Madness In The Middle Kingdom

With one comment on the Monty Hall Problem I can see that my efforts to stimulate the viewers with some good ol’ mathematical problems have fallen short of expectations. Nevertheless I will continue to press on in my efforts to train your brain.

So here is the solution which most people find surprising.

It is better to switch than to remain with your original door. The common thinking (and I this is what I thought as well) is that since there are two doors we don’t know and one of them contains the car then there is 1/2 chance that the car is behind one of the doors. Sounds nice but unfortunately it doesn’t work quite that way.

When you chose the door initially there was only a 1/3 chance you chose the car and a 2/3 chance that it was one of the other doors. When the host opens a door (not containing the car) he has changed nothing because he knows which door has the car and he purposefully shows you a goat. There is still only a 1/3 chance you chose the car and a 2/3 chance you did not. It is more likely the car was behind a door you did not choose to begin with - remember the cars and goats do not move.

If this doesn’t sit right imagine there are 100 doors and you choose one at random. The host then shows you 98 goats and leaves you 2 doors to choose from - yours and another. You only had a 1/100 chance to pick the door correctly at the start and since the car doesn’t move then you still only have a 1/100 chance of winning so you should definitely switch.

Click here for a website to play the game. Check you results after playing many trials. Was it better to stick or switch?

There is still a few more days to tackle that Birthday Problem