't know in the original problem statement), you still jump at the opportunity to switch because you never gain by not switching and you might gain big time by switching. Heck, even if you still believe that switching and not-switching always end up of the same you ought to switch because of the small chance you are wrong and the, so-called, experts are right - stranger things have happened. Of course, if you have formed a strong attachment to the door you chose at the outset or if you have a lurking suspicion the host knows you picked the right door and is trying to trick you into switching, then defy the odds and stick with it. Imagine the mental anguish if you chose the winning door originally and switched - the experts haven't factored that into their analysis!Millbast5 (talk) 00:47, 28 September 2008 (UTC)

Correction! While its true that you never gain on your 1/3 chance of winning by not switching, its possible to fall below winning 1/3 of the time by switching - whenever the host is allowed the option of not opening a door at all. He may, for example, open a door only when your first guess was correct. The general rule governing it is: if the host opens a door known to hide a goat P percent of the time when he knows you've picked the winning door and does the same less than P/2 percent of the time when he knows you've picked a losing door, then the switch strategy wins less than 1/3 of the time.Millbast5 (talk) 05:39, 28 September 2008 (UTC)

I agree that a more convincing explanation is required. Why not add your suggestion to this page user:Martin Hogbin/Monty Hall problem (draft), where it can be discussed and improved with a view to adding it here when it is ready and acceptable.Martin Hogbin (talk) 09:13, 28 September 2008 (UTC)

I see that you have.Martin Hogbin (talk) 11:05, 28 September 2008 (UTC)

Reversion of good faith edits marked as minor

I have seen at least two occasions where good faith edits to this article have been immediately reverted and the reversion marked as a minor edit. This is extremely rude and contrary to WP policy. I understand that editors wish to maintain the quality of this article but this is not the way to do it. Martin Hogbin (talk) 08:34, 3 October 2008 (UTC)

I suppose you refer to edits like this and this (perhaps also this). You're absolutely right. Except, it's maybe a little bit rude to call others extremely rude because they made a slight mistake. (And now I nearly did the same!)--Noe (talk) 11:10, 3 October 2008 (UTC)

I didn't deliberately mark my revert as minor -- looks like Twinkle did that for me, and it appears Huggle does the same. I'll change my settings. I explained my reasons for the revert in the summary line -- the edit was pure OR, something this particular article suffers from a lot. Jonobennett (talk) 12:50, 3 October 2008 (UTC)

Generally, reverting an edit occurs because of vandalism, so all software features dealing with this automatically mark it as a minor edit. This is true of the admin rollback button too. Now, whether these software features should be used to revert good faith edits is a different issue. Mindmatrix 13:38, 3 October 2008 (UTC)

OK, I accept the points made and maybe overstated my case a little, but I first came to this page in response to an RFC which claimed that the current editors were being overly protective of the page. Immediate reversion of somebody's work, with that reversion being marked as minor, is not welcoming to new potential contributors. As I have said before, I believe that the very strict enforcement of no OR might be relaxed a little to the benefit of the article. Martin Hogbin (talk) 16:45, 3 October 2008 (UTC)

Problems with the current Souces of Confusion section

The first two paragraphs give a reasonable overview of what other people have said about the sources of confusion for this and similar problems.

The first sentence in paragraph three is: "A competing deeply rooted intuition at work in the Monty Hall problem is the belief that exposing information that is already known does not affect probabilities (Falk 1992:207)." This gives the impression that exposing information that is already known does affect probabilities, while the truth is that our deeply rooted intuition, in this case, is right on target. The Falk paper gives the same (false)impression but whenever the idea was used the problem turned out to be a mistaken belief that no new information was or would be exposed. In most cases the author pointed out the new information himself but in no case did exposing only known information change a probability.

