Monty Hall/Bayes' Theorem conflict?

Question

Given the Monty Hall problem:

Assuming Player chooses door A and Monty opens door B, what is the probability that the car is behind door C?

The following calculation can be found in many places in the web (always assuming that player has chosen A):

P(Car_c|Open_b) = P(Car_c) * P(Open_b|Car_c) / P(Open_b)

P(Open_b|Car_a) = 1/2

P(Open_b|Car_b) = 0

P(Open_b|Car_c) = 1

The probability P(Open_b) is the sum of P(Car_x) * P(Open_b|Car_x)

P(Open_b) = 1/3 * 1/2 + 1/3 = 1/2.

This gives P(Car_c|Open_b) = 1/3 * 1 / 1/2 = 2/3 as expected.

Assuming we slightly change the rules so that Monty always has to open door B if the Car is behind door A.

This would make P(Open_b|Car_a) = 1, leading to P(Open_b) = 2/3.

Now P(Car_c|Open_b) = 1/3 * 1 / 2/3 = 1/2.

This does not seem correct, since the situation is still the same: The probability that the car is in one of the doors that the player did not choose is 2/3, and Monty opens one of those doors that does not hide the car. So the probability that the car is behind the remaining door must always be 2/3.

So the assumptions leading to the formula must be wrong, but where is the problem?

Answer 1

In the standard version of the Monty Hall problem the the probability that the first door you pick has the car behind it is $\dfrac{1}{3}$. Since Monty will always open a door without the car behind it the probability that you will win if you switch is $\dfrac{2}{3}$ and if you stick $\dfrac{1}{3}$.

In your modified version where Monty's move is forced, then things change

If you always pick A then Monty can pick B if the car was behind A or C.

If Monty picks C however, then you know the car can't have been behind A otherwise he would have been forced to pick B. So it must be behind B.

So an optimum strategy is that if Monty picks C Change to B otherwise it makes no difference if you swap or not.

Answer 2

The calculation of the probability of the car being behind door C under the new rules is correct, and there is no contradiction, because the rule that Monty has to open door B if the car is behind door A will cause door B to have a higher unconditional probability of being opened than door C. From the contestant's perspective, this means that if Monty opens door B under the new rule, there is a higher posterior likelihood that the car is behind door A compared to the old rule, in which the equal likelihood of opening door B or C if the car is behind door A is not informative about whether the car is behind A.