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Zach Wegner
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Oct0502, 02:20 PM (EST) 

"Monty Hall problem... again"

There are two solutions that are suggested. Each solution depends upon the representation of data. For 1/2From contestants POV, without 'perfect information'where the tree doesnt rely upon knowing the prize. All outcomes equally likely. For 2/3From MH POV, with perfect information. Different outcomes have different chance. Thats why ALL the simulations show 2/3. No one thought of a simulation without perfect information. I made one, and it'shows 1/2. It does this by choosing a picked door, then choosing a door to reveal. The prize is then picked, and it can have 2 places because it cant be in the revealed door. This goes along with if you have a 100 door problem, or say an n door problem. if you take away all but two, the probability of winning by switching is 99/100 or n1/n. If you take away one, then it is 1/99, or 1/n1. If you have 3 doors, taking away one and taking away all but two yields the same result, so the amount of doors opened doesnt matter. Which would the final probability be? n1/n or 1/n1? Of course, with this problem, n1/n = 2/3 1/n1 = 1/2 If you say perfect information, then the number of doors able to open is either 1 or 2 (or n1 and n2, because you know if the prize was picked or not), and it gives 2/3. If you say no perfect information, then the number of doors able to open is 2 (or n1 because you dont know if the prize is unpicked or not), and it is 1/2.Since the problem states that it is from the contestants POV, it is 1/2. I would like to see a counterexample. The perfect information seems to be in the mindset of all who say 2/3, and that is the flaw on their part. Zach


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Jack Wert
guest

Oct0602, 07:40 AM (EST) 

1. "RE: Monty Hall problem... again"
In response to message #0

Perfect  schmerfect. Who defines the type of information in a simple problem such as the Monty Hall dilemma? Just review the exact way the problem was stated, and there is only one solution  2/3!! Only one time out of three does the contestant pick the prize door, and in that case, if he/she loses if a "switch" is decided. Two times out of three the contestant picks the "goat" door, and in these two cases, he/she wins when the decision is to switch. What does "perfect" information have to do with such a simple situation? Jack Wert 

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Zach Wegner
guest

Oct0702, 04:50 PM (EST) 

2. "RE: Monty Hall problem... again"
In response to message #1

Who defines the information? The problem of course. It'says you are the contestant. Heres a real tree: You pick A B C Monty reveals B C A C A B Because prize was in C A A B B C A B B C A C To say 2/3 would be to say that in the two leftmost branch leaf nodes 'C' is more likely than 'A'. Its not. The people arguing for 2/3 are saying that if you chose a door, it would be twice as likely for him to reveal a door because the prize wasnt selected. Supposing you pick A, a chart: Prize Reveal Win by switching A B N A C N B C Y C B Y If youre saying it is 2/3, then you know that picking B or C in the first case is less likely because you know the prize. You dont, youre obviously the contestant. To answer your question, in this situation perfect information says you know the prize but the pick is random. Every tree Ive seen with the prize first is nonsense. You dont know the prize until youve made your final deciscion. 

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G Kossinets
guest

Oct0802, 04:50 PM (EST) 

3. "RE: Monty Hall problem... again"
In response to message #1

It is indeed difficult to see where the symmetry breaks; a careful reading of the game rules might be helpful. I will use the Bayes theorem to support the argument about incomplete/complete information. Suppose that you _never_ change your mind. What are your chances of winning the prize? The prior probabilities of winning the prize versus losing are P{Win}=1/3 and P{Lose}=2/3, respectively. The conditional probability that the host has revealed a goat given that you win is unity: P{Goat_revealedWin}=1. That is because if you never switch and you have chosen the prize door, there are only goats left. Now the tricky part. Assuming that you have picked a goat, can the host open the door with the prize and say "You're lost"? If so, then the probability that a goat was revealed given that you lose is P{Goat_revealedLose}=1/2 (perhaps  if the host choses "randomly". That is, provided that you have chosen a goat, the host reveals the remaining goat or the prize with probability 1/2. Using the Bayes formula, P{WinGoat_revealed} = P{Goat_revealedWin}*P{Win} / (P{Goat_revealedWin}*P{Win}+P{Goat_revealedLose}*P{Lose}) = (1*1/3) / (1*1/3 + 1/2*2/3) = 1/2 However, assuming that the host always reveals a goat and never the prize, the probability that a goat was revealed given that you lose is unity: P{Goat_revealedLose}=1. Consequently, P{WinGoat_revealed} = P{Goat_revealedWin}*P{Win} / (P{Goat_revealedWin}*P{Win}+P{Goat_revealedLose}*P{Lose}) = (1*1/3) / (1*1/3 + 1*2/3) = 1/3 The conclusion is: if you know that the host opens the door with a goat "on purpose" then you should switch. If the host acts "randomly", then it does not matter whether you switch or not. It is probably a naive question, but does anybody know if Monty Hall ever ended the game revealing the prize after the participant had made the wrong choice?