In this problem the question comes up relative to host intentionally opening a door he knows conceals a goat. The old information was that there was a door, unpicked by the contestant, which hid a goat, and the new information is the location of one of the goats - which has no effect on the probability of the contestants initial choice being right but a great effect on the probability of winning by switching. In the Falk paper referenced (2nd reference) the author does point out that in the variant games, alluded to in paragraph 3, the mistake was in failing to recognize new information was imparted, not that the deeply rooted intuition was wrong. The current section does not mention that and thus solidifies the false impression given before.

Try to imagine the computation of a probability that you would agree would be different if information you knew hadn't been given to you a second time. Yeah, it fairly boggles the mind that anyone would think such a thing is possible. (Just imagine how repeated exposures to same information might give a whole sequence of different answers - it would never be safe to compute a probability!)[This paragraph added in edit]Millbast5 (talk) 02:04, 9 October 2008 (UTC)

The fourth paragraph is all about variant games - those in which the host has a preference for which door to open when there are goats behind both doors the contestant didn't pick. In the Falk paper it is made clear that for that preference to have any effect it must be known to the puzzle solvers - which is not mentioned in the current section. Since no such preference is mentioned in the puzzle, it has no effect whatsoever - that is to say, the whole subject is irrelevant. Clearly no confusion about the puzzle this article is concerned with is going to be eliminated by the fourth paragraph but, since it fails to clearly delineate its subject matter, more confusion can be added.

By the way, if the information were added it would not change the odds of winning by switching, it would only increase the odds in some cases and decrease them an exactly compensating amount in other cases. That is just how irrelevant the whole subject is to the current problem.68.2.55.158 (talk) 01:35, 9 October 2008 (UTC)

I'm not sure why but the preceding edit identifies me by number. In case you're wondering, my name in here is Millbast5.68.2.55.158 (talk) 01:41, 9 October 2008 (UTC) Okay, now I see... it didn't automatically log me in like it did other timesMillbast5 (talk) 01:44, 9 October 2008 (UTC)

I think the point is that if we view the action of opening a door as providing no additional information pertaining to the initial pick (i.e. we already know one of the two unpicked doors is a goat so knowing which does not change the probability of our initial choice being the car) then we're being misled, at least in some circumstances, just as much as we're misled by the "n doors means the probability must be 1/n" intuition. I forget the examples Falk uses for this, but this "no new information" intuition certainly can lead to incorrect conclusions - for example in the (Morgan et al.) case where the host has a preference for one door over another. The main point she (Falk) is making is that people have intuitions about probability, but these intuitions often interfere with the actual solutions and, conversely, solutions appealing to these intuitions are generally inadequate.

The fourth paragraph is not about variant games, but about the nature of the question. The variants are introduced only to show the difference between the two potential questions that might be what is being asked. Many solutions address the unconditional question (ignoring which specific door the host opens, as if the question is raised before the initial pick and before the host opens a door) even though the general statement of the problem clearly (to some) is a conditional question (specifically, with knowledge of which door the host has opened). This particular "confusion" is the main topic of the Morgan et al. and Gillman papers.

If you can clarify these points, please do so. -- Rick Block (talk) 02:58, 9 October 2008 (UTC)

I am more than happy to clarify matters.

Here is an example taken from Falk's article - 'second' refers to our "no-news, no-change" belief, as she calls it. It makes my point very clearly that the error is always in evaluating 'no-news' and never in the intuition/belief itself.

I agree, it is not always obvious what information is gained by a particular action and it is is easy to think that it is none when that is not, in fact, the case.Martin Hogbin (talk) 20:22, 9 October 2008 (UTC)

It is often the case, as in the problem below, that no new information is given about one thing but there is about something else. Someone focussed on the first thing might well jump to the conclusion there is no new information at all. I'm sure that happens with MH, too - host opening a door to disclose a goat reveals no new information regarding the contestants chosen door hence it contains no new information.

