Talk:Monty Hall problem/Archive 2

message: I've moved the existing talk page to Talk:Monty Hall problem/Archive1, so the edit history is now with the archive page. I've copied back the most recent thread. Hope this helps, Wile E. Heresiarch 04:40, 10 Aug 2004 (UTC)

message: :Bayes' Theorem Approach in the main page:
:P(C2|03)= P(03|C2)*P(C2) / P(03) (True)
:But
:P(C1|03)= P(03|C1)*P(C1) / P(03) (True also)
:and what do we have: if P(C2|O3)=1*1/3 / P(03) were 2/3, in the same way P(C1|03)= 1*1/3 / P(03) = 2/3. Summ is greater than 1. Impossible!
:Let's proove that P(C2|03)= P(C1|O3)= 1/2
:P(C2|O3)-P(C1|O3) = (P(03|C2)*P(C2)-P(03|C1)*P(C1)/P(03))=0
:because P(O3|C2)=P(O3|C1)=1; P(C2)=P(C1)=1/3
:so P(C2|O3)=P(C1|O3)
What's wrong?
:P(C2|03)= P(03|C2)*P(C2) / P(03) = 1*1/3 / P(03),
:but P(03) is not 1/2 as it says on the main page. Imho it is 2/3.
:P(03)=P(E3)*1+P(C3)*0 (~)
:where P(E3) - 3rd is empty, if it is, so Monty opens (1) it
:if there is a car (P(C3)), Monty doesn't (0) open it.
:and P(E3)=2/3, just because we are in the begining of the show and no box is opened yet. 
:so P(C2|O3)=1/2=P(C1|03)
Am I right or Am I right or Am I right?
:The only problem here can be only in (~). Am not really good at mathematicain english to proove now that it's okay. But can 
try to do it later.
::Best regards, Fade <fade@list.ru>
--213.148.18.178 10:57, 17 Jan 2005 (UTC)
thread_title:  IMHO - Just Mathematics - probability is 1/2 

message: It's critically important to the problem that Monty knows what's behind each door.  In the actual correct problem situation, there are three scenarios for what's behind each door:
Scenario  	Door 1	Door 2	Door 3
Scenario A	Car   	Goat  	Goat
Scenario B	Goat  	Car   	Goat
Scenario C	Goat  	Goat  	Car
Let's presume that the player always first chooses Door 1 (a safe assumption, since we can always just choose how the doors are numbered after the door is chosen.)  If we assume that Monty knows what's behind all doors, and opens the door he does because it has a goat behind it, then our three scenarios look like this (with the revealed goat crossed out because the player knows to avoid it):
Scenario  	Player's first choice	Door 2	Door 3
Scenario A	Car                  	Goat  	Goat
Scenario B	Goat                 	Car   	Goat
Scenario C	Goat                 	Goat  	Car
If on the other hand we leave the problem too vague, by just saying "Monty opens a door that has a goat behind it", we may create the impression that it is just chance that, this time, the door had a goat behind it -- and this changes the problem completely.  Because it simply makes it look as if some of our original scenarios could not have happened:
Scenario  	Player's first choice	Door Monty opens	Door 3
Scenario A	Car                  	Goat            	Goat
Scenario B	Goat                 	Car             	Goat
Scenario C	Goat                 	Goat            	Car
The faulty logic thus goes "If Scenario B had been correct, there would have been a car behind the door Monty opened; but there wasn't, so therefore Scenario B is not correct.  The only remaining scenarios are A, where the player's first choice contains the car, and C, where the player's first choice contains the goat; therefore the chances are 50/50 whether the player switches or stays."  -- Antaeus Feldspar 20:44, 21 Nov 2004 (UTC)
thread_title: Monty's knowledge is critical to the problem

message: The standard solution seems illogical for 2 reasons. First, probabilities describe chances, and since Monty has removed one of the uncertainty by opening up a door, he changes the conditions which the original set of probabilities described. This is to say, the 2/3 probability should not be transferrable, and the new probability designed to describe the new conditions should state that the car has a 50-50 chance of being behind either of the unopened door. Second, say the standard solution is logical, and say the car is behind door 1, and the player will always choose between door 1 or 2, and Monty will always open up door 3: if the player had originally chosen door 2, then he should change his selection to door 1; if the player had originally chosen door 1, then he should change his choice to door 2: the results of the 1st situation contradicts that of the 2nd, and vice versa; the contradiction shows that the standard solution is wrong, and it doesn't make a difference whether to stick to the original choice or not.
:There's no contradiction. Door 1 and Door 2 are just labels, without any further information they are possesed of exactly the same qualities. Whichever door he chooses he should switch. Swap the labels on the doors and the solution is the same. Once more, by far the simplest way to look at the problem is to recognize that if the player randomly chooses a goat, and switches, he will win the car every time, and he has a 2/3 chace of choosing a goat. Now I wish someone could explain the two envelope problem to me. Mintguy (T) 17:22, 25 Aug 2004 (UTC)
replies: text: I'd love to explain the two envelope problem to you, but the author of that article won't let me, as it seems. :-( --INic 22:37, 14 Jan 2005 (UTC)
thread_title:  Standard Solution is Illogical

