# The Monty Hall Problem: Some Explanations

If you don't know what the Monty Hall "Let's Make a Deal" problem is, start here.

You'll do better on average if you switch. There is a 2/3 chance that the prize is behind the door you switched to, and only 1/3 that prize is behind your original door.

Before I try to explain why, there are a few things will remain true throughout the following discussion:

• The host and assistants never lie or cheat.
• The prize is placed behind one door and not moved.
• You do not know where the prize is.
• Random numbers are truly random.
• Dice are fair.

## Two Common Misconceptions

There seem to be two reasons for thinking the chance of picking the door with the prize is 50%.

1. After one door has been revealed, there are two doors left, and the prize is behind one of them. Hence there must be a 50/50 chance of it being behind either unopened door.

2. There are four possible outcomes (to be shown later). Two are winning outcomes if you stay, two if you switch. So there are 2 out of 4 winning cases if you either stay or switch. That means there is a 50/50 chance of winning with either strategy.

I'll tackle these misconceptions in order.

### Does having two choices mean there's a 50/50 chance for each one?

Let's try a thought experiment. There are two doors. The host has hidden a prize behind one of them. He asks you to choose one. If the prize is behind it, you win. 50/50 chance of winning, right?

You choose a door. Before the host opens it, however, he tells you one more thing. Before you entered the studio, he rolled a six sided die. If the topmost die face was 1, the prize was put behind the left door. If it was 2, 3, 4, 5, or 6 the prize was put behind the right door. The host lets you choose again. Which door would you choose, left or right? Does it matter?

It should be clear that you have a 1 in 6 chance of winning if you pick the left door, and a 5 in 6 chance of winning if you pick the right door. This demonstrates that just because there are two choices, you cannot assume the odds of either one being correct are 50%.

You might object that in the original 3-door problem, the chances are equally likely that it's behind a given door, so this weighted game doesn't apply. That's good thinking, and it's correct. The three door problem isn't 50/50 for a different reason, but this "equal weight" misconception plays a role in the next section.

### Looking at the cases

Let's go back to the classic three door problem, and make an assumption that simplifies the problem greatly: the prize is behind door 1. The host knows this, but you don't. (The other cases, where it's behind door 2 or 3, are analyzed in just the same way. If you don't like door 1, just swap all three numbers consistently.)

There are now three cases, each of which is equally likely:

1. You choose door 1
2. You choose door 2
3. You choose door 3

Let's look at cases 2 and 3. If you choose door 2, the host must reveal door 3's booby prize (since he can't show door 1; it has the prize). Likewise, if you choose door 3, he must reveal door 2. Let's update our list to reflect this.

1. You choose door 1
2. You choose door 2: door 3 is revealed
3. You choose door 3: door 2 is revealed

What happens if you choose door 1? The host can do two things. Since both doors 2 and 3 have booby prizes, he can reveal either one. Let's call these cases 1a and 1b.

1. You choose door 1
1. door 2 is revealed
2. door 3 is revealed
2. You choose door 2: door 3 is revealed
3. You choose door 3: door 2 is revealed

We can now determine the winning strategies. They're actually the same as we would have chosen all along.

1. You choose door 1
1. door 2 is revealed : win = stay
2. door 3 is revealed : win = stay
2. You choose door 2: door 3 is revealed : win = switch
3. You choose door 3: door 2 is revealed : win = switch

"There", you say. "This is just what I was saying all along! There are two ways to win if I stay, and two ways to win if I switch. So it doesn't matter if I do either, since I'll win 50% of the time with either strategy."

And it does look like that, doesn't it? It's true there are four cases, and each strategy has two wins and two losses. What we haven't done yet is the missing piece: calculating the probability of each case occurring.

We agreed that the chance of first choosing any door is equally likely: 1 in 3. I'll write the probability of each case in square braces:

1. You choose door 1 [1/3]
2. You choose door 2 [1/3]
3. You choose door 3 [1/3]

Now let's add in cases 1a and 1b, where the host could choose to reveal either door 2 or 3. Assume he flips a coin to choose which. It doesn't really matter; they both hide booby prizes.

Also, we note which door was revealed.

1. You choose door 1 [1/3]
1. door 2 is revealed [1/2]
2. door 3 is revealed [1/2]
2. You choose door 2: door 3 is revealed [1/3]
3. You choose door 3: door 2 is revealed [1/3]

We're almost done. We just have to add the winning strategies, which haven't changed.

1. You choose door 1 [1/3]
1. door 2 is revealed [1/2] : win = stay
2. door 3 is revealed [1/2] : win = stay
2. You choose door 2: door 3 is revealed [1/3] : win = switch
3. You choose door 3: door 2 is revealed [1/3] : win = switch

Now we calculate the probabilities of each winning strategy. For cases 2 and 3, it's straightforward. Each case has a 1/3 chance of happening, so the chance of winning with a switch is 1/3 + 1/3 = 2/3. What about door 1?

(If you're clever, you might note that since all the probabilities must sum to 1, if the probability of switching and winning is 2/3 then the probability of staying and winning is only 1 - 2/3 = 1/3.)

Analyzing door 1 a little more complex. There's a 1/3 chance that you chose it, then there were two equally probable subcases (1a and 1b). The probability that you chose door 1 and the host revealed door 2 is 1/3 × 1/2 = 1/6. The calculation is the same for case 1b. The results are in the curly braces.

1. You choose door 1 [1/3]
1. door 2 is revealed [1/2] : win = stay {1/6}
2. door 3 is revealed [1/2] : win = stay {1/6}
2. You choose door 2: door 3 is revealed [1/3] : win = switch {1/3}
3. You choose door 3: door 2 is revealed [1/3] : win = switch {1/3}

So now we just add up the cases. What's the probability of a win if you stay? That's case 1a + case 1b = 1/6 + 1/6 = 1/3. If you switch, that's case 2 + case 3 = 1/3 + 1/3 = 2/3. So you win twice as much if you switch, rather than stay.

It's not necessarily intuitive, but it's true. The key is that each of the two winning cases where you stay are only half as probable as the winning cases where you switch. That's because the host can show you either of two doors if you choose the prize door, whereas he has no choice if you choose a non-prize door. That 50/50 choice affects the probability of winning.

Note also that the probability is the same as this:

1. You choose door 1 [1/3] : win = stay {1/6 + 1/6 = 1/3}
2. You choose door 2 [1/3] : win = switch {1/3}
3. You choose door 3 [1/3] : win = switch {1/3}

Here's another way to think of it: if you happened to choose the prize door (1/3 probability), you win if you stay. It doesn't matter what happens after that. That means the other 2/3 of the time you'll win if you switch.

If that doesn't convince you, take a look at some more approaches to this problem.

Last updated 29 June 2004