Now we just have to bolster your, and others', faith in your intuition that no new information invariably means no change in probabilities!Millbast5 (talk) 22:07, 9 October 2008 (UTC)

>>The second, mentioned in passing, is embedded in Dick’s attempt to induce the warden to tell him the name of one of the other prisoners who will be executed. Dick argues that he already knows that at least one of Tom or Harry is to be executed. Therefore, he maintains, he will receive no information about his own fate by getting one of these two names. If nothing new has been learned, no change in the probabilities should ensue. This is the “no-news, no-change” (or “I’ve known it all along”) belief. The uniformity belief is seldom questioned. It appears to be second nature for many naive, as well as statistically educated solvers. As for the “no-news” argument, it has convinced not only the warden, but most of the readers of the three-prisoner problem as well. It leads to a probability of pardon of l/3 for Dick (the correct answer for this particular problem)...<<

Dick tells the warden he will receive no information about his own fate and he was quite right about that, whether by accident or design. However it is quite clear that the warden did communicate some information and that information can be used to change the calculation of both of the others chances for a pardon - Harry's cut from 1/3 to 0 and Tom's raised from 1/3 to 2/3.

My point is that while we do make mistakes about whether new information has been communicated that in no way indicates that our belief that "no-news, no-change" is not precisely correct. When Falk states that the warden and problem readers were convinced by the "no-news" argument, isn't it clear that they erred in believing that no news was communicated?

When Falk says, "If nothing new has been learned, no change in the probabilities should ensue." she is entirely correct. Since a change in probabilities did ensue that meant something new had been learned and if one goes through the calculation of the new probabilities you can see exactly what that new information is - the identity of one the prisoners to be executed. (Not that it wasn't obvious beforehand.)

There are subtlties in evaluating information but our intuition that "no new information means no change in probabilities" is as solid as a rock.

While it might seem that Falk is indicting our intuition in this regard, on page 207 she correctly points out that the problem lies with the mistaken assumption that nothing new has been learned - she is discussing a different puzzle at that point but the same principle is at work. Without that I might have been concerned that she was as confused as the attributions used in the current "Sources of Confusion" make her seem.

Rick, I will promise you that if you bring forth any purported example where no new information changed probabilities, I will point out the new information. Heck, if you go through the two probability calculations yourself, its hard to believe you won't spot it right off.

Yes, we do have intuitions about probability that lead us astray but bringing up one that never does and treating it like it was a myth or an old wives tale is a terrible way to clear up confusion. Frankly, I think it would be very encouraging to the readers to point out that some of their intuitions are dead right.

I do have to agree with you that conditional and unconditional statements of the puzzle are part of the subject matter of the fourth paragraph. I should have indicted that aspect of it as well. The goal of the section is to reduce confusion about the central problem of the article and dragging in conditional and unconditional probability without a thorough discussion of what they mean is guaranteed to increase confusion - in virtually any context. Perhaps in some remote corner of the article, with warnings posted and a thorough introduction to the concepts one might profitably broach the subject - though, I doubt it.

I stand by what I said regarding the variant games - I assume their introduction was well intentioned but bringing up a variant of the game then proceeding to say and I quote, "In its usual form the problem statement does not specify this detail of the host's behavior, making the answer that switching wins the car with probability 2/3 mathematically unjustified. Many commonly presented solutions address the unconditional probability, ignoring which door the host opens; Morgan et al. call these "false solutions" (1991)." makes it sound like the solutions given in the article are erroneous.

How can that do anything but create more confusion? Solving the variant games was not hard and after doing it, I still had no idea what these statements meant. Exactly what is mathematically unjustified? "false solutions"? I suppose I could somehow dredge up Morgan et al and see what they were talking about but after doing it with Falk and finding that it did not support what "Sources of Confusion" purported, I figured that would be a similar waste of time. And on a subject that is far better left out altogether.Millbast5 (talk) 13:51, 9 October 2008 (UTC)

Of course if no new information has actually been imparted the probabilities cannot change, just as lacking other information about which of N alternatives might be the correct one makes the probability of each 1/N. These are BOTH "true" intuitions but they resonate so powerfully that they are easily misapplied.