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jman_red
Member since Jul202

Oct0902, 04:50 PM (EST) 

4. "Another unbeliever... again"
In response to message #0

The type of information used is the same in all simulations, and the simulations all give 2/3 as the answer as part of a worldwide conspiricy. They say 2/3 because 2/3 is CORRECT. With probability 1/3, you initially pick the correct door. If you switch after the revelation of an incorrect door, you will lose. With probability 2/3, you pick an incorrect door. There is only one incorrect door left for the host to eliminate as a winning door. Thus, with probabilty 2/3, you will have eliminated both incorrect doors (the host's and your own original pick), and if you switch the only door to switch to is correct door. Switching Wins 2/3, loses 1/3. jman_red 

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Zach Wegner
guest

Oct0902, 08:14 AM (EST) 

5. "RE: Another unbeliever... again"
In response to message #4

>The type of information used is the same in all simulations, >and the simulations all give 2/3 as the answer as part of a >worldwide conspiricy. > >They say 2/3 because 2/3 is CORRECT. Then why does mine say 1/2? Look at it this way. Assuming you pick A. The host can either open B or C, because you dont know if the prize is in either one. If he opens B, it is because the prize is in A or C. Why would C be more likely than A? If you say that you opened either A or C to monty, he opens B. Then which door would be more likely? Neither. They are equal. So if it was because the prize was in A, you lose by switching. If it was in C, you win. Probablility of A versus C in this case is equal, so winning by switching is 1/2. 

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Eric Shrader
guest

Oct1002, 09:49 PM (EST) 

6. "RE: Another unbeliever... again"
In response to message #5

>>The type of information used is the same in all simulations, >>and the simulations all give 2/3 as the answer as part of a >>worldwide conspiricy. >> >>They say 2/3 because 2/3 is CORRECT. > >Then why does mine say 1/2? > >Look at it this way. Assuming you pick A. The host can >either open B or C, because you dont know if the prize is in >either one. If he opens B, it is because the prize is in A >or C. Why would C be more likely than A? If you say that you >opened either A or C to monty, he opens B. Then which door >would be more likely? Neither. They are equal. So if it was >because the prize was in A, you lose by switching. If it was >in C, you win. Probablility of A versus C in this case is >equal, so winning by switching is 1/2. Zach, Your simulation says 1/2 because you are modeling a completely different game. In your simulation, the contestant chose a door, then Monty chose a door, and only then was the prize placed (with equal probability) behind one of the doors that Monty didn't choose. Of course this simulation yields 1/2! But, it is in no way a simulation of the Monty Hall game. In the Monty Hall game, the prize is placed *first*, then the contestant chooses, and then Monty reveals a door. Order matters! If you redo your simulation in the right order, you will get the right answer.  Eric


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Zach Wegner
guest

Oct1302, 10:28 PM (EST) 

8. "RE: Another unbeliever... again"
In response to message #6

>Your simulation says 1/2 because you are modeling a >completely different game. In your simulation, the >contestant chose a door, then Monty chose a door, and only >then was the prize placed (with equal probability) behind >one of the doors that Monty didn't choose. Of course this >simulation yields 1/2! But, it is in no way a simulation of >the Monty Hall game.Of course it is. My simulation exploits the fact that you dont know where the prize is. If the prize is selected last, how would you know? If he opened B, then the prize could be in A or C. It makes no difference if when the prize door was selected or even which of the other doors you selected. Most say that if you pick the incorrect door then Monty can only pick one door. Well how do you know if you picked it? In any case he can open both doors, period. 