message: I found the section on "Assumptions" to be a little confusing and to obscure the true spirit of the puzzle/problem. I have therefore removed that section and reworded the problem in the introduction to make it clearer that the host knows what is behind the doors and that he always opens a door with a goat. Then the assumptions are redundant and the intentions of the host, benign or otherwise, are not a factor. I think it is obvious that the contestant will want a car, not a goat, and there is no need to state this. I have also reworded the answer to try to make it clearer and more concise.
P  S  K  --213.122.39.201 17:00, 25 Oct 2004 (UTC)
replies: text: I'm in agreement with the changes you made; the article was wandering into original research on variations of the problem. I think your recent edits have substantially tightened and clarified the article. By the way it would be terrific if you would make a user name and log in. Regards & happy editing, Wile E. Heresiarch 21:40, 25 Oct 2004 (UTC)
thread_title:  Remove "assumptions" and rewording for clarity 

message: I have written up what I think is a clearer explanation of the solution, at Monty Hall problem/temp1.  I'd like others' feedback, to find out if it's as clear as I think it is and can go in the article. -- Antaeus Feldspar 01:38, 1 Dec 2004 (UTC)
thread_title:  A clearer explanation? 

message: The Monty Hall problem is not a problem in people's skill at using probability and mathematics. It is not a problem of gullible people falling prey to tricks. It is not even a paradox. It is a problem in communication and rhetoric.
As commonly stated, the Monty Hall problem is ambiguous as far as how the host chooses the second door.
Whether information about the third door can be inferred by the host's actions, depends on whether the host uses information about what's behind the third door to make his decision.
If the host chooses the second door randomly among the two doors not already selected by the player, and it has a goat behind it, then there is no advantage to the player switching.
If the host chooses the second door because it has a goat behind it, then the probability that the third door has a car behind it jumps to 2/3, and the player should switch.
If someone were to ask me the Monty Hall problem, and it were not clear whether the host used information from behind the third door (or, equivalently, information from behind both the first and second doors) to make his decision, then I would come back and say that the problem was ill-defined. Neither of the two common solutions would be acceptable. 
Actually, a third scenario is possible: If the host chooses the second door because the third door contains a goat, then the player should never switch.
The point is that you cannot infer information about the third door unless you can infer information about the host's decision-making process. Unless this is spelled out clearly (which it most often isn't), the problem is ill-defined.
Neither side in this "debate" has done bad mathematics, and mathematicians who righteously scoff at those who express contrary views, calling it bad mathematics, or mathematical illiteracy, add nothing to our understanding.
I'd like a humanities professor's analysis of this problem, because it is not fundamentally mathematical. I'd like to see a detailed analysis of all of the different precise phrasings of this problem, and how they parse against accepted English grammar. What assumptions are socially "valid"?
An analysis of rhetoric is more important than an analysis of probability, in this problem.
(FWIW, I'm a computer scientist who writes mathematical software.)
Sorry, but I can't force myself to agree wtih your solutution. When the host opens the door he not only skips over the dorr you choose, but the other door. There is a 2/3 chance it was the goat...right? But no remove a goat, just remove the door the host opened. You would not, in your right mind, choose a goat against a car. Now there are two doors left. One is your door with car or goat, and other a door with same. The problem with most of the tables and charts here are that they use goat twice in two rows, making it two chance. Now, goat one and two are interchangable:
Player Chooses Goat One:Remaining Choice: Car
Plyaer Chooses Car: Remaining Choice: Car
You should not add:
Player Chooses Goat 2: Remaining Choice Car
That would not happen. There are three doors, of which one of them is open, which goat DOES NOT MATTER EVERYONE. There will always be two doors left open, not matter if the players chooses goat one or two. There will alwys be a car and a goat remaining. If the player choose goat one and goat 2 is opened, you can't make another choice be player chooses goat two and goat one is opened. With this amount of doors they result in the smae thing unless you did it more often. Try it.
Door One:                          Door 2:                         Door 3:
Goat                               Goat                            Car
Now  to help you if you still don't get it: Door One or Door two can be opened and reslult in the smae outcome. Which door they open does not matter.                               
Goat
:Ah, thank you very much. Your explanation made this clear to me. -- PenguinPirate 23:45, 24 Jan 2005 (UTC)
thread_title:  Everyone's making a mountain out of a molehill, or a Lexus out of a goat 

