3 ms·
Divide the doors into two sets, the door that you chose and those that you did not. There is a 1/3 chance that the correct door is in the first set. There is
by yarg 9y ago
Divide the doors into two sets, the door that you chose and those that you did not.
There is a 1/3 chance that the correct door is in the first set.
There is a 2/3 chance that the correct door is in the second set.
You are then told that one of the doors in the second set is not the correct door.
This new information has no impact on the probabilities associated with the complete sets - it only impacts the probabilities within the second set.
Previously both doors (in the second set) had a 1/3 probability, now one has been eliminated and the remaining has a 2/3 probability.
- valleyer 9y agoI've seen this explanation a million times, and I don't think it's very good. Here is one that sounds almost as plausible yet yields the opposite result: 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. You are then told that one of the doors in the first set is not the correct door ... I find this line of thought much more intuitive: there is a 2/3 chance your original guess was wrong. If your original guess was wrong, the other door (the one that is not the one you chose and is not the one that was opened) is the right door. So there is a 2/3 chance the other door is the right door.
- danbruc 9y agoDivide 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.
- deleted 9y ago[deleted]
- fhars 9y agoYour first arguement assumes that you can choose the door you pick initially and the door with the goat that I will open once you heve picked your first door before I have opened that second door. That is not even remotely plausible. The second argument is of course correct.