The Monty Hall Problem
I came across this oddity while reading The Curious Incident of the Dog in the Night-time. From Wikipedia, the problem can be stated as follows:
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?
Common sense might tell you that because there are two unopened doors left, you have a 50-50 chance of picking the one with the car so it doesn’t matter if you switch doors. But in fact if you always switch doors, you will win a car 2 out of 3 times. To understand this you can look at the possible outcomes:
| You originally picked a door with a goat | You originally picked a door with a goat | You originally picked a door with a car | |||
| Stay | Switch | Stay | Switch | Stay | Switch |
| You win a goat | You win a car | You win a goat | You win a car | You win a car | You win a goat |
If you stay, you will only have a 1 in 3 chance of winning a car. But if you switch doors, you have a 2 in 3 chance.