message: It has nothing to do with what Monty knows.
Like the solution states, there's a 2/3 probability that the car is behind one of the doors you didn't pick. It doesn't matter if the doors are purple, have aliens behind them, are invisible or any of that other nonesense. There's a 2/3 probability that the car is behind one of the doors you didn't originally pick.
:Wrong.  The fact that Monty knows is vital to the problem.
:Let's define whichever door the player picks as door P, whichever door Monty picks as door M, and whichever door is left over as door L.  For clarity, let's also distinguish all three values for what's behind the door:  C for the car and G1 and G2 for the goats.
:Now, just looking at prizes and doors, and not about the constraints of the problem, there are six possibilities for how three prizes can be arranged one-to-one behind three doors:
P 	M 	U
C 	G1	G2
C 	G2	G1
G1	C 	G2
G1	G2	C
G2	C 	G1
G2	G1	C
:Now, if we add the constraint that Monty knows what is behind each door, we can see that even if the player chooses wrongly at first and leaves Monty both a door with a car and a door with a goat, Monty can always choose a door with a goat:
P 	M 	U
C 	G1	G2
C 	G2	G1
G1	G2	C
G1	G2	C
G2	G1	C
G2	G1	C
:But if you establish that Monty does not know what is behind each door, and then say that he picks a door with a goat behind it, there is only one way to interpret that:  that there are six possibilities for how the prizes are distributed behind the doors, but we are only considering those four where the condition that "Monty's door has a goat behind it" is true:
P 	M 	U
C 	G1	G2
C 	G2	G1
G1	C 	G2
G1	G2	C
G2	C 	G1
G2	G1	C
:Just as in the old Lewis Carroll problem about "a regular coin and a double-headed coin are tossed in a bag; you take out one coin and see a heads; what is the chance that the other side is heads?" if we say that Monty does not know which door contains the car but somehow always avoids it, we can only conclude that we are artificially removing those possible outcomes where he did choose a car.  This leaves us with the unsurprising result that if the player receives no information whatsoever about what is behind any of the doors, his chances stay the same no matter which door he picks.  It is only because Monty knows what is behind each door, and eliminates a false choice for the player, that switching becomes an option more likely to be rewarded with success. -- Antaeus Feldspar 08:22, 24 Jan 2005 (UTC)
thread_title: Simple Problem

message: I realize that I am probably wrong. I mean, if this phenomenon weren't mathematically solid it wouldn't have survived for this long, but something just occurred to me.
How can my picking a certain door affect it's probability?
Consider the following:
It's the exact same set up, with three doors etc. etc.
In my mind I pick door 3, but I tell the host I pick door 2 and he believes me. In his mind I've picked door 2.
He opens door number 1 and it is a goat.
Which door has the two third probability and which has the one third?
How can the exact same scenario bring up two probabilities based on the door I "picked?" Picking out one door in my mind doesn't mathematically substantiate anything as it has no relevance to anything mathematically.
:A good question.  The answer is that the significance of "picking" a door and communicating that pick to the host, is that the host will not give you any information about that door.  In your example, you say that you pick door 3 in your mind, but tell the host you pick 2, and he opens door 1 to reveal a goat.  This could happen, but it could also happen that he opens door 3 to reveal a goat; he has no way of knowing that you "really" picked door 3.  But whichever door you tell him you picked, that has the effect that the host will not give you information that affects the probability of what's behind that door.  In the example you give, if it is door number 1 the host opens, the door with the 2/3rds probability must be door 3, and the door with the 1/3rd probability is door 2, because the host thought you picked that one.  -- Antaeus Feldspar 05:08, 3 Feb 2005 (UTC)
thread_title:  Could somebody answer this question, please? 

message: You, the candidate, pick a door. Probability dictates that in 1 case out of 3, you picked the car. This case will now be called (C). In 2 cases out of 3, you will have picked a goat. This will now be called (G).
(C): The host will open one of the two remaining doors, revealing a goat. Changing doors will let you loose, as you picked the door with the car already; STAYING with the first choice will always win in case (C).
(G): The host will have no choice which of the remaining two doors two open, as only one contains another goat, which he is bound to show. So in case (G), CHANGING the door will always win, while staying with the goat is a clear loss.
Conclusion: It can be seen easily that in 1 out of 3 cases, staying with your originial choice will be a sure win, while in more probable 2 out of 3 cases changing doors will be a sure win.
790 12:52, 19 Feb 2005 (UTC)
thread_title:  give me a try explaining 

message: Here, for those of you who don't get the math (or think you do but still think there's a 50/50 chance at the end): Monty Hall applet. Play the game over and over. It keeps track of your wins, and whether you won by switching or staying. Still not convinced? Try the iterative applet. Run it for a few thousand iterations. The statistics are borne out. -- Wapcaplet 00:20, 2 Mar 2005 (UTC)
thread_title:  Empirical demonstration 

message: The issue of the Marilyn vos Savant article was not that she was "wrong" but rather that she was mis-stating the problem. The key word she used was "guess" in regards to making the 2nd selection. By definition, "guessing" requires one to choose without making calculations. When one calculates before "guessing" it's no longer a mere guess, but an educated guess.
The other problem with her article was that the concept of odds has a dual layer in this "Monty Hall" problem which is often overlooked. The odds of finding the car do increase by switching. But the statistical likelyhood that it exists, does not. Additional information can always increase the likelyhood of a successful choice, but it does not affect the underlying statistical facts. If there are 100 piles of horseshit in your backyard and a single can of caviar buried in one of them, the singleness of the can of caviar never changes, only the odds of finding it. Suffice it to say, I have found that most people who argue about Monty Hall problems, overlook this point. Please see this variant  of the Monty2.gif from the article page.  I suggest that this puzzle would be less puzzling if the location of the car was referred to as "odds of finding card here" rather than "odds of car being here". 216.153.214.94 06:04, 24 Mar 2005 (UTC)
thread_title:  Marilyn vos Savant 

