This post on EvolutionBlog brought back to mind one of the classic non-intuitive mathematics problems, the Monty Hall problem. The problem goes like this-- you are on a Let's Make a Deal-style game show, where you have three doors to choose from. One of the doors has a car behind it, and the other two have goats. There is an equal 1/3 probability for the car to be behind each of the doors, and the game show host, Monty Hall, knows which door contains the car. You choose Door 1. Monty Hall, wanting to maximize suspense, reveals that Door 2 contains a goat, and asks if you would like to switch to Door 3. Intuition tells you that there is nothing gained by switching doors, since each door is equally likely to contain the car. But in fact, your probability of winning by sticking with Door 1 is 1/3, while your probability of winning by switching to Door 3 is 2/3. How can that be? If each of the doors started off with equal probability of being a winner, how can the doors suddenly change their probabilities mid-game? There are a couple ways to view the situation. Probably the most straightforward is that your door has a probability of winning of 1/3, while the other two doors have a combined probability of winning of 2/3. When Monty reveals that a goat is behind Door 2, that door suddenly has a probability of winning of 0, meaning that now Door 3 has a probability of winning of 2/3. Another way to look at it is in terms of information. Monty knows which door contains the car, and he can share that information with you in various ways. When you choose a door, you restrict Monty from opening the door, so you restrict him from revealing any information (directly, at least) about the door you have chosen. So the information he can give you is about the other two doors. To better illustrate the situation, imagine that there are four doors, each with a 1/4 probability of containing the car. But before you choose a door, Monty tells you that Doors 1 and 2 both contain goats. We are assuming that Monty is always honest with us, so what he has done is told us that Doors 3 and 4 now have a 1/2 probability of containing the car. But let's say that we know Monty will never tell us anything about Door 4. All we know is that at the beginning of the game each of the doors has an equal probability of being a winner. So, if he tells us before we choose that he knows that, among Doors 1, 2, and 3, Doors 1 and 2 contain goats, we know that Door 3 now has a 3/4 probability of containing the car. This is because he will never reveal anything about Door 4 to us, but the fact that he DIDN'T tell us that Door 3 contains a goat tells us volumes. Since we assume that Monty is randomly choosing which goat doors to reveal to us, there is very low probability that Monty would randomly not tell us that Door 3 contains a goat if it did, but much better probability that Monty would avoid telling us about Door 3 because it contains a car. Once again, he will never say anthing about the door we chose, Door 4 in this case. If you're thoroughly confused by my analysis, you can fairly easily go through the possibilities in the 3-door problem and convince yourself that you're better off switching. Assume you choose Door 1, which is a winner 1/3 of the time, then the possibilities are:

1. If Door 1 is a winner (1/3 of the time), then Monty will reveal the goat behind Door 2 1/2 the time and the goat behind Door 3 the other 1/2 of the time.
2. If Door 2 is a winner (1/3 of the time), then Monty will always reveal the goat behind Door 3.
3. If Door 3 is a winner (1/3 of the time), then Monty will always reveal the goat behind Door 2.

We can calculate the various probabilities, and rewrite the list like this:

1. Door 1 will be a winner and Monty will reveal Door 2 1/6 of the time.
2. Door 1 will be a winner and Monty will reveal Door 3 1/6 of the time.
3. Door 2 will be a winner and Monty will reveal Door 3 1/3 of the time.
4. Door 3 will be a winner and Monty will reveal Door 2 1/3 of the time.

So, if you've chosen Door 1 and Monty chooses Door 2 (possibilities 1 and 4 in the above list), then we can see that there is twice the probability that Door 3 will be the winner than that Door 1 will be the winner. Similarly, if you choose Door 1 and Monty chooses Door 3 (possibilities 2 and 3 in the list), you will have twice the chance of winning by switching to Door 2. More on this tomorrow.

Posted at 8:27 AM | Permalink | Comments (2) | TrackBacks (0)