Person One enters the Monty Hall problem as usual with usual rules. Of the three doors A, B and C, One chooses B, Monty opens C (goat) and One switches to A calculating that the probability of a car behind A is 2/3.
Before One is allowed to open door A he is asked to sit down and remain silent. He is told that he will get one more chance to switch but not until later. Monty closes door C. The car and goats remain where they are.
Person Two enters the same Monty Hall problem. Of the three doors Two chooses A, Monty opens C (same goat) and Two switches to B calculating that the probability of a car behind B is 2/3.
Before opening door B Monty tells Two he will be allowed one more switch of choice in a few minutes.
Before One and Two have to make their final choice they are allowed to discuss the situation. Both One and Two are brilliant statisticians and 100% honest people and both know that about each other.
One explains what he did and that he reached the conclusion to open A as with the information he had at the time that gave the probability that the car is behind B is 2/3. Two, having been watched by One, does not have to explain that he has calculated the probability that the car is behind A is 2/3.
One and two now get into a discussion to determine the probabilities of the car being behind A or B.
What conclusions do One and Two reach?
My first thoughts are that
P(Car behind A) + P(car behind B) = 2/3 + 2/3 > 1 impossible.
Now too much contradictory information, so now in a situation of a fifty fifty choice.
Any better thoughts?
EDIT 1 Following David K's (quite correct) request for assumptions to be made explicit here they are
- Monty Hall is 100% honest
- The goats and car are place behind A, B and C at random.
- Monty Hall will always open a door with a goat behind it never the car
- Monty Hall will always open the door that gives away the least information, when equal he will chose a door at random.
EDIT 2 Given these assumptions I believe Alexander Geldhof provided the solution. In order to clarify my thoughts I re-write that as
P(X) means the probability that the car is behind door X.
When One enters he calculates P(A) = P(B) = P(C) = 1/3
Once Monty opens C he knows P(A) = 2/3, P(B) = 1/3 and P(C) = 0;
Two enters and makes his choice and decides P(B) = 2/3.
One provides the following argument to Two
When you entered you calculated that P(B) = 2/3, this was based on what you knew and the logical assumption that P(A) = P(B) = P(c) = 1/3. However my and Monty's previous actions determined that when you entered the P(A) = 2/3, P(B) = 1/3 and P(C) = 0 and I can give you that information.
Once your first choice was A, Monty could only open door C. If the car was behind B he could not open it so had to open C. If the goat was behind B then when he opened it we would know that the goats were behind B and C and so the car was behind A. So he had to open C to give us the least information. In fact this provided no new information since P(C) = 0 is already known.
So given all the information we are currently provided with it remains P(A) = 2/3, P(B) = 1/3 and P(C) = 0. So the best chance of the car is for us to open door A
Best Answer
At the start, there are three equally likely car-distributions: car behind A, car behind B, car behind C.
Now One comes in and chooses door B. Monty is now forced to open a goat-door which is not B, as by the hidden rule of this problem. In the A-situation, he'd be forced to open C, in the C-situation, he'd be forced to open A, and in the B-case, he can choose doors.
Here's where the trouble starts. By naming Monty's door C, you've merged the A- and C-situations; both are equally likely, so now there's a 2 in 3 chance of the car being behind A, 1 in 3 chance of the car being in B, and ZERO chance of the car being in C.
Two enters. He chooses door A, where the car is going to be two times out of three. What happens after this does not really matter - swapping is only going to net him the car one out of three times. With the extra information of A, this would also be the conclusion of B, being a perfect statistician.
Note that if Two would choose a random door, and Monty would again be forced to open a goat-door, without extra information, swapping would give him the car two times out of three! You just hand-picked the worst option for Two (door A), and by doing so fudged the answer