message: I agree the probabilities are 1/2 when picking the second door, and it doesn't matter why monty picks his door because if he does reveal the prize then the game ends with you winning anyway.
:From the article (emphasis mine):
::". . . after Jane has selected a door but before she actually opens it, the host (who knows what is behind each door) opens one of the other doors to show that there is a goat behind it . . ."
:Monty opening a door with a goat is part of the premise.  That is why a) there is no provision for what happens if Monty opens a car door (which he won't), and b) the odds are 2/3rd.
:I had trouble understanding this myself, until I looked it as a simple 'one of three' problem.  There are three possible situations for the first choice: You pick the car (Monty picks randomly); you pick goat #1 (Monty picks #2); you pick goat #2 (Monty picks #1).  In the first case, if you switch, you lose.  In both the other two cases, if you switch, you win.  Hence, out of three possible scenarios, switching will win two.  That's two thirds, and that's why you should switch. -- Wisq 16:57, 2005 Apr 28 (UTC)
::Why are the best explanations at the discussion page and not in the article? ;-) --Abe Lincoln 17:42, 28 Apr 2005 (UTC)
thread_title:  1/2 

message: Section "The problem" says "Nydick." This author is listed in "References" as "Nudick, Robert L." Which is correct? Graue 01:47, 25 May 2005 (UTC)
replies: text: Nydick.  He's collaborated with Liberatore on other publications.  There's a reference to the paper from here. -- Rick Block 03:07, 25 May 2005 (UTC)
thread_title:  Nydick/Nudick 

message: I remember once reading that this riddle/theorem whatever is much older then ever mentioned in the article. i understand it was atelast from the 19:th century.
I can't verify this with google though. Does anyone have any better knowledge of the issue or better googling-luck?
Gillis 08:13, July 23, 2005 (UTC)
:I've never even heard of it existing previously than specified... no google-luck here.  By the way, I took the liberty to sign your message for you.  In the future, you can (and should) sign your messages yourself by placing four tildas ( ~~~~ ) after your message. Fieari 08:16, July 23, 2005 (UTC)
:Actually, it seems i found it with another attempt at google, i added in the new information, here is the source http://barryispuzzled.com/zmonty.htm. SSomeone could do some verifying... and ok i'll try to remember always signing my posts Gillis 09:47, 23 July 2005 (UTC)
thread_title:  Theorem history 

message: I would argue it's a problem in probability. Rich Farmbrough 12:44, 23 July 2005 (UTC)
replies: text: I have switch the statement back because it is certainly a problem of probability. Apparently MathWorld categorizes the problem under game theory, but in my opinion the connection is tenuous. In the standard statement of the problem there are no "conflicting interests" because the host is not an active player in the sense that he cannot make choices that affect the outcome. Because the problem clearly could generalize into a game theoretical topic, I would have no objections to a Wiki categorization as such. In the text, however, this would require some explanation, so in my opinion it isn't appropriate for the first line. Certainly it does not supersede probability. Davilla 16:21, 23 July 2005 (UTC)
thread_title:  Game Theory? 

message: Can showing you that one of the three candidates in a tight race is a loser make you change or question your vote?
I happen to believe that people love to feel included by voting for the popular candidate.
Let's say people were given a choice to vote for three candidates A B C.  After the election, you find out that one of the candidates was a loser and the two candidates were so close that a special runoff election was required.  Would you be more likely to be with the majority by switching your vote on the runoff?
Consider this election .. 
Dino Rossi (R) and Christine O. Gregoire (D) in 2004 ran one of the tightest governor races in history.  Christine Gregoire won by a mere 129 votes.     
The Republicans wanted a runoff election to solve the inaccuracies.
Do party affiliations play in this problem?
:There are quite a number of things to say about a situation like this, as per the mathematical treatment of social choice functions. Your slant is new to me, and interesting if not a bit offensive. I don't think it's necessarily wrong to be in the minority, although probability-wise you may have a point. I'm not much to speculate about the psychological aspects. Davilla 16:36, 23 July 2005 (UTC)
thread_title:  Monty Hall Problem in Politics / Elections 

message: This poorly-titled section formerly said this:
:When you pick a door (A), the probability of having chosen a door with a car is 1/3 since there is only 1 car and 3 doors. When the host eliminates a door (B), the probability of the car being in A remains constant, but now there are only two possible doors to pick from: A and C. Since one door's probability is 1/3, the other one is 1-1/3 = 2/3. Therefore, if the car is in C, it means it's not in A, and so the probability of the car being in C is the probability of the car being in A negated.
This is circular logic.  In support of the idea that door A ends up with a 1/3 probability of having the car, it makes the statement that "the probability of the car being in A remains constant" without any support.  One may reasonably ask, why doesn't the probability of the car being in C remain constant instead?  This explanation provides no answer.  (It also ends with a spectacularly muddled sentence that adds nothing but confusion.)  I have remove the section.  --Doradus 16:29, July 23, 2005 (UTC)
:The statment is, of course, correct, but I see your problem with it, especially the unsupported claim. Rather than deleting it, maybe it could be fixed without making the explanation too verbose. Davilla 16:43, 23 July 2005 (UTC)
thread_title:  Removed "probability laws" section 

message: Probability is 2/3 if all decisions are made before any doors are opened.
Probability is 1/2 if a fresh decision independent of history is made after Monty reveals a goat.
This misunderstanding is why the problem seems counterintuitive to people.
So should you switch if Monty gives you the choice AFTER he reveals a goat?  Doesn't matter.  Should your predetermined strategy for playing these games be to always switch?  Absolutely.
:Are you sure you understand the problem? You don't have to decide to switch before the choice is offered. A person who switches, even if they don't have a strategy or understand the mathematics, has a 2/3 chance of winning.
:Statements like the above exemplify that problem is counter-intuitive. Of course, I don't of know anyone who completely understood the problem the first time they heard it. Davilla 16:48, 23 July 2005 (UTC)
thread_title:  1/2 or 2/3 depending... 

