Consider the following problem

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. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

.

(Formulation by Marilyn vos Savant at Whitaker, Craig F. (1990).)

.

So what is the answer? Let’s see a simple solution. The following table shows the three possible arrangements of two goats and one car behind three closed doors and the result of switching or staying after initially picking Door 1

.

Door 1 |
Door 2 |
Door 3 |
Outcome if switching |
Outcome if staying |

Car |
Goat |
Goat |
Goat |
Car |

Goat |
Car |
Goat |
Car |
Goat |

Goat |
Goat |
Car |
Car |
Goat |

.

Which means that in 2 out of 3 cases we would win the car by switching doors. Hence it would make perfect sense to always switch. To understand why this is happening do take into consideration that the event “staying” is composed by only one element while its complement by two (note that all elements are equiprobable).

.

Frequency of winning when staying/switching doors in a sequence of "Monty-Hall" games