The Monty Hall Problem, Simplified

By Cian Costello, Year 12

You’re on a game show. The prize for winning is a brand new car. The way the game show works is that the game show host (in this case it’ll be me) is standing in front of you, with three doors.

Behind one of those doors is the car you’ve always wanted. Behind the other two doors are goats. I know where the car is, but I won’t tell you. Suppose I ask you: “Pick a door! Any door!”. You choose a door. I tell you that that’s a good choice, but that I want to make this more interesting. I walk over to the doors and open one of them to reveal a goat.

I announce that you may switch from your door to the other door — if you’d like. Now before we proceed, what are the chances of you getting the car if you switched or if you stuck to your door? 50/50 would seem right as it looks like just a 1 in 2 chance of being one car richer, however, the odds of getting the car are actually ⅔ if you switch.

This problem was submitted to a maths newspaper called The American Statistician by the author Steve Selvin, who stated that the odds of getting the car after switching were ⅔. This caused many readers to write in and complain about the mistake made, as they believed it was a coin toss, in other words, a 50/50 chance. It was after this controversial theory that the problem became so popular.

To understand why it’s ⅔, it’s helpful to map out the possible outcomes. As you read the tree diagram, it starts to make sense why you should always switch, and why it is much more complicated than just a coin toss.

goat car

More intuitively, 2/3 of the time you will end up picking one of the doors with the goats. As the other goat is revealed, the remaining door must have the car behind it. Thus, 2/3 of the time, switching doors is the better decision.