The Morgan et al. paper is available online (for a fee) but should also be available in nearly any university library. They argue 1) the problem as generally stated should be interpreted as a conditional problem, and 2) since the problem statement (per vos Savant) doesn't specify the host opens one of two goat doors with equal probability we must treat the host's preference for one door over the other as a variable p. Given these assumptions, the general 2/3 answer is either non-responsive (answering the unconditional question, not the conditional question) or incorrectly assumes a constraint on the host that is not imposed by the problem statement. They argue the "correct" answer to the vos Savant version of the problem (granting that the host must open a door, and won't reveal the car) is that the probability of winning by switching is 1/(1+p), i.e. somewhere between 1/2 and 1 depending on the host's preference for the door that the host opens (in the case where the host has a choice). Relating this back to Falk, this is a case where the "no news" intuition (nothing the host does can affect your initial chance of winning the car, which must remain 1/3) leads to an incorrect answer.

Are there specific changes you're suggesting should be made to the article? -- Rick Block (talk) 14:11, 10 October 2008 (UTC)

Naturally I do have some suggestions for what might reduce the confusion of the readers though I do believe we can do a much better job of providing solutions and try to eliminate confusion that way. The primary source of confusion about this problem is that readers have a very strong intuition that it doesn't matter whether the host purposefully reveals a goat he knows is behind a door or he opens a door at random and by chance reveals a goat. The latter leaves the switch and non-switch strategies equally good - they can reason that out quite well - so how could the former not do the same? Anyway, I will address that source of confusion in a short while.

For the moment I will delve into the issues of paragraph 4.

Recall that the problem posed in "Ask Marilyn" did not state that the contestant picked door number 1, it said "say No. 1" which means that the particular door is not significant in the English speaking world and he's choosing a particular one just for purposes of illustration. Likewise the host opening door No. 3 is purely illustrative. That means the intent of the question is not what is the probability of the contestant winning with different strategies in the case he opened door 1 and the host opened door 3, the question is clearly what are his chances of winning the game with either of two strategies. Not only is that question answerable, it is answerable in case the host has a preference for some doors over others and the answer is the same!

Clearly Morgan et al interpreted the question to be about a particular pair of doors because to fully specify the hosts door opening preferences 3 parameters like p are needed - one for each possible pair of doors he may have to choose between, 1 & 2, 1 & 3, and 2 & 3. Why they choose that interpretation is anybody's guess, but mathematicians do things like that - take a different view and you might have another interesting problem. (I did the same thing you might recall - I chose an interpretation where the puzzle had to solved exactly as stated, without any knowledge or assumptions about the hosts behavior beyond what the puzzle statement supplies. I knew that that was not the intent of the question - for one thing its far too difficult to ask of a general audience. It was too complex to be included in this article, too - it was considered "OR", I believe.)

For illustrative purposes say the contestant picked door 1 and p is the probability the host will pick door 2 to open given the choice between doors 2 and 3. Indeed the probability of winning using the switch strategy with door 2 being opened is 1/(1+p) and the corresponding chance of winning with door 3 being opened is 1/(1+(1-p)). So what are the chances of the switch strategy winning? All we have to is multiply the two probabilities above by the probability of each of the doors being opened and add the two products together - childs play. Door 2 is opened with probability 1/3 + p/3 and door 3 is opened with probability 1/3 + (1-p)/3 thus the chances of winning via the switch strategy is: (1/3 + p/3) * 1/(1+p) + (1/3 + (1-p)/3)*(1/(1+(1-p))) - the algebra is easy and the answer is 2/3, as we knew it would have to be.

Door 1 was just illustrative and since the answer is independent of probability p, the answer will be the same for any other choice by the contestant. So the answer to the problem is that the contestants chances of winning by switching are 2/3 - versus 1/3 by not switching.