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Eric Shrader
guest

Oct1402, 12:18 PM (EST) 

10. "RE: Another unbeliever... again"
In response to message #8

>> is in no way a simulation of >>the Monty Hall game. > >Of course it is. My simulation exploits the fact that you >dont know where the prize is.The contestant doesn't know where the prize is, but Monty does! And that's the point that you are missing. Sometimes Monty's choice is forced. You seem to be saying that since the contestant doesn't know that Monty's choice is forced, it isn't forced, but that of course is nonsense. Monty's choice can be forced and the contestant can not know that it was forced. > If the prize is selected last, how would you know? The rules of the game are such that the prize is selected first. The contestant can reasonably assume that the game is being played according to the rules, and therefore it is not selected last. I agree with you that if the prize is selected last, then the probabilit's are equal. But that's a different game! In fact, it's a rather uninteresting game where Monty serves no real purpose. Can you explain to me why you switch the order of the operations? The game is welldescribed with events happening in a certain order. When you modeled the game, you switched the order. Furthermore, everyone who models the game in the stated order gets an answer different from yours. So, prima facie, you are modeling a different game. > If he opened B, then the prize could be in A or C. Yes, but not with equal probability!


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Zach Wegner
guest

Oct1402, 09:11 PM (EST) 

11. "RE: Another unbeliever... again"
In response to message #10

>> If he opened B, then the prize could be in A or C. > >Yes, but not with equal probability! Yes it is. I didnt even tell you which door was opened. How would it not be? 

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Hemulen
guest

Jan0204, 05:09 PM (EST) 

13. "RE: think this way"
In response to message #8

If there was 1 000 000 doors, all with goats, exept one and you had to chose one door. Then Monty would open 999 998 doors all with goats, which one would you chose? Of course you would switch, because the chanses are 1/1 000 000 vs. 999 999/1 000 000.


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Mark Huber
guest

Jan0604, 05:34 PM (EST) 

16. "RE: think this way"
In response to message #13

The problem is that the originator of this thread did not define which version of the Monte Hall problem they are using. The problem as submitted to Parade magazine (see here) was ambiguous, and so multiple solutions emerge. If the problem statement says that Monte *always* opens a door with a goat, then the 2/3 solution is the only one, but if Monte is allowed to change his behavior when the game is played (and that's how the actual "Let's Make a Deal" was set up) the probability under switching can be made to go anywhere from 0 to 1. Mark 

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jman_red
Member since Jul202

Oct1002, 09:49 PM (EST) 

7. "RE: Another unbeliever... again"
In response to message #5

<< Look at it this way. Assuming you pick A. The host can either open << B or C, because you dont know if the prize is in either one. If he << opens B, it is because the prize is in A or C. Why would C be more << likely than A? If you say that you opened either A or C You did not tell Monty your initial pick was A or C. You picked *one* door and only one door. << to monty, he opens B. Then which door would be more likely? << Neither. They are equal. So if it was because the prize was in A, << you lose by switching. If it was in C, you win. Probablility of A << versus C in this case is equal, so winning by switching is 1/2. Follow my argument. Tell me which step is false. Step 1: Your initial pick has a one third chance of being the correct door. Step 2: Thus, If you switch, you will lose. Step 3: Your initial pick has a 2/3 chance of being an incorrect door. Step 4: If your initial pick is incorrect, there will only be one incorrect door left that Monty can open and reveal to be a loser. Step 5: Monty will show you an incorrect door, and if your initial pick was incorrect (with probability 2/3), the only door left to switch to will be the winning door Step 6: With probability 2/3, you win while switching. With probability 1/3, you lose when switching. jman_red 

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Zach Wegner
guest

Oct1302, 10:28 PM (EST) 

