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 $2/3$.

**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 $1/2$.
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 $2/3$

### Generalization

The probability of being initially wrong is the same as the probability of being correct after switching. We can generalize this to $n$ doors. The probability of winning the game by switching after the host has opened $n-2$ doors is $(1 - 1/n)$

### Script

Not convinced? Simulate using this Colab Script