message: This whole thing seems to hing around independent steps.
50/50 result. Some condition is set up; in this case chose 1 of 3; chance of selecting the correct outcome is exactly one third. BUT the next INDEPENDANT step is offered after the removal of one of the incorrect outcomes. What is left ? 2 options; of which one and only one is correct.Proberbility of picking the correct outcome is 1 out of 2 or 1/2.
Please demonstrate very clearly and precisely how any event prior to being offered the 2nd step choice can possibly influance the outcome of that choice. 
Or to restate that - you are left with a choice of one out of two.
Somebody mentioned tossing coins; it doesn't matter how many sides previous coins had; or how many coins were tossed prior; the next toss of a double sided coin has what probability of producing a head...
yes; exactly 1/2 as PRIOR conditions have zero effect. I presume at this stage no one would bet a million pounds/dollars/euros that following 1 million heads a tail would occur ? (and if you believe it MUST be a tail then you need to re-vise your basic proberbility theory
Please cut all the waffle and explain how any preceeding action affects the 1 out of 2 selection left at the end !
I'm stuck behind a dynamic isp so can only sign Pete-dtm
:You've got it exactly backwards: claiming that the chance is 1/2 requires that you explain how the preceding actions do not affect the outcome. See the section below with a million doors: are you saying that after picking a door (with just a 1/1000000 chance of being right) and Monty closing 999998 wrong doors, the chance that your door is still right at the end is suddenly 1/2?! No. It started off as 1/1000000, and it still is. Since we know either this door or the other is right, there's a chance of 999999/1000000 that it's the other. And you'd like to claim tossing a coin would be the same? That's true only if the candidate is stupid and deliberately chooses at random. But we're assuming the candidate knows what Monty's doing. In that case the candidate had better use his brain, not a coin, and pick the other door.
:What you're saying is that this is exactly the same as being dragged from your bed and being asked to choose between two doors, with no other information at all, except that one of them is definitely right and the other definitely isn't. That's not what's happening here. Statistically we only know that the chance of one of these doors being right is 1. You're making the mistake of assuming that means that it must be evenly split between the two doors. It's not. Tossing coins has nothing to do with this, as each coin toss, as you say, is independent of the other. But Monty closing all but one door and giving you a very good chance of getting the right one by switching is as dependent as it gets. 82.92.119.11 19:31, 23 July 2005 (UTC)
replies: text: Well put. hydnjo talk 20:39, 23 July 2005 (UTC)
thread_title:  Multi step problem 

message: Jeesh, I can't make up my mind as to which is more interesting (revealing), the article or this talk page! hydnjo talk 18:26, 23 July 2005 (UTC)
thread_title: Which would win?

message: The corrolation between gravity and time becomes evident at a black hole.
* What Monty is really doing 
:Guaranteeing that the correct door is one of the two left.
* Bigger
:* One million doors, pick one.
:* Monty opens 999998 and leaves one closed.
:* Monty guarantees the correct door is either the original you picked or the one he left closed.
:* Do you switch?
::Contributed by User:Tailpig
::*And now what are the odds of winning by the switcher? hydnjo talk 18:58, 23 July 2005 (UTC)
replies: text: *I don't know, pretty good. Basically, the odds that he got the correct door on his first guess were pretty small. I'm trying to write an explanation without resorting to fractions.--Tailpig 19:28, 23 July 2005 (UTC)
replies: text: *My question was rhetorical and intended for the non-believers out there. Your expansion of the problem should help them to see the error of their perception (or maybe not). Anyway the odds in favor of the switcher are obviously about 999999:1 but I'm sure that there are some that will still argue for 50%. It turns out that the odds in favor of the switcher are exactly the same as the odds against his making the winning choice at the outset.  BTW, your example is an excellent expansion of the problem and should help someone to understand the underlying principle; exaggeration can sometimes be illuminating.  Perhaps it should go into the main article. hydnjo talk 19:45, 23 July 2005 (UTC)

text: Yes it is right here using 100 doors (not 1000000). I missed that part. Thanks. hydnjo talk 22:08, 23 July 2005 (UTC)
thread_title:  Blow up the problem 

message: Kinda looks like this talk page. I guess there's no hope, they're everywhere.  I'm going to rv the link as it doesn't add anything new to the article except controversy. hydnjo talk 00:50, 24 July 2005 (UTC)
thread_title: [http://www.straightdope.com/classics/a3_189.html Straight Dope article] link added to main page