9. "RE: Another unbeliever... again"
In response to message #7

><< Look at it this way. Assuming you pick A. The host can >either open << B or C, because you dont know if the prize is >in either one. If he << opens B, it is because the prize is >in A or C. Why would C be more << likely than A? If you say >that you opened either A or C > >You did not tell Monty your initial pick was A or C. You >picked *one* door and only one door. Bad wording on my part. Say I was on the show and was telling someone that I picked a door other than B. I was merely explaining the equal chance of the prize being in the doors otehr than the selected one. >Follow my argument. Tell me which step is false. This one: >Step 4: If your initial pick is incorrect, there will only >be one incorrect door left that Monty can open and reveal to >be a loser. In this case you know that Monty can only open one door. Why would you? The only door you can exclude is the one you picked. You might as well say step 4 as: If you initially picked, there will be only two doors Monty can reveal and be a loser. You can make no statement about how teh number of doors available to open changes from case to case, for you would know the prize. 

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Whymme
guest

Nov1402, 08:00 PM (EST) 

12. "RE: Another unbeliever... again"
In response to message #9

Let's see if I can have success in explaining. Let's say that you pick door A. We can distinguish three different situations: either the car is behind door A, door B or door C. Each situation has a chance of 33% of occurring. If the car were behind door A, Monty would have had the choice of opening either door B or door C. Half the time he would open door B and half the time he would open door C. in total the chance that the car is behind door A and that Monty opens door B is 17%, and the same probability exist of the car being behind door A and Monty opening door C. If the car were behind door B, however, Monty is forced to open door C. He has no choice. The reverse is true for C. Each of these chances is 33%. Monty opens door B. Of the situations described above, the situations where he would open door C disappear. What is left is:  The car is behind door A, Monty opens door B (17%x2 = 33%)  The car is behind door C, Monty opens door B (33%x2 = 67%) If you don't agree, please tell me where I made a mistake. I don't think that I did.
Whymme


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Rod H
guest

Jan0604, 05:34 PM (EST) 

14. "RE: Another unbeliever... again"
In response to message #12

Very interesting discussion. Reminds me of the fact that you can't teach physics to a dog. Unless Monty is acting randomly, switching will win two out of three times. Clearly, you can't teach an old Zack new tricks. 

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A physics nerd
guest

Jan0604, 05:34 PM (EST) 

15. "RE: Monty Hall problem... again"
In response to message #0

See if you follow this: I'll give you 100 bucks if you pick the ace of spades out of a deck without looking at the cards. You pick one. I then look at all the remaining cards and throw away 50 cards that are not the ace of spades. I then ask you if you would want to switch cards with me. You do so  and most likely win. Same principle as the doors. It's really, really simple. 

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Graham C
Member since Feb503

Jan1104, 02:33 PM (EST) 

17. "RE: Monty Hall problem... again"
In response to message #0

Just to join in. Someone correctly pointed out that in the original game and in the original statement of the problem, it was *not* specified that Hall always opened a door. In fact no strategy was given for Hall at all. Answers that depend on Hall always opening a door (or always opening a door with a goat) are therefore not answers to the original problem. A fortiori, if Hall does have a strategy, the contestant doesn't know whether he has one or not. So, back to the question, 'should the contestant switch?' It has two 'at least) possible meanings. 'Would switching improve his chances of winning?' 'What should a rational person do in the contestant's situation?' In the first case, the answer depends on complete knowledge of Hall's strategy. In the second case, the question is not a probability question but a game theoretic one, and essentially the contestant has to judge what Hall's most likely strategy is. Is he deliberately trying to assist the contestant? Is he deliberately trying to get him to look stupid )game show hosts do that)? Is he acting randomly (making his own pick, so that he might have revealed a car)? Or what? In the first case, he should switch. In the second he should stay. In the third it doesn't matter. In the rest...? I'd love to play poker with all the people who say the answer is certainly he should switch (and probably own a lot of swampland in Florida). 

