The Monty Hall Problem

Three doors and one car

A car is hidden behind one of three closed doors, and a goat stands behind each of the other two. The car is equally likely to be behind any of the three doors. You pick a door, say door 1.

Your first pick hides the car with probability $\frac{1}{3}$1/3, and a goat with probability $\frac{2}{3}$2/3.

The first pick finds the car one time in three, and a goat two times in three.

What the host must do

The host knows where the car is. He always opens one of the two doors you did not pick, he always shows a goat, and he always offers you the chance to switch to the last closed door.

If your first pick is the car, both of the other doors hide goats, and the host can open either one. If your first pick is a goat, the other two doors hide the car and the second goat. The host may not open the car, so he has no choice: he opens the door with the second goat.

List every case

You pick door 1. If the car is behind door 1, the host opens door 2 or door 3, and switching loses. If the car is behind door 2, the host must open door 3, and switching to door 2 wins. If the car is behind door 3, the host must open door 2, and switching to door 3 wins.

The three cases are equally likely, and switching wins in 2 of them. So the probability of winning by switching is $\frac{2}{3}$2/3.

The same is true whichever door you pick first. With 3 places for the car and 3 first picks, there are 9 equally likely cases. In 3 of them the first pick is the car, and switching loses. In the other 6 the first pick is a goat, the host is left with one goat door to open, and switching wins: $\frac{6}{9}=\frac{2}{3}$6/9 = 2/3.

Down the side is the door that hides the car, and across the top is your first pick. In each of the 6 colored cells the first pick is a goat, the number is the door the host must open, and switching wins. In the 3 blank cells the first pick is the car.

One case, two doors for the host

When the car is behind door 1, the host can open door 2 or door 3, and he picks one of them at random. That splits the case into two halves of $\frac{1}{3}\times \frac{1}{2}=\frac{1}{6}$1/3 × 1/2 = 1/6 each, and switching loses in both halves. The case does not turn into two whole cases.

Add up where switching wins: $\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$1/3 + 1/3 = 2/3. Staying wins in the two halves: $\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$1/6 + 1/6 = 1/3. The four paths add to $\frac{1}{6}+\frac{1}{6}+\frac{1}{3}+\frac{1}{3}=1$1/6 + 1/6 + 1/3 + 1/3 = 1.

You pick door 1. The first branches end on the door that hides the car, the second branches end on the door the host opens, and each path ends on its probability. On the colored path the car is behind door 2, the host must open door 3, and switching wins with probability $\frac{1}{3}$1/3.

The $\frac{2}{3}$2/3 moves to the other door

Staying wins only when your first pick was the car, so the probability of winning by staying is $\frac{1}{3}$1/3. Opening a door does not change that, because the host can always show a goat, whichever door you picked. Seeing the goat tells you nothing new about your own door.

The car is behind your door or behind the other closed door, so the two chances add to 1. The other door carries $1-\frac{1}{3}=\frac{2}{3}$1 − 1/3 = 2/3. Switching wins exactly when your first pick was wrong.

Each bar is cut into the three equally likely places for the car. Staying wins in 1 of them and switching wins in 2.

A host who does not know

The host's knowledge is what makes switching better. Suppose instead that he does not know where the car is, and opens door 2 or door 3 at random. You pick door 1, so there are 6 equally likely games: the car behind door 1, 2 or 3, and the host opening door 2 or door 3.

In 2 of those games he opens the door with the car. In the other 4 he shows a goat. Two of the 4 have the car behind door 1, where staying wins. In the other two the car is behind the door he did not open, where switching wins. So when this host happens to show a goat, staying and switching each win with probability $\frac{2}{4}=\frac{1}{2}$2/4 = 1/2.

You pick door 1. Down the side is the door that hides the car, and across the top is the door this host opens at random. In the 2 crossed cells he shows the car. Of the other 4, switching wins in the 2 colored cells and staying wins in the 2 blank ones.

A hundred doors

With 100 doors, your first pick is right with probability $\frac{1}{100}$1/100. The host, who knows where the car is, opens 98 of the other 99 doors and shows 98 goats. If your first pick was wrong, the car is behind one of the 99 other doors, and the host had to leave that door closed.

So switching wins with probability $\frac{99}{100}$99/100, and staying wins with probability $\frac{1}{100}$1/100.

The usual mistakes

Saying that two closed doors are left, so each has a chance of $\frac{1}{2}$1/2. The two doors are not alike: the host could never open yours, and he chose the door he opened knowing where the car is.

Swapping the two answers. Staying wins only when the first pick was right, $\frac{1}{3}$1/3. Switching wins when the first pick was wrong, $\frac{2}{3}$2/3.

Counting the case where your first pick is the car as two cases, because the host can open either of two doors. It is one case with probability $\frac{1}{3}$1/3, split into two halves of $\frac{1}{6}$1/6.

The game show

In the application below, part (a) lists every place the car can be and follows what the host must do in each case. Part (b) plays the same game with 10 doors.