message: One of Wikipedia’s policies is No original research.  So, please do not try to prove to me what the answer is with your own calculations.  Show me your source.
The following web sites claim that the answer is 2/3 probability if you switch:
* http://www.straightdope.com/classics/a3_189.html, assertion by Cecil Adams
* http://econwpa.wustl.edu/eps/exp/papers/9906/9906001.html, assertion by professors Peter R. Mueser and Donald Granberg of the University of Missouri–Columbia
* http://www.cut-the-knot.org/hall.shtml assertion by Alexander Bogomolny
* http://www.letsmakeadeal.com/problem.htm, assertion by Monty Hall – Yes, the game show host after whom this paradox is named
For those that assert that the probability is 1/2, please site your source.  -- JamesTeterenko 03:01, 24 July 2005 (UTC)
thread_title: No original reasearch

message: Part of what messes up intuition is how to make use of the partial information you have when deciding whether or not to switch.  Here's another way to look at it that might help (this is sort of a rewording of "Best, most concise explanation yet") :
Assume the contestant always switches, i.e. ignore the second decision.  The contestant's first choice is either a) car (probability 1/3) or b) goat (probability 2/3).  In (a), contestant loses, in (b) contestant wins.  Thus 1/3 probability of losing, 2/3 probability of winning.
Now assume the contestant always stays. The contestant's first choice is either a) car (probability 1/3) or b) goat (probability 2/3).  In (a), contestant wins, in (b) contestant loses.  Thus 1/3 probability of winning, 2/3 probability of losing. Peter Grey 04:17, 24 July 2005 (UTC)
thread_title:  Think through the problem backwards 

message: * Let X1 be the initial choice of the player.
* Let M be the what Monty eliminate.
* Let X2 be the final choice of the player.
* Let C, G1, G2 be the car, goat_1 and goat_2.
P(X1)         	P(M l X1)           	P(X2 l X1, M)               	.	P(X1, M, X2)
P(X1=C) = 1/3 	P(M=G2 l X1=C) = 1/2	P(X2=C l X1=C, M=G2) = 1/2  	.	P(X1=C, M=G2, X2=C) = 1/12
P(X1=C) = 1/3 	P(M=G2 l X1=C) = 1/2	P(X2=G1 l X1=C, M=G2) = 1/2 	.	P(X1=C, M=G2, X2=G1) = 1/12
P(X1=C) = 1/3 	P(M=G1 l X1=C) = 1/2	P(X2=C l X1=C, M=G1) = 1/2  	.	P(X1=C, M=G1, X2=C) = 1/12
P(X1=C) = 1/3 	P(M=G1 l X1=C) = 1/2	P(X2=G2 l X1=C, M=G1) = 1/2 	.	P(X1=C, M=G1, X2=G2) = 1/12
P(X1=G1) = 1/3	P(M=G2 l X1=G1) = 1 	P(X2=C l X1=G1, M=G2) = 1/2 	.	P(X1=G1, M=G2, X2=C) = 1/6
P(X1=G1) = 1/3	P(M=G2 l X1=G1) = 1 	P(X2=G1 l X1=G1, M=G2) = 1/2	.	P(X1=G1, M=G2, X2=G1) = 1/6
P(X1=G2) = 1/3	P(M=G1 l X1=G2) = 1 	P(X2=C l X1=G2, M=G1) = 1/2 	.	P(X1=G2, M=G1, X2=C) = 1/6
P(X1=G2) = 1/3	P(M=G1 l X1=G2) = 1 	P(X2=G2 l X1=G2, M=G1) = 1/2	.	P(X1=G2, M=G1, X2=G2) = 1/6
The probability of winning (not Monty Hall problem) is:
* P(X2=C)
: = P(X1=C, M=G2, X2=C) + P(X1=C, M=G1, X2=C) + P(X1=G1, M=G2, X2=C) + P(X1=G2, M=G1, X2=C)
: = 1/12 + 1/12 + 1/6 + 1/6 = 1/2
The probability of winning without switching is:
* P(X2=C l X2=X1)
: = P(X2=C, X2=X1) / P(X2=X1)
: = (P(X1=C, M=G2, X2=C) + P(X1=C, M=G1, X2=C)) / (P(X1=C, M=G2, X2=C) + P(X1=C, M=G1, X2=C) + P(X1=G1, M=G2, X2=G1) + P(X1=G2, M=G1, X2=G2))
: = (1/12 + 1/12) / (1/12 + 1/12 + 1/6 + 1/6) = 1/3
The probability of winning with switching is:
* P(X2=C l X2<>X1)
: = P(X2=C, X2<>X1) / P(X2<>X1)
: = (P(X1=G1, M=G2, X2=C) + P(X1=G2, M=G1, X2=C)) / (P(X1=C, M=G2, X2=G1) + P(X1=C, M=G1, X2=G2) + P(X1=G1, M=G2, X2=C) + P(X1=G2, M=G1, X2=C))
: = (1/6 + 1/6) / (1/12 + 1/12 + 1/6 + 1/6) = 2/3
The answer to the Monty Hall problem:
* Since probability of winning with switching is greater than probability of winning without switching, the player's strategy should be to switch.  In other words, the player should be free to make any initial choice (X1), but must make sure that the final choice (X2) is different in order to maximize the probability of winning. 
aCute 05:22, 24 July 2005 (UTC)
thread_title: Full Mathematical Solution