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Jack Wert
guest

Jan1304, 01:33 AM (EST) 

18. "RE: Monty Hall problem... again"
In response to message #17

>Just to join in. > >Someone correctly pointed out that in the original game and >in the original statement of the problem, it was *not* >specified that Hall always opened a door. In fact no >strategy was given for Hall at all. > >Answers that depend on Hall always opening a door (or always >opening a door with a goat) are therefore not answers to the >original problem. Because there have been so many interpretations of the "original" problem, I hereby offer it for your perusal: Dear Marilyn: Suppose you're on a game show, and given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you , "do you want to pick door number 2? Is it to your advantage to switch your choice of doors? Craig F. Whitaker, Columbia, Maryland That is a very simple and direct situation, and not really open for much interpretation. The host definitely opens a door behind which is a goat. We are not supposed to second guess his strategy or motives, only to come up with a simple solution to a simple probability situation  which is what Marilyn did  and which started the turmoil in the mathematical elit'st community. Enough said. Just look at the original statement of the problem and review your thinking. > >A fortiori, if Hall does have a strategy, the contestant >doesn't know whether he has one or not. > >So, back to the question, 'should the contestant switch?' > >It has two 'at least) possible meanings. 'Would switching >improve his chances of winning?' 'What should a rational >person do in the contestant's situation?' > >In the first case, the answer depends on complete knowledge >of Hall's strategy. > >In the second case, the question is not a probability >question but a game theoretic one, and essentially the >contestant has to judge what Hall's most likely strategy is. >Is he deliberately trying to assist the contestant? Is he >deliberately trying to get him to look stupid )game show >hosts do that)? Is he acting randomly (making his own pick, >so that he might have revealed a car)? Or what? > >In the first case, he should switch. In the second he should >stay. In the third it doesn't matter. In the rest...? > >I'd love to play poker with all the people who say the >answer is certainly he should switch (and probably own a lot >of swampland in Florida).
Jack Wert


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Graham C
Member since Feb503

Jan1404, 11:53 AM (EST) 

19. "RE: Monty Hall problem... again"
In response to message #18

> >Because there have been so many interpretations of the >"original" problem, I hereby offer it for your perusal: > >Dear Marilyn: > >Suppose you're on a game show, and given the choice of three >doors. Behind one door is a car, behind the others, goats. >You pick a door, say number 1, and the host, who knows >what's behind the doors, opens another door, say number 3, >which has a goat. > >He says to you , "do you want to pick door number 2? Is it >to your advantage to switch your choice of doors? > >Craig F. Whitaker, Columbia, Maryland You make my point for me. It was a oneoff situation with no indication as to why Hall opened the door. To decide whether it is to your advantage to switch, you have to make some assumption about his motive (or the statistical likelihood of his doing what he did in different circumstances). If his motive (or the likelihoods) was stated in the problem then the answer would indeed be clearcut. Take the (simple) Bayesian view. A=contestant chose a car. B=Hall would open a door. C=Hall's door would reveal a goat. We're asked, essentially, to find: P(AB&C) = P(B&CA)*P(A)/P(B&C) P(A)=1/3, OK. But where do you go from there? P(B)? Was it 1? The contestant has no idea. P(C)? Was is 1? Again, no idea. Was it 2/3? It would be if he chose at random, but we don't know that. P(B&CA)? If B and C are independent of A, and P(B)=P(C)=1, then it is 1/3. But the contestant doesn't know P(B) or P(C) or whether B or C is dependent on A. So with the information available the solution is not calculable. Assume P(B)=1 and B,C independent of A, then P(C)=2/3, P(B&C)=2/3=P(B&CA) and P(AB&C)=1/3 and yes the contestant should switch. But there is no reason at all given for making those assumptions. Actually the critical factor is whether B and C are independent of A. As long as they are, then P(B&CA)=P(B&C) and P(AB&C)=P(A)=1/3 and the contestant benefits from switching. But if he only opened the door because the contestant had chosen a car, then P(B)=1/3, P(C)=1 and P(AB&C)=1*(1/3)/(1/3)=1 and he should definitely not switch. We are in decisionmaking under uncertainty territory, not decisionmaking under risk. It's a gametheoretic problem, not a probability one. > >That is a very simple and direct situation, and not really >open for much interpretation. The host definitely opens a >door behind which is a goat. We are not supposed to second >guess his strategy or motives, only to come up with a simple >solution to a simple probability situation  which is what >Marilyn did  and which started the turmoil in the >mathematical elit'st community. > >Enough said. Just look at the original statement of the >problem and review your thinking. > > >> >>A fortiori, if Hall does have a strategy, the contestant >>doesn't know whether he has one or not. >> >>So, back to the question, 'should the contestant switch?' >> >>It has two 'at least) possible meanings. 'Would switching >>improve his chances of winning?' 'What should a rational >>person do in the contestant's situation?' >> >>In the first case, the answer depends on complete knowledge >>of Hall's strategy. >> >>In the second case, the question is not a probability >>question but a game theoretic one, and essentially the >>contestant has to judge what Hall's most likely strategy is. >>Is he deliberately trying to assist the contestant? Is he >>deliberately trying to get him to look stupid )game show >>hosts do that)? Is he acting randomly (making his own pick, >>so that he might have revealed a car)? Or what? >> >>In the first case, he should switch. In the second he should >>stay. In the third it doesn't matter. In the rest...? >> >>I'd love to play poker with all the people who say the >>answer is certainly he should switch (and probably own a lot >>of swampland in Florida). > >Jack Wert


