| easy | probability |
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. Now, do you want to pick door No. 2? What is the probability to win the car if you switch?
The host always opens a door with a goat.
2/3
The probability that your initial choice did not have a car is indeed .
Initial Misstep: After one door is opened, there are exactly two doors left, and one of them has a car. So the probability that the car is behind either door is . This is incorrect because the host knows which door has a car and which door has a goat. The host always opens a door with a goat.
Correct Solution: The following table shows which Door the host might open. Assume that the car is behind Gate #1, and we randomly choose one door.
| Initial choice: | Door #1 | Door #2 | Door #3 |
|---|---|---|---|
| Reality | Car | Goat | Goat |
| Host opens: | #2 or #3 | #3 | #2 |
| Remaining: | #3 or #2 | #1 | #1 |
| Good to switch: | No | Yes | Yes |
We see that at the end, the remaining unopened door is Door #1 if we start with Door #2 or Door #3. This means that in 2 out of 3 cases, we started with the incorrect Door (#2 or #3) and we got the option to switch with the correct door at the end (#1).
So you should switch to the other door, and win the car with a probability of
Generalization
The probability of being initially wrong is the same as the probability of being correct after switching. We can generalize this to doors. The probability of winning the game by switching after the host has opened doors is
Script
Not convinced? Simulate using this Colab Script