I suppose by their peculiar lights it isn't mathematically justified to say the probability of winning by switching is 2/3 when it varies according to which door is opened. Of course, if you do not know what value p has, then you only know one door may be better than the other but not which and by how much. Yet you still know the probability of winning by switching is 2/3 regardless of p - what are they going to say to that? What could they say? "Yeah, but if you don't know p you can't tell what are the chances when Door 3 is opened and thats what the puzzle asked and intended to ask." Frankly, I'd disagree with them if the puzzle had been stated without the writer making it clear the mention of doors was only for illustrative reasons - and so would almost everyone else. I suppose if one of them were contributing here we could have a philosophical-mathematical discussion about how to state puzzles so only one interpretation is possible.

Also note that the puzzle as stated here only asked if switching was better and thats true as long as p is not 0 or 1 and the appropriate door is opened with each value - door 3 when p=0 and door 2 when p=1. In those extremal case not switching is equally as good, not better. If you don't know what p is then you're clearly better off switching since you can never do worse than by not switching and you may do vastly better.

How are the readers going to be helped by going through a mess like this? None of them interpret the problem the way Morgan et al did so one would have to have a very good reason for delving into it. Yes, you could explain that if the contestant did know the value of probability p for the pair doors he did not pick, he'd have a better idea of whether he won or not depending on which door the host opened for those few seconds before the door he settled on was opened to make the probability either 1 or 0. You would also explain that there was no way for the contestant to make use of his knowledge of probability p - except if p was 1 and door 2 was opened or if it was 0 and door 3 opened - then he could not switch and win with the same probability as switching (1/2). And then you could reassure him that the contestants chances of winning by switching were still 2/3, regardless of the value of p.Millbast5 (talk) 13:23, 11 October 2008 (UTC)

Your second to last paragraph above is exactly the same as the conclusion Morgan et al. come to - i.e. if you don't know p you might as well switch since by doing so your chances of winning don't go down but may go up considerably (and, people who a priori decide to switch before seeing which door the host opens win with 2/3 probability). I don't think anyone takes "let's say" as anything other than illustrative - on the other hand, I think the point Morgan et al. make is valid. Their point is that the player is not making a decision at the beginning of the game, before picking a door and before knowing which door the host opens, but after picking the initial door (illustratively, door 1) and after the host has opened a door (illustratively, door 3). At least one of the experimental psychology papers about this problem indicates that this aspect of the problem (putting the player in front of two closed doors and one open door at the point of attempting to evaluate the probability of where the car is), is what makes it resonate so strongly with the "must be 1/N" intuition. Roughly a year ago (in the talk archives) there was an extended discussion here about this, which led to changes to the article including introducing the paragraph you're objecting to. To a large extent, the issue is not what we think but what the published sources say. We could certainly rephrase things for better clarity. Is this what you're suggesting at the very end of your post above? -- Rick Block (talk) 15:27, 11 October 2008 (UTC)

I agree with Millbast5that the issues addressed by the likes of Morgan et al are essentially a diversion from the main question. The current article has too much about 'versions' of the problem but does not provide a convincing resolution of the central paradox. Martin Hogbin (talk) 20:07, 11 October 2008 (UTC)

I'm glad to hear that Morgan et al recognized that short of knowing 'p' took the value 1 or 0, their analysis is entirely without benefit to the contestant and that in no case could he actually gain from it. Of course that means that while the contestant doesn't have to make a decision until the door is opened, he can't do any better than he could by deciding whether to switch or not before the game started. If one looks at their paper in that light perhaps there is a place for a reference to it - in case someone wondered if there wasn't some way putting off the decision might help the contestants chance of winning. However, to appear in the 'Sources of Confusion' section it needs to be enlightening to the puzzle in general, and I see the objections I made originally as still being in full force.