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Jody
guest

Jan2004, 09:02 AM (EST) 

20. "RE: Monty Hall problem... again"
In response to message #17

>Someone correctly pointed out that in the original game and >in the original statement of the problem, it was *not* >specified that Hall always opened a door. In fact no >strategy was given for Hall at all. But it did mention that he knew what was behind the doors, which implies that he was opening the one with the goat "on purpose". >Answers that depend on Hall always opening a door (or always >opening a door with a goat) are therefore not answers to the >original problem. > >A fortiori, if Hall does have a strategy, the contestant >doesn't know whether he has one or not. Maybe it helps to have a little background. I happened to watch Let's Make a Deal a lot as a kid. Monty often opened a door (but not always). In all the times I saw him, he NEVER revealed a car, always a goat (or giant mousetrap or some such joke prize). Also, there was no apparent "strategy" to when he opened a second door. If there had been, people likely would've figured it out and then used that to their advantage. So... if you know that Monty always reveals a goat, never a car, then the answer is indeed 2/3. But I agree that the original problem, as stated, is not entirely clear. Jody


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Graham C
Member since Feb503

Jan2204, 07:45 AM (EST) 

21. "RE: Monty Hall problem... again"
In response to message #20

>>Someone correctly pointed out that in the original game and >>in the original statement of the problem, it was *not* >>specified that Hall always opened a door. In fact no >>strategy was given for Hall at all. > >But it did mention that he knew what was behind the doors, >which implies that he was opening the one with the goat "on >purpose". I don't think that follows, not in the oneoff case. He might have been, eg, opening at random, and would have said 'sorry, bad luck' if it'showed a car. Of course if you'd seen the show many times before and every time he'd shown a goat, then you would assume he was doing it deliberately. > >>Answers that depend on Hall always opening a door (or always >>opening a door with a goat) are therefore not answers to the >>original problem. >> >>A fortiori, if Hall does have a strategy, the contestant >>doesn't know whether he has one or not. > >Maybe it helps to have a little background. I happened to >watch Let's Make a Deal a lot as a kid. Monty often opened a >door (but not always). In all the times I saw him, he NEVER >revealed a car, always a goat (or giant mousetrap or some >such joke prize). > >Also, there was no apparent "strategy" to when he opened a >second door. If there had been, people likely would've >figured it out and then used that to their advantage. True. This was my point. They would have been able to assign values to the variables in my Bayesian equation. > >So... if you know that Monty always reveals a goat, never a >car, then the answer is indeed 2/3. The critical condition I pointed to was that the probability he opened a door be independent of whether the contestant had chosen a car or not. I slipped a little in saying that the probability of him opening a door and revealing a goat also had to be independent. This is untrue since the probability of revealing a goat is obviously dependent: it is 1 if the contestant chose the car. So, essentially, what you're saying is a special case of what I was saying, which is that whether he opens the door does not depend on the contestant's choice. > >But I agree that the original problem, as stated, is not >entirely clear. > It is a classic example of the value of Bayes' theorem, certainly.


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