message: Note: It is my observation to lead me to conclude the following:
* It is often the case that some people argue about the final event (X2) to the exclusion of other factors, which is P(X2=C l X1=*, M=*) = 1/2, and although it is correct, but sadly it is NOT the Monty Hall problem.
* It is also often the case that some people argue about the winning-probability of player in the entire game, which is P(X2=C) = 1/2, and although it is correct, but sadly it is also NOT the Monty Hall problem.
* The Monty Hall problem is to compare the strategy of switching, which is P(X2=C l X1<>X2) = 2/3, versus the strategy of not switching, which is P(X2=C l X1=X2) = 1/3, and to conclude switching is better than not switching.
The Monty Hall problem is NOT to determine the winning-probability of the last event or the entire game, but rather to determine which of the two strategies will maximize winning-probability.  The problem is like comparing the subset cases in the manner similar to the Simpsons paradox.  This is also like trying to find a winning strategy in Blackjack by counting cards already played previously as additional information (sort of like Monty Hall revealing a goat) in order to adjust subsequent choices to gain an advantage over the house. aCute 06:49, 24 July 2005 (UTC)
thread_title: What is and what is NOT the Monty Hall problem

message: In "the formal probability diagram" within the project page of the Monty Hall problem, the leaves of the diagram tree has the values "1/3 1/3 1/3 1/3 1/6 1/6 1/6 1/6" which cannot possibly exist since the sum of these values canNOT exceed 1.
left
:The letter "Y" should be "1/2"
:The letter "N" should be "1/2"
:The final line should be "1/6 1/6 1/6 1/6 1/12 1/12 1/12 1/12"
Please someone change them ASAP.
aCute 05:22, 24 July 2005 (UTC)
thread_title: Diagram Problem

message: "Never argue with someone who knows that he's right" hydnjo talk 05:27, 24 July 2005 (UTC)
thread_title: Never Argue

message: When this doesn't work well, it's time to go on to something more productive. Buh bye now.  hydnjo talk 05:45, 24 July 2005 (UTC)
thread_title: [[Reductio ad absurdum]]

message: Ok, so stupid me, but I have to do this. Lets take this down to such a basic level that even I understand.
*Three doors and I pick one of them. 
*Monty, by the rules, gets the other two doors.
:*Forget this step as Monty will always reveal a goat door whether he has one or two goats thus this step makes no difference. In "magic" performance this would be called misdirection as it has nothing to do with the outcome but it will distract you. The hidden rule is that Monty will never open a car door (this is what throws some folks off the trail) - Monty doesn't make a random selection; he knows what is behind every door after all so why bother with this step at all.
*So now, would you trade your one door selection for Monty's two door leftovers or not (remembering that he hasn't even gone to the trouble of opening a (for sure) goat door).
:*n.b.: My youngish grandchildren all elected to switch.
Respectfully, hydnjo talk 21:18, 24 July 2005 (UTC)
thread_title: One More Time

message: The game is fixed from the moment that you pick one door and Monty gets two doors. All else is just fluff and misdirection to make the game last longer. You hold a 1/3 chance of winning and (because he has two doors) Monty has a 2/3 chance of winning. Swap with him at your earliest opportunity. Please. hydnjo talk 02:10, 25 July 2005 (UTC)
thread_title: Look at it this way

message: You get one of three possibilities and Monty gets two of the three possibilities. The fact that you chose first changes nothing. If given the opportunity to swap with Monty well, just jump at it. He did, after all, try to screw you at the very outset. And, if Monty tries to confuse you with more doors, well then hold fast. When he offers a swap (which he will) just take it. Don't forget, you got to pick one door and Monty gets all the others. Lucky lucky him he would like you to believe! hydnjo talk 02:48, 25 July 2005 (UTC)
thread_title: I don't understand the controversy