I just had a cool idea - a way that someone who didn't believe switching would improve his chances of winning could hedge, just in case he was wrong. Suppose he thinks door 2 is the door with the car behind it. Instead he picks door 1, intending to switch to door 2 when the host gives him the option to switch. The only way this play could go awry would be if the host opened door 2 to reveal it as a loser - and then he'd have to be glad he didn't pick it and stick with it. Of course, with door 2 out of play he has a real choice about switching. Heck, why not rank all three doors as to his subjective feel of their chances of concealing the prize and he intially picks the worst of the three so he will get to switch to one of his first two choices for sure. He could explain that rationale to his friends and family if they started to tease him about switching from the winning door - and they might even buy it. Or tell them his plan ahead of time and then he probably wouldn't panic under the gun and not switch - lets face it, switching is wussy.

I'm not sure it belongs in the Sources of Confusion section but I've been thinking that someplace in the article we should bring up the idea that people do not like to look indecisive and the switch strategy does give that appearance. Few would fail to switch on that basis if they did believe switching doubled their chances of winning but it probably is part of the reason why everyone who believes the two strategies are equally good elects not to switch.

Sorry, I'm getting off topic here.

Rick, I'm in full agreement that the contestant being faced with two closed doors stimulates the automatic response that they are equally likely to hide the prize. I can't remember if I explained that I believed one reason the urn and balls version of the puzzle is easier to understand is because that image of the contestant is not called to mind. Also it does not appear that the person who knowingly removes the white ball from the urn, after the subject blindly picks a ball, is an adversary while the host is still rather suspicious in that regard.

Published sources should attributed wherever their ideas and results are used but, outside of history and 'related topics', we should not look to refer to them simply because they exist if they do not help us achieve our goals.

I also believe that in an article that is about a popular topic we should have a goal of making the article interesting or even entertaining to read because people will not be enlightened if reading the article is so boring they stop reading. A little lattitude with OR ought to allowed, too, when the subject is just not that serious. Something publishable in a learned journal is not what I mean, of course, but 'novel and creative' enhance an article.

With popular topics, such as this, an article that one has to find the disclosing authority for and then pay thirty bucks to eyeball it, is about as close to useless as you can get - again outside of history and related topics. So if you can't quote enough of it to be helpful you should try to take a course that avoids it altogether - paraphrasing is rife with pitfalls, especially when it is done with an eye to keeping it short. Look what happened with Falk and Morgan et al in the current section - Falk didn't conclude what the section implied and the paraphase of Morgan highlighted their conclusions that made it sound like the puzzle solution was chock full of errors. As I mentioned above the important conclusion from Morgan is that knowledge of the hosts preference for opening doors doesn't affect the chances of the contestant winning, yet that part got left out.

I disagree with the tone of the article (and many of the references) in regard to people's intuition about probability - they are, in fact, generally quite good and we should try to explicitly use them in presenting solutions. Thus they will be nodding their heads and thinking "yes, I do think just that way" instead of feeling like they are being treated as dolts for thinking/believing as they do.Millbast5 (talk) 23:23, 11 October 2008 (UTC)

Isn't the first part of the solution section about as simple as you can get? Three equally likely scenarios, switching wins in two and loses in one, ergo 2/3 chance of winning by switching. The subsequent deeper analysis, using the Morgan et al. reference, addresses a mathematically significant criticism of this solution which is that the solution doesn't utilize the "equal goat door" constraint imposed on the host. Without this constraint, the simple solution may be incorrect making it at least incomplete. You might argue that no one would interpret the problem to mean that the host exhibits a preference for one door over another, but if you don't include this constraint that's exactly what a mathematician would do (find a hole in the argument). If you want the answer to be 2/3 chance of winning by switching, and you want the decision to switch to be made after the host has opened a door, you have to include the "equal goat door" constraint and the solution has to use this constraint. Anything less and you're either asking a different question (like what's the probability of winning if you decide to switch before the host opens a door knowing that the host will open a door), or 2/3 isn't always the right answer. -- Rick Block (talk) 04:30, 26 October 2008 (UTC)

Proposed substitute for the last two paragraphs of Sources of Confusion section

Just for the sake of variety, you get to be the contestant for a while.

