Divide the doors into two sets. The first set is the door you chose and the door that was opened; the second set is the remaining door. There is a 2/3 chance th

Divide the doors into two sets. The first set is the door you chose and the door that was opened; the second set is the remaining door. There is a 2/3 chance the correct door is in the first set and a 1/3 chance the correct door is in the second set. The error is assuming that the first set has a 2/3 probability of winning just because it contains 2 out of 3 doors while the way the set is constructed ensures that one of the doors, the door that is opened, never is the winning door. Including or excluding this door from a set must not change the probability you assign to this set containing the winning door, i.e. you can freely move this door between two sets or drop it altogether without affecting probabilities.