Since there's ambiguity, you should assign prior probabilities to the two types of behavior Monty can have. Then the overall probabilities are the average of th

Since there's ambiguity, you should assign prior probabilities to the two types of behavior Monty can have. Then the overall probabilities are the average of the probabilities for each of the possible cases, weighted by this prior. So of you think there's 1/2 chance Monty always reveals a goat (in which case swiching wins 2/3 of the time) and a 1/2 chance Monty picks randomly (in which case switching wins 1/2 the time), then your overall chance of winning if you switch is 1/2×2/3 + 1/2×1/2 = 7/12. So you should still switch. Indeed you should switch if you think there's any chance at all that Monty always opens doors with goats. (Unless of course you think that there's also a chance that Monty always reveals cars whenever he can, but that wouldn't make for very good TV.)