There is no question that the most difficult thing to evaluate is the effect of the host intentionally opening a door he knows hides a goat after you've made your initial pick. Everyone can figure out the effect if the host just picks one of the two available doors at random and opens it. A third of the time the action will reveal a car and you'll know immediately that your door and the switch door each hide the car with probability 0, and two thirds of the time the action reveals a goat leaving your door and switch door each harboring the car with probability 1/2. Our intuition is very good about that. If the host jumped ahead of you and opened a door at random before you could pick the outcome is exactly the same. Our intuition is all over that one. Suppose the host didn't wait for you to pick but opened the door he knew hid a goat before you picked. That would be great, you'd win half of the time instead of a third of the time. Also the 'switch' door would hide the car half of the time. Our intuition nails that in milliseconds!

One thing that has been true in every case is that your picked door and the 'switch' door have hid the car with equal probability so it is very tempting to assume the same holds when the host opens the known goat door after you pick and your intuition sends you a strong 'buy' signal on that assumption. Still the smart thing to do is check it out before you buy.

One good way to get a handle on it is to imagine the game being played 9 times all at once. First you make your nine choices. But then you're allowed to send an observer to look behind the doors - she reports that there are nine cars altogether and 3 of them are behind doors that you picked. Now the host goes through all nine games opening a door he knows conceals a goat. The observer does her thing again and reports all the cars are just where they were before - no funny business going on. So there are 18 doors that are closed and three of the cars are behind your 9 of them - that leaves six cars behind the other nine doors. You do the math again to make sure. It doesn't help - there twice as many cars behind the 'switch' doors as behind your doors. Is that darn observer being truthful? What she says makes sense - 3 cars out nines tries is normal and you really didn't expect any more cars to move behind your doors so it sure seems like she is telling the truth. There's no way out of it - the 'switch' doors are twice as likely as yours to harbor a car.

Why does our intuition get this last case wrong? In every case the image we have in our minds is the same - we are standing before two closed doors and one open door, and leaving out the random cases where the open door reveals the car, there is a car behind one of the two closed doors and we don't know which. Generally speaking, when we don't know which door its behind its equally likely to be behind either of the closed doors. It seems only fair - heck it seems right - that it should be equally likely. Even when it is not equally likely but we don't know which door is more likely, its the same as it being equally likely as far as we're concerned. This particular case is the oddball - the doors are not equally likely and we know which one is more likely. It just isn't apparent until we exercise our imagination a bit.

We still can't really fault our intuition because it led us through the 9 games demo just fine, taking it one small step at a time.Millbast5 (talk) 04:44, 13 October 2008 (UTC)

Here is another go it it, where I focus on tracking down the source of confusion

Another possibility for the Sources of Confusion

Comment: This version directly shows "why the probability isn't 1/2" so it replaces the section by that name as well.

The source of confusion is not the puzzle itself. It arises from contemplating what would happen if the host opened one of the 'other' doors at random. The solution in that case is gotten right by virtually everyone: the contestant picks the winning door 1 time in 3, the host opens a randomly chosen door to reveal the car 1 time in 3 and the unpicked and unopened door hides the car 1 time in 3. Now comes the confusion: if you forget about the 1 time in 3 the host reveals the car then the contestant wins 1 time in 2! Of course, if you eliminate half the losing choices for whatever reason you want, the contestant would have winners half the time. Its just simple numbers. It means nothing and means no more if you phrase it this way: when the host reveals a goat by chance the contestant wins half the time. Clearly that is just another way to say you are eliminating half his losing choices. Now the source of confusion leaps into full view - if you attribute the change in frequency of winning to the revelation of a goat instead to the fact that you're disregarding half his losing choices you have the makings of a real misconception.

In the actual problem only goats are ever revealed but if one believes that the revelation of goats causes the contestants chances of winning to increase, it ought to do it even more when more of them are revealed. And so it does, according to the faulty belief - instead of the contestant winning 1 time in 2 when half his losers are removed, his original pick wins 1 time in 2 overall.

