Monty Hall problem

The Monty Hall problem, also called the three doors problem, is a probability puzzle that has confused plenty of people.

The problem

Let’s say there are three doors. The host, Monty Hall, puts a car behind one door and a goat behind each of the other two. Your task is to find and pick the door with the car behind it. He invites you to choose one of the three doors, which we will call A, B and C. You pick one door, say B. Now the host comes into play. From the remaining doors, that is A, C, he opens a door with a goat behind it – so he shows you a door that is definitely not the right one. Say he opens door A.

The point of the whole problem comes in the next step – the host offers you the chance to change your choice. You first chose door B, then the host told you that the car is not behind door A and offers you the chance to switch to door C. But you can also stick with your original choice, B.

What do you decide? Is the car more likely to be behind door B or door C? Or does it make no difference?

The solution

You have three doors. One door is out of the game, so you really only have two doors left. The car is definitely behind one of these two. You know nothing about the two remaining doors and have no further clues. So the probability is, of course, 50% for each door.

That is roughly how a large proportion of people think, so don’t worry if you thought the same. But of course the solution is wrong :-).

The correct solution

It pays to switch to the door that the host did not eliminate. To understand why, we can write out the probabilities from the beginning. When you first choose a door, the probability that the car is behind it is 1/3. We have no further clues, so the probability is spread evenly, 1/3 for each door.

So if you choose door B, you have a 1/3 chance that the car is behind it. At the same time, there is a 2/3 probability that the car is behind door A or C. So it is more likely that the car is behind a door we did not choose. This is important.

In the next step the host comes into play and opens one of the doors with no car behind it. The host opens door A. So the car is not behind door A. What happens to that 2/3 probability? We know that the probability that the car is behind door A or C is 2/3, and at the same time we know it is not behind door A. So the whole 2/3 probability moves to door C. The chance that the car is behind door C is 2/3.

But our chance for door B stays the same – it has not changed at all. So we have a 1/3 chance that the car is behind door B, which we chose, and a 2/3 chance that it is behind door C, which we can switch to. So it makes sense to switch.

More doors

The example may be easier to understand if we increase the number of doors and change the problem like this: suppose there are ten doors. At the start the contestant picks one door, and the host opens all the doors except the one we picked and the one with the car behind it. The problem is practically the same as in the previous example, except that there, when the host opened the doors, he opened just one door because he could not open any more.

At the start you pick door 1. The chance that the car is behind it is 1/10. There is a 9/10 probability that the car is behind one of the other doors. Then the host removes 8 doors that definitely have no car behind them, and again you are left with just two doors. The chance that the car is behind the door the host left closed is 9/10, while the chance that it is behind the door you picked is still 1/10.

Further sources

• Wikipedia