The Monty Hall Problem

Wacky game show host Monty Hall is giving you a chance to win a Ferrari. He has hidden the Ferrari behind one door, a goat behind another door, and a goat behind yet another door. He then asks you to choose a door, and you do so. Instead of opening the door you picked, Monty opens a different door, one with a goat behind it. (Monty never opens a door with a Ferrari behind it.)

Now he gives you a choice: keep what's behind the door you picked or choose the other unopened door. What should you do? It turns out that you should change doors, because doing so increases your odds of winning the Ferrari. The puzzle is explaining why this is the case when it looks as if your odds are 50/50. After all, there are now only two doors and a Ferrari behind one--that’s one desired outcome and two possible outcomes. Why does it matter which one you choose?

The reaction to this puzzle is perhaps more interesting than the puzzle itself. Most people, academics and non-academics alike, insist on the intuitive explanation, i.e., that it makes no difference whether you switch or not. One of Marilyn vos Savant's most famous columns discussed the puzzle, and she received a heated response from her audience for saying that you should switch. However, it can be demonstrated empirically that the probability of winning after switching is higher than the probability of winning when sticking to your original choice. There are computer programs showing that you win more times by switching than you do by sticking. So we need an explanation.

Let's start by altering the story a bit--but this won't change anything substantial. We're going to name each goat. Let's call one George and the other Harry. So now we have a Ferrari, George, and Harry, each behind a door. The probability of choosing any one of them is 1 in 3. The probability of choosing the Ferrari or George or Harry is 1; in other words, there is a 100% chance that you are going to pick a door, so long as you play the game. That sounds obvious, but it's going to help drive the explanation home. Another thing to keep in mind is that your chance of winning (i.e., picking the Ferrari) is 1/3 but your chance of losing (i.e., picking a goat) is 2/3. Let’s recap:

• Pr of choosing F = 1/3
• Pr of choosing G = 1/3
• Pr of choosing H = 1/3
• Pr of choosing any door = 1
• Pr of choosing either G or H is 2/3

Now you choose a door. Monty goes to work and opens one of the doors you didn't pick. George is behind it munching on some hay. So the possibility of choosing George has been eliminated. Nonetheless, you're given a second choice to make: stick or switch. Note that the probability of choosing a door is still 1. Hence, one way of stating the puzzle is this: how do we keep this number constant, since it has to remain constant? Let’s illustrate this by removing George from our list:

• Pr of choosing F = 1/3
• Pr of choosing H = 1/3
• Pr of choosing any door = 1

The problem is that 1/3 and 1/3 don’t make 1. So the probabilities must be recalculated. But how? The intuitive solution is this:

• Pr of choosing F = 1/2
• Pr of choosing H = 1/2
• Pr of choosing any door = 1

Because 1/2 and 1/2 do make 1, it looks as if we've solved the problem. However, according to the empirical evidence, this is the solution:

• Pr of choosing F = 1/3
• Pr of choosing H = 2/3
• Pr of choosing any door = 1

This gives us the same result: it keeps the probability of choosing a door at 1. It is also supported by the empirical evidence. So what has gone wrong with the former solution? In a nutshell, it neglects to take into account new information that is revealed when Monty opens a goat door.

Consider again: you initially have a 1/3 chance of winning the Ferrari and a 2/3 chance of not winning it; you have a low chance of winning and a high chance of losing. To put this into sharper contrast, suppose that there are a hundred doors. Now you have a 1/100 or 1% chance of winning and a 99/100 or 99% chance of losing. That means that the winning door is most likely in the set of doors you didn't choose. This fact doesn't change when Monty begins opening doors. What does change is that the cardinal number of the set of closed doors--it gets smaller. (A set's cardinal number is the number of members in the set.) But remember, the probability of the winning door being in the set of doors you didn't choose stays the same. To make this clearer:

• Let X = set of doors you didn’t initially choose
• Let Y = set of open doors
• Let Z = set of closed doors

X and Z initially have the same cardinal number: 999. Monty opens one of the doors. Z now has 998 doors and Y has 1 door. The cardinal number of Y continues to increase, and the cardinal number of Z continues to decrease. The probability, as we've seen, that the winning door is in X is 99%. But the cardinal number of X is simply the sum of the cardinal number of Y and the cardinal number of Z; the cardinal number of X hasn't changed, and hence neither has the 99% probability that X has the winning door. What has changed is the number of doors in Z, a number which gets smaller and smaller as Monty opens more goat doors. Eventually Z is left with one member, and since that member is the only closed door in a set of doors having a 99% chance of having the Ferrari door, it has a 99% chance of being the Ferrari door.

Therefore, you should switch when Monty gives you the chance. The same is true when there are only three doors, although the chance of winning is 2/3 rather than 999/1000--still good odds in your favor. The heart of the paradox is that Monty always opens a goat door, and in so doing gives us information about the remaining closed doors belonging to the set of unpicked doors. Monty is telling us: "The likelihood of your door being the winning door is lower than the likelihood of the winning door being in the set of doors you didn't pick. But I’m going to help you out. I’m going to eliminate possible winners out of that high-probability set, and what's left over is still going to have the high probability of the entire set. So the smart money is on the door that I don't open, and the safe bet is to switch your allegiance to that door.”

Further reading

"The Monty Hall Problem" by Dr. Math.

"Game Show Problem" by Marilyn vos Savant.