I don’t understand your Monty Hall example. If Monty always reveals the leftmost non-car door that you didn’t pick (which I guess is what you mean), then he will reveal either (A) the leftmost door that you didn’t pick or (B) the rightmost door that you didn’t pick. In case (B) you can be sure that switching will result in a win, because if Monty passed over the leftmost door then it must be because it contained a car. On the other hand, in case (A) the chance is actually 50% that you win by switching, because Monty’s algorithm gives no information as to whether the rightmost (unrevealed) door contains a car. In both cases the conclusion is different from the standard (random) Monty Hall problem, where you have a 2⁄3 chance to win by switching. What am I missing?
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.
If Monty always chooses the left most door that doesn’t have the car behind it, then, if he chooses, out of the doors you didn’t pick, the left one, then you should switch. If he chooses the right door you shouldn’t.
This is true. However, if you don’t know in advance if Monty has the rule:
‘open the leftmost door that doesn’t have a car behind it’
or
‘open the rightmost door that doesn’t have a car behind it’
Then even if you knew that Monty had one of those rules, then which door Monty opens doesn’t tell you anything (until after you find out the results of your choice).
I don’t understand your Monty Hall example. If Monty always reveals the leftmost non-car door that you didn’t pick (which I guess is what you mean), then he will reveal either (A) the leftmost door that you didn’t pick or (B) the rightmost door that you didn’t pick. In case (B) you can be sure that switching will result in a win, because if Monty passed over the leftmost door then it must be because it contained a car. On the other hand, in case (A) the chance is actually 50% that you win by switching, because Monty’s algorithm gives no information as to whether the rightmost (unrevealed) door contains a car. In both cases the conclusion is different from the standard (random) Monty Hall problem, where you have a 2⁄3 chance to win by switching. What am I missing?
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.
There are 3 doors1, 2, 3You pick one of them.(1), (2), (3)If Monty always chooses the left most door that doesn’t have the car behind it, then, if he chooses, out of the doors you didn’t pick, the left one, then you should switch. If he chooses the right door you shouldn’t.This is true. However,if you don’t know in advance if Monty has the rule:‘open the leftmost door that doesn’t have a car behind it’
or
‘open the rightmost door that doesn’t have a car behind it’
Then even if you knew that Monty had one of those rules, then which door Monty opens doesn’t tell you anything (until after you find out the results of your choice).