Subtle Twist on the Monty Hall Problem---Does It Make a Difference? In the Monty Hall problem, when the host picks a door and reveals an goat, does it make any difference if he did not know which door the real car was behind, and he just happened to pick a door with a goat?

This has been said in various ways in other answers/comments, but I think this is a clearer way to explain it that requires very little theory.
The most obvious difference is that sometimes when the host opened that door, the car would be there and the game would be over. This would happen 1/3 of the time. Also 1/3 of the time the car would be behind the door the player picked initially, and the other 1/3 of the time the car would be behind the door the player had the option of switching to.
If the question is "What is the probability of winning if you switch doors?" well it depends on whether you are including the instances where the contestant does not have that opportunity (on account of Monty opening the winning door by accident). If so the answer is 1/3. If, on the other hand, you are only looking at when Monty doesn't show the car prematurely (maybe the contestant gets to go again?) well then 1/3 out of 2/3 is 1/2.
