It’s Christmas, and you know what that means – gifts for everyone! My gift to all of you will be… the gift of knowledge!

That’s actually too cheesy, so I’ve gone for a gift-related maths problem (I’m really pushing it here to keep it related to Christmas to be honest) – the infamous Monty Hall problem. Loosely based on the American game show *Let’s Make a Deal* and named after its original host Monty Hall, the problem is the following.

You’re on a game show and you’re given the choice of three doors, which we’ll number door #1, #2 and #3. You take home whatever is behind the door you choose. The game show host tells you that behind one door is a car, and behind the other two doors are goats. There’s no trick to figuring out which door has the car behind it, so your guess is entirely random – i.e. there’s a 1/2 chance of choosing the door with the car behind it.

So there’s nothing you can do but choose a door at random, say door #1. However, the game doesn’t end at that. The host opens a different door which has a goat behind it, say door #3. Now he asks you whether you want to stick with your original choice, door #1, or to switch to door #2. What should you do if you want to increases your chances of taking home the car?

The answer at first appears to be that it doesn’t matter whether you stick or switch. One of the doors has a car behind it, and the other has a goat behind it, so either way, the probability of choosing the car should be 1/2. This isn’t correct, however, as we’ll see when we look at the problem with a hundred doors.

Once again, you pick one of the hundred doors at random, say door #1. You think to yourself, ‘how could I possibly win this, there’s such a low probability of getting the car’. But then the host opens doors #2 to #99 to reveal goats, leaving doors #1 and #100 closed. Now he asks if you want to stick with door #1, or to switch to door #100.

Here it becomes clear that it is advantageous to switch to door #100. The original door you chose had a 1/100 chance of having a car behind it. This probability does not change when the host reveals the ninety-eight goats, so therefore switching results in there being a 99/100 chance of receiving the car. Because the door you chose originally had a low probability of having the car in the first place, the door you can swap to must have a high probability of having the car.

Now when we reduce the number of doors to three, the same logic applies. The door you stick with has a 1/3 chance of having a car behind it, and the door you switch to has a 2/3 chance. So if you ever find yourself in this situation (basically never), always switch!

You get a car! And you get a car! And you don’t get a car! Two-thirds of you get a car! And have a very merry Christmas!

*Yanhao*

The same scenario was used in a subplot in a Brooklynn NineNine episode. That is how I know of the same problem

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