The Monty Hall Problem

This is a strange one. I can't think of another theoretical maths problem quite like it. The answer is so counter-intuitive that generations of gifted mathematicians have got it wrong and undoubtedly generations to come will too. It goes like this:

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. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 3 and the host opens Door 1, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

So is it better to stick with your original choice or switch to the other unopened door? At first it seems the obvious answer is that it makes no difference what you do. Its a 50/50 choice, right? Well, no, wrong actually. You are always better off switching to the other unopened door. You might not believe it but it's been tested time and again and the conclusion is always the same.

Switch.

I know what you're thinking. I thought the exact same thing. But the fact remains that I was wrong, and so are you. Here's why:

When you first make your choice it's a 1/3 chance that you have picked the door with the car behind it. When Monty opens one of the other doors and offers you the switch, by sticking with your original choice nothing has changed, so your odds are still 1/3. However, if you switch you are effectively getting both the opened door and the other unopened door, meaning yours odds have become 2/3.

Again, I know what you're thinking. "If I stick it's also a 2/3 chance because I now know what's behind one of the doors". But again, you would be wrong. Another way to think of it is to imagine 1,000,000 doors instead of just three. When you choose one door you have a 1/1,000,000 chance of winning. Monty then proceeds to open 999,998 of the other doors, all of which show goats, leaving just one other unopened door. If you now stick, your odds are still 1/1,000,000 because the new information makes no difference to the choice you originally had. If you switch your odds jump to 999,999/1,000,000, because you essentially have been given every door except your original choice. The same rationale holds for 3 doors.

The below figure should explain it:

Hopefully it now makes sense. If not, don't worry. You are in good company. As cognitive psychologist Massimo Piattelli-Palmarini says "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer."