# That scene in that movie

1. Sep 14, 2008

### chhitiz

in 21, when kevin spacey plays 'the 3 doors' with the lead character in a classroom, the guy says something like:"before you opened the door, the probability of each door having a car was 33.33% but now since door 3 has a goat and i have door1, the probability of door 2 adds up to 66.66% so i will take door 2 now." shouldn't the probability have become 50% for both doors?

2. Sep 14, 2008

### mathman

No! When door 1 was chosen, it had a probability of 1/3. Opening door 3 doesn't change that. Therefore door 2 has a probability of 2/3.

3. Sep 14, 2008

### Ygggdrasil

It sounds like you're referring to the classic Monty Hall problem. The answer is indeed 2/3 but it assumes that you know the host's behavior exactly.

4. Sep 27, 2008

### ssd

Actually, door (2 or 3) had 2/3 probability. Now that door 3 has 0 prob after openning, prob of door 2 becomes 2/3.
To remove your confusion, think of 1000 doors behind one which there is a prize. You select one door and then 998 of the remaining doors are openned which are empty. Should you not then switch your choice?