Numbers · 3 min read
The Monty Hall Problem, Explained So It Finally Clicks
Three doors, one prize, and a host who opens an empty door. Why switching wins two times in three, explained four different ways until one of them lands.
By the Cognosc team ·
The Monty Hall problem is famous for two reasons: the answer is simple, and almost nobody believes it the first time.
Try it
Play the Monty Hall game
Pick a door. The host, who knows where the car is, opens an empty one. Then stick or switch, and keep score.
Pick a door. Sticking has won no games yet; switching has won no games yet.
The problem
You’re on a game show. There are three doors. Behind one is a car; behind the other two are goats.
- You pick a door, say door 1.
- The host, who knows where the car is, opens one of the other doors, say door 3, and shows you a goat.
- He offers you a choice: stick with door 1, or switch to door 2.
Should you switch?
The answer
Yes. Switching wins the car 2 times in 3. Sticking wins only 1 time in 3.
Most people feel it should be 50/50: two doors left, one car. When the columnist Marilyn vos Savant published the correct answer in 1990, she received thousands of letters telling her she was wrong, many from people with PhDs.
Here are four ways to see why it isn’t 50/50. Stop reading when one of them clicks.
Way 1: your first pick doesn’t improve
When you first pick, you have a 1 in 3 chance of choosing the car. Nothing the host does changes that: he will always open an empty door, whatever you picked. So your door is still a 1 in 3 shot.
The remaining 2 in 3 has to go somewhere. It was spread over the other two doors; the host has shown you which of those two is empty. So all of it now sits on the one door he left closed.
Way 2: list every case
Say the car is behind door 1, 2 or 3 with equal chance, and you always pick door 1.
- Car behind door 1: the host opens door 2 or 3. Switching loses.
- Car behind door 2: the host must open door 3. Switching wins.
- Car behind door 3: the host must open door 2. Switching wins.
Switching wins in two of the three equally likely cases.
Way 3: switching means “my first pick was wrong”
If you plan to switch, you win exactly when your first pick was a goat. Your first pick is a goat 2 times in 3. So switching wins 2 times in 3.
Way 4: make it 100 doors
Imagine 100 doors, one car. You pick door 1. The host, who knows where the car is, opens 98 of the other doors, all goats, leaving just door 57 closed.
Do you really think your first pick, a 1 in 100 guess, is as good as the one door he carefully avoided opening? Switching now wins 99 times in 100.
Why intuition fails
The mistake is assuming that two options must be equally likely. That’s true only when nothing about how you got to those two options favours either one.
Here, the host’s choice isn’t random: he knows where the car is and will never reveal it. His action carries information about the other doors, but none about yours. That’s the heart of conditional probability: how the way you learned something changes what it tells you.
What if the host didn’t know?
If the host opened a door at random and it happened to show a goat, the odds really would be 50/50. The difference is entirely in what the host knows and how he chooses. Same doors, different information, different answer.
Try it yourself
Take three playing cards, one of them an ace, and have a friend play host twenty times. Keep score of sticking versus switching. Most people are convinced within a few rounds.
The Monty Hall problem is one of ten classic traps in the free probability test.