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.

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_revealed|Win}=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_revealed|Lose}=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{Win|Goat_revealed} = P{Goat_revealed|Win}*P{Win} / (P{Goat_revealed|Win}*P{Win}+P{Goat_revealed|Lose}*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_revealed|Lose}=1.

Consequently,

P{Win|Goat_revealed} = P{Goat_revealed|Win}*P{Win} / (P{Goat_revealed|Win}*P{Win}+P{Goat_revealed|Lose}*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?

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

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.

Mark

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

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.

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

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

You make my point for me. It was a one-off 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 clear-cut.

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(A|B&C) = P(B&C|A)*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&C|A)? 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&C|A) and P(A|B&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&C|A)=P(B&C) and P(A|B&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(A|B&C)=1*(1/3)/(1/3)=1 and he should definitely not switch.

We are in decision-making under uncertainty territory, not decision-making under risk. It's a game-theoretic 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

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-

I don't think that follows, not in the one-off 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.