Ahh, the famous game show problem (also known as The Monty Hall Problem).
This is a probability puzzle you’ve heard of:
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 #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, “Do you want to pick door #2?” Is it to your advantage to switch your choice of doors?
Marilyn Vos Savant, one of the “smartest” people in the word, offers her answer:
When you switch, you win 2/3 of the time and lose 1/3, but when you don’t switch, you only win 1/3 of the time and lose 2/3. You can try it yourself and see…..
The winning odds of 1/3 on the first choice can’t go up to 1/2 just because the host opens a losing door. To illustrate this, let’s say we play a shell game. You look away, and I put a pea under one of three shells. Then I ask you to put your finger on a shell. The odds that your choice contains a pea are 1/3, agreed? Then I simply lift up an empty shell from the remaining other two. As I can (and will) do this regardless of what you’ve chosen, we’ve learned nothing to allow us to revise the odds on the shell under your finger.
The benefits of switching are readily proven by playing through the six games that exhaust all the possibilities. For the first three games, you choose #1 and “switch” each time, for the second three games, you choose #1 and “stay” each time, and the host always opens a loser.