As usual on a Sunday night, I was watching the weekly episode of the Mythbusters TV show, and to my surprise they were testing the Monty Hall problem. I had thought that by now it was so well explored that it wouldn't qualify as a myth, but then I realized that most people are still unaware of this curious example of deceptive statistics. And so for that reason, although the TV show did an excellent job of demonstrating this problem, I will give an overview here.

The Monty Hall problem is very simple to state. You are on a gameshow, and shown three doors which I will call A,B and C. Behind one of them is a prize, which you will win if you guess the correct door. You win nothing for guessing wrong.

But there is a catch in the game, meant to build drama and suspense. After you make you selection, the host opens one of the doors that you didn't pick. (They often claim that this is to see if you guessed wrong right off, but in truth the host knows that they are selecting a losing door.). Then you have one winning door and one losing door, and the host gives you a chance to change your selection or stick with your first choice.

Most people, even those with mathematics training in many cases, figure that it is a 50% chance either way and so it makes no difference. Psychologically most people will stick with their first choice.

And yet changing your pick actually gives you a 67% chance of winning, while staying with the first pick is only right 33% of the time!

This seems counterintuitive, because it is designed to be deceptive. But because the host has information on where the prize is, and because he uses that information, the odds are changed.

Consider a similar game without the door opening.  When you first choose, there is a 33% chance that you have guessed correctly and a 67% chance that you are wrong. You are then allowed to choose the one door that you picked, or go with both doors that you didn't pick.

After you have made that choice, the host opens a losing door but it is too late to change. The odds now are clearly in favour of switching when given the chance.

Or another way of viewing the problem that makes the information exchange more transparent. Suppose that after your first choice you somehow know for a fact that it is wrong (in reality there is a 67% chance of being wrong, but suppose that is raised to 100%).  Then the gameshow host, knowing where the prize is, tells you one more wrong choice. You know one of the losing doors, and the host tells you which is the other losing door! That means you are now 100% certain of the winner. In the real situation, you are 67% certain that your choice is wrong, and the host tells you which other door is 100% certain to be wrong, so that now you are 67% certain that the remaining door is the winner.

And of course the final way of displaying this statistical fact is to simply write down all the possible ways that you can select one door, be shown another that is a loser, and then see how many let you win by changing your mind and how mind let you win by sticking with your first choice. (Just be careful to add up the probabilities correctly). If you do this right, then again you will see a 67% win rate for switching.

So that is the infamous Monty Hall problem. A simple game in which it appears to be even odds of winning, but for which there is a simple but clear strategy for doubling your chances. Statistics can be very deceptive if not viewed correctly...