I might not have described the original debate very clearly. My claim was that if Monty chose “leftmost non-car door” you still get the car 2⁄3 of the time by always switching and 1⁄3 by never switching. Your conditional probabilities look correct to me. The only thing you might be “missing” is that (A) occurs 2⁄3 of the time and (B) occurs only 1⁄3 of the time. So if you always switch your chance of getting the car is still (chance of A)*(prob of car given A) + (chance of B)*(prob of car given B)=(2/3)*(1/2) + (1/3)*(1) = (2/3).
One difference (outside the bounds of the original debate) is that if Monty behaves this way there are other strategies that also give you the car 2⁄3 of the time. For example, you could switch only in scenario B and not in scenario A. There doesn’t appear to be any way to exploit Monty’s behavior and do better than 2⁄3 though.
I might not have described the original debate very clearly. My claim was that if Monty chose “leftmost non-car door” you still get the car 2⁄3 of the time by always switching and 1⁄3 by never switching. Your conditional probabilities look correct to me. The only thing you might be “missing” is that (A) occurs 2⁄3 of the time and (B) occurs only 1⁄3 of the time. So if you always switch your chance of getting the car is still (chance of A)*(prob of car given A) + (chance of B)*(prob of car given B)=(2/3)*(1/2) + (1/3)*(1) = (2/3).
One difference (outside the bounds of the original debate) is that if Monty behaves this way there are other strategies that also give you the car 2⁄3 of the time. For example, you could switch only in scenario B and not in scenario A. There doesn’t appear to be any way to exploit Monty’s behavior and do better than 2⁄3 though.
Ah, I see, fair enough.