message: (as opposed to the correctness of the mathematical answer)
It is not enough to describe why the mathematically derived solution is correct. To resolve the paradox to the satisfaction of all, one must also describe why the intuitive solution is wrong. I think this requires three steps.
1)	an understanding that the "intuitive solution" is just a dismissive term for a first analysis that turned out to have a flaw
2)	a description of the logical steps that were employed in coming to the intuitive solution
3)	an analysis of that logic, to find its flaw(s)
(1)
I not argue the point, but rather just hope that people agree with it.
(2)
I think that the intuitive solution takes the following steps:
* transform the game into a simpler yet completely isomorphic game
* calculate (or intuit) the probabilities for that game
* extrapolate the answer back to the original Monty Hall game
Here is the game that I believe is used intuitively (I’ll call it the Silly game):
* There are three doors, A, B, and C.
* Door A is open and has a goat behind it.
* Doors B and C are closed, and there is a goat behind one and a car behind the other.
* The contestant just guessed that door B has the car behind it, and is now being given a chance to change his mind.
Should he change it or not?
Clearly the answer to this is that the chance is 50% either way, and it does not matter whether the contestant changes his mind.
Now compare the Silly game to an original Monty Hall game at a point half way through, in which the contestant has guessed door B and Monty has opened door A.
* There are three doors, A, B, and C
* Door A is open and has a goat behind it.
* Doors B and C are closed and there is a goat behind one and a car behind the other.
* The contestant just guessed that door B has the car behind it, and is now being given a chance to change his mind
It seems that the Silly game is exactly the same as the original Monty Hall game at this point. To a viewer who has just ‘tuned in’ and does not know what has previously happened, the games look identical. Therefore, logic would dictate that the answer to the original question is that it doesn't matter whether the contestant changes their mind; the probability is 50% either way.
(3) It turns out, of course because the two games are not completely isomorphic. There is a crucial piece of information missing from the Silly game that is needed to make it isomorphic with the Monty Hall game, as follows: Monty (who knows where the car is)
* was required to open a door to reveal a goat,
* was given an opportunity to open door C,
* chose not to open door C.
Monty has thus said something concrete about door C (“I didn’t choose it, perhaps because I couldn’t choose it”), but nothing about door B. This is the source of the asymmetry between doors B and C, and the reason that door C is more likely to not have a goat behind it.
Happyharris 20:57, 25 July 2005 (UTC)
replies: text: Perhaps, what we need in the main project page is an explanation to "why the intuitive answer is NOT the Monty Hall problem".  Something like: Extrapolating the probabilities of an isomorphic game back to the Monty Hall game is the cause of much of the controversy of the game, since answering the probabilities of the game with randomized strategy (1/2) is NOT answering the probabilities of each strategy of switching (2/3) and not switching (1/3).  aCute 08:50, 27 July 2005 (UTC)
thread_title: Analysis of the errors in the intuitive answer

message: For the people who still find it hard to see why the probability is not 50/50 when there are two doors left, I hope the following illustration may help.  I'm going to show that the Monty Hall problem is a specific case of a more general game; I'll call all the games that differ in their parameters "Hall games" for ease of use.
The basic idea behind a Hall game is this:  We start with one large set of secret-hiding items -- they can be doors to be opened, or cards to be turned over, it doesn't actually matter.  What matters is that there are n cards, but only one of them is the Prize card.  The rest are Null cards.
Step One: The cards are divided into two hands.  Each hand must have at least one card.  For simplicity's sake, we call the number of cards in the first and second hands h1 and h2.
Step Two: Someone who can see which cards are Nulls can discard some number of Nulls from one hand, the other, or both.  We call the number of cards remaining in each hand after the discarding of Nulls r1 and r2 (like h1 and h2, they must be at least one.)
Step Three: The player makes a guess at which of the two hands contains the Prize.
Step Four: The player makes a guess at which card out of the hand he selected is the Prize.
It's clear that to win the game, the player has to make correct guesses in both Step Three and Step Four.  What are the chances of picking the correct hand?  The first hand is correct h1/n of the time; the second hand is correct h2/n of the time.  If we want, we could enumerate the cases:  if h1=2 and h2=5, then the Prize could be the first card of the first hand, the second card of the first hand, the first card of the second hand ... et cetera, et cetera.
Now this is the part that many people find counter-intuitive.  If we enumerate the cases, and then we reduce Nulls to meet any legal value of r1 and r2, we find that in no case can the removal of Nulls switch the Prize from one hand to the other.  This means that the chance of the card being in the first hand or the second always stays at h1/n and h2/n -- even if the sizes of the hands do not stay at h1 and h2.  If it's dealt to that hand, it stays in that hand; therefore the chance of it being in a particular hand is always equal to the chance that it was dealt to that hand.
What about the removal of Nulls?  Does it affect anything after all?  Yes, it does -- it affects the player's chances in Step Four.  The chance that the Prize is in a particular hand is dependent upon h1 and h2 -- how many cards each hand started with.  But the chance of finding the Prize in a hand (assuming it's the right hand) is based on how many cards that hand contains after Nulls have been removed -- since there's only one Prize, the chance of finding it in a hand of r1 cards is 1/r1.
Now, what are the chances of making both guesses correctly?  If no Nulls get removed from either hand, then the chances of picking the Prize are either h1/n x 1/h1 (since r1 is equal to h1 when no Nulls have been removed) or by similar logic h2/n x 1/h2, which also multiplies out to 1/n.  If, however, one of the hands -- say, hand 1 -- has been reduced down to one card (r1 = 1), then the chances of finding the Prize in that hand is h1/n x 1/1 -- if you've correctly guessed that the Prize is in that hand, you have a 100% chance of finding it in that hand when it's the only card in the hand.
With this being the general structure of the Hall game, we can see that the Monty Hall problem is really just the case where n=3, h1=1 and h2=2, r1=1 and r2=1.  The chance that the Prize is in the player's hand is h1/n -- 1/3.  The chance that it's in Monty's hand is 2/3.  Before one Null is removed from Monty's hand, the chance of finding the Prize in his hand is 2/3 x 1/2 -- i.e., 1/3.  But when the Null is removed, the chance is now 2/3 x 1/1 -- i.e., 2/3! -- Antaeus Feldspar 03:44, 26 July 2005 (UTC)
thread_title:  Breaking it down into steps