This belief gets a boost from another source - the mental image of the contestant having to pick between two closed doors, one of which hides the car. The contestant doesn't know which one, and generally this means that the car is behind either door 1 time in 2 - perfect corroboration! The belief becomes so solid that nothing can shake it. (This is not speculation as a number of the referenced sources demonstrate the absolute conviction of many believers - some of them get irate if one continues to challenge the belief.)

One way to combat the belief is to consider a number of plays of the game simultaneously so as to make the original picks more concrete. Imagine for example, 99 plays of the game in which the contestant picks 33 winners and 66 losers - which is, statistically, the most likely outcome. Now its clear that if no cars or goats are shuttled to and fro behind his chosen doors one can do anything one wants with the 198 other doors and the contestants 33 winners and 66 losers will not be altered. One can open those doors at random to ones hearts content or look behind them and expose goats till the cows come home and the contestants 33 winners and 66 losers sit there undisturbed.

Phrased in frequency terms - the contestant picks the winning door 1 time in 3 and the host purposefully revealing goats does not change that fact. It applies to the imagined contestant choosing between two closed doors, too. True, he doesn't know for certain which door hides the car but he does know his initially chosen door still hides the car 1 time in 3 so, inevitably, it is behind the other door 2 times in 3.

As is apparent at this point, the solution to the game is extremely simple if one just proceeds without doubting that the initial pick remains a winner 1 time in 3 after the host opens a door to reveal a goat. One easy logical step shows the remaining door, the 'switch to' door, hides the car 2 times in 3. This simplicity probably works against it as far as engendering belief - the problem baffles our immediate intuition so it surely it can't be as simple as subtracting 1/3 from 1!Millbast5 (talk) 11:07, 18 October 2008 (UTC)

I think you are quite right in where the confusion lies, but I am not sure that you have found a way of making that clear to the reader.

As you point out at the end, if we take what has been described here as the unconditional case, then there is a a very simple and convincing argument - you have a 1/3 chance of initially choosing the car and you always get the opposite if you swap. With the reasonable assumptions that most people make (host aways offers swap, aways opens a goat door) the same argument 'happens' to apply for the conditional case (since the host action does not change the probability that the player has chosen a car). I therefore suggest that the conditional/ unconditional issue is something of a red herring and should take a secondary place in the article. In the game, as understood by most, the conditional version of the problem has the same odds as the unconditional one. It might have been otherwise, but it was not (as most people understand the problem).Martin Hogbin (talk) 12:27, 18 October 2008 (UTC)

I inserted a new phrase into the middle of the third paragraph for increased emphasis.

I'm not too happy myself with the clarity of the what I wrote. That first paragraph is too long and I don't get to the actual statement of the source of confusion until the end of it. I've also considered starting with the other half of the false argument - the contestant faced with two doors, etc, but I believe the confusion does start with contemplating the effect of opening a door at random. I imagine it varies from person to person where it starts. In any case, if you have any thoughts about how to improve it, I'd be glad to hear them.

You know you don't have to sell me on the idea of eliminating any mention of conditional and unconditional probability from the article. Of course, I'm using the idea myself in talking about the odds under the condition that a goat was revealed but people are quite used to the idea as we phrase things in that way in our everyday speech. "If TMac stays healthy the Rockets have a good shot at the title." We understand perfectly that different conditons change the odds but developing the formalism is out of place in an article like this and using the terms 'conditional' and 'unconditional' without developement is worse. Frankly, I'm not sure at all what you (and others) mean when you use those terms in these discussions. Imagine how someone feels who is a bit shaky when the word probability is used all by itself.

Yes I agree. I was not really making my conditional/non-conditional point to you but to others who might be reading. I think this article suffers from what might be called 'wiki correctness'. In other words it keeps to all the rules but does a bad job. Martin Hogbin (talk) 23:26, 18 October 2008 (UTC)