3 ms·
> Either of your interpretations of Monty picking a door at random lead to a conditional probability of 1/2 that your chosen door has the car given that he reve
by henrikschroder 7y ago
> Either of your interpretations of Monty picking a door at random lead to a conditional probability of 1/2 that your chosen door has the car given that he revealed a goat.
No, if Monty picks a door by random, but the show shuffles the doors until he gets it "right" and picks a goat, that is exactly the same scenario as if Monty knows where the goats are and picks one straight away, which means my odds of winning when switching is 2/3.
So that is a scenario where "Monty picks a door by random" doesn't lead to the odds changing to 50/50.
- daveFNbuck 7y agoIf by shuffle you mean that Monty just gets to keep trying until he picks a goat, you're right. If instead you mean that they randomize all three doors and start over, the probability is 1/2 because half the time that the player has a goat you start over and get to pick again, which perfectly cancels out the double probability of having a goat. This can be analyzed pretty easily using the three possibilities you outlined earlier. Suppose the player gets door 1 and Monty always picks door 2. Each time they shuffle, you get one of the following with equal probability: A. car-goat-goat B. goat-car-goat C. goat-goat-car A means the player wins, B means we re-shuffle, C means the player loses. It's pretty easy to see that A and B are equally likely outcomes, so they become 50/50 odds after all the re-shuffling is done.