> therefore when they switch, they are choosing between one wrong and one right door, every time, regardless of what came before. (p=1/2) I get why you intuit

> therefore when they switch, they are choosing between one wrong and one right door, every time, regardless of what came before. (p=1/2) I get why you intuit this - I had a lot of trouble with it at first as well. Common sense is lying to you, however, because you are still dealing with the original odds (1/3 vs 2/3). Monty's choice is always to choose a Goat (probability 1.0), therefore his effect on the probabilities will always be with a weighting of 1.0. You end up with a probably spread of (1/3 * 1.0 vs 2/3 * 1.0). Therefore it is always in your interests to choose Monty's remaining door.