Tuesday, August 7, 2007
I promised a follow-up to my Friday post on the Monty Hall problem, so here it is. We found on Friday that, even though the three doors from which we have to choose have equal probabilities of being a winner, after Monty reveals a door, we are twice as likely to win if we switch to the remaining door. So, basically, instead of the two remaining doors having a 50/50 probability of revealing a car, they have a 67/33 probability, with the door you chose having the 33% probability, the same probability it had when you originally chose it. So we found that, counterintuitively, the probability stays the same and does not go up even after Monty reveals a goat behind one of the other doors. But can we get even more counterintuitive? Can the probability of our door go down AFTER Monty reveals another door contains a goat. It certainly doesn't seem possible. If we have three or four or five doors to choose from, and Monty essentially reduces the number of doors that could be a winner by revealing that one contains a goat, how could the probability of our door being a winner go down? In fact, that is what Jason Rosenhouse found when he analyzed a five-door Monty Hall problem in the post I linked to on Friday. In his analysis, you start out with five doors, and you choose, for example, Door 1, and Monty reveals a goat behind Door 2. You then, unwisely as it turns out, switch to Door 3. At this point, Door 3 has a 27% (4/15) chance of being a winner. If Monty then opens Door 1, the door you originally chose, and reveals a goat, it turns out that the new probability for Door 3 to be a winner is 25%. That is, the probability has gone down even though Monty has eliminated one of the possible doors. Rosenhouse's analysis, based on a probability theorem called Bayes' theorem, is complicated and hard to follow, so let's go back to the three-door problem to see a more straightforward illustration of this weird effect. Let's now assume that the game is fixed. A trusted friend of ours who works on Monty's staff tells us that the car is most definitely not behind Door 1. So we go into the game knowing that Door 1 has zero probability of winning, and Doors 2 and 3 each have a 50% probability of winning. Our best strategy in this case is to choose Door 1, forcing Monty to reveal whether Door 2 or Door 3 contains the goat. If he opens up Door 2 and shows it has a goat, we now know that Door 3 has a 100% chance of winning, so we will switch to that door, absolutely certain that we will win the car. But let's suppose we flub things and choose Door 2 initially. Monty then opens Door 1, which we know already will not be the winner. So Monty has given us no new information, meaning that the probabilities of Doors 2 and 3 are unchanged, right? Wrong. Let's look at the scenarios, remembering that we have chosen Door 2:
1. If Door 2 is the winner (1/2 the time), then Monty will reveal the goat behind Door 1 1/2 the time and the goat behind Door 3 the other 1/2 of the time.
2. If Door 3 is the winner (1/2 the time), then Monty will always reveal the goat behind Door 1.
Calculating each of the probabilities, we can rewrite the list like this:
1. Door 2 is the winner, and Monty will reveal the goat behind Door 1 1/4 of the time.
2. Door 2 is the winner, and Monty will reveal the goat behind Door 3 1/4 of the time.
3. Door 3 is the winner, and Monty will reveal the goat behind Door 1 1/2 of the time.
So, numbers 1 and 3 in the above list correspond to the situation where you choose Door 2 and Monty opens Door 1. We see that for those list items, it is twice as likely that Door 3 is the winner than that Door 2 is the winner. In other words, Door 3 has a 2/3 chance of being a winner and Door 2 has a 1/3 chance of being a winner. But remember that Door 2 initially had a 1/2 chance of being a winner. So, amazingly, the probability of Door 2 being a winner went down when Monty revealed the goat behind Door 1. The reason for this is that Monty gave us a lot of information by not choosing to reveal a goat behind Door 3. If Doors 1 and 3 both contain goats and Door 2 is the winner, then Monty will be equally likely to open either door. The fact that he chose Door 1 tells you that there is a good possibility that he avoided Door 3 because it is the winner. In effect, Monty raised the probability of Door 3 by opening Door 1. Since only Doors 2 and 3 had any probability of winning, if the probability of Door 3 goes up, the probability of Door 2 must go down. Even though we analyzed the simple 3-door problem, it will always be the case that, when the doors have unequal probabilities, Monty will lower the probability of your chosen door by opening a low probability door.
1. If Door 2 is the winner (1/2 the time), then Monty will reveal the goat behind Door 1 1/2 the time and the goat behind Door 3 the other 1/2 of the time.
2. If Door 3 is the winner (1/2 the time), then Monty will always reveal the goat behind Door 1.
Calculating each of the probabilities, we can rewrite the list like this:
1. Door 2 is the winner, and Monty will reveal the goat behind Door 1 1/4 of the time.
2. Door 2 is the winner, and Monty will reveal the goat behind Door 3 1/4 of the time.
3. Door 3 is the winner, and Monty will reveal the goat behind Door 1 1/2 of the time.
So, numbers 1 and 3 in the above list correspond to the situation where you choose Door 2 and Monty opens Door 1. We see that for those list items, it is twice as likely that Door 3 is the winner than that Door 2 is the winner. In other words, Door 3 has a 2/3 chance of being a winner and Door 2 has a 1/3 chance of being a winner. But remember that Door 2 initially had a 1/2 chance of being a winner. So, amazingly, the probability of Door 2 being a winner went down when Monty revealed the goat behind Door 1. The reason for this is that Monty gave us a lot of information by not choosing to reveal a goat behind Door 3. If Doors 1 and 3 both contain goats and Door 2 is the winner, then Monty will be equally likely to open either door. The fact that he chose Door 1 tells you that there is a good possibility that he avoided Door 3 because it is the winner. In effect, Monty raised the probability of Door 3 by opening Door 1. Since only Doors 2 and 3 had any probability of winning, if the probability of Door 3 goes up, the probability of Door 2 must go down. Even though we analyzed the simple 3-door problem, it will always be the case that, when the doors have unequal probabilities, Monty will lower the probability of your chosen door by opening a low probability door.