The Monty Hall Problem: Why Switching Wins 2/3 of the Time
In the Monty Hall problem, switching doors wins the car two-thirds of the time. See the counterintuitive probability explained step by step, with its assumptions.
The Monty Hall problem is a famous probability puzzle. You face three doors: behind one is a car, behind the other two are goats. You pick one door. The host, who knows what is behind every door, opens a different door to reveal a goat and asks whether you want to switch to the remaining closed door. You should switch. Switching wins the car with probability 2/3, while staying wins only 1/3. Most people expect 50-50 odds, which makes it a classic example of faulty reasoning about conditional probability.
Where does the name come from?
The puzzle is named after Monty Hall, the host of the American television game show Let's Make a Deal. The statistician Steve Selvin posed it in a letter to The American Statistician in 1975. It became famous in September 1990, when Marilyn vos Savant answered it in her Parade magazine column and said that switching was correct. Thousands of readers wrote in to say she was wrong, many of them with doctorates. She was right. According to a widely told account, even the prolific mathematician Paul Erdős remained unconvinced until he watched a computer simulation.
Why does switching win?
Your first pick has a 1 in 3 chance of hiding the car, which means there is a 2 in 3 chance the car is behind one of the other two doors. The host's action does not change your door's chance, because he is not opening a door at random: he always opens a goat door among the two you did not choose. So the whole 2/3 probability that the car was somewhere else is concentrated on the one other door that is still closed. Switching wins exactly when your first pick was wrong, and that happens two times out of three.
Suppose you pick door 1. The three equally likely cases are:
| Car is behind | Host opens | If you stay | If you switch |
|---|---|---|---|
| Door 1 (chance 1/3) | Door 2 or 3 | Win | Lose |
| Door 2 (chance 1/3) | Door 3 | Lose | Win |
| Door 3 (chance 1/3) | Door 2 | Lose | Win |
Staying wins in one case out of three; switching wins in two. The puzzle becomes obvious with 100 doors: you pick one, and the host opens 98 goat doors, leaving yours and one other. Your door had a 1 in 100 chance, so the other door holds the car 99 times out of 100.
What assumptions does the puzzle need?
The answer depends on how the host behaves. The standard version assumes the host knows where the car is, always opens a door you did not pick, always reveals a goat, and always offers the switch. Change these and the answer changes. If the host opened a random door and it merely happened to show a goat, switching and staying would be equal at 50-50. If the host offered a switch only when you had picked the car, switching would always lose. The puzzle is really about how much information the host's action carries.
The Bayes' theorem calculation
This is a textbook case for Bayes' theorem, the rule for updating probabilities when you get new evidence. You pick door 1 and the host opens door 3. The chance that he opens door 3 depends on where the car is:
- If the car is behind door 1, he picks door 2 or 3 at random, so the chance of door 3 is 1/2.
- If the car is behind door 2, he must open door 3, so the chance is 1.
- If the car is behind door 3, he cannot open it, so the chance is 0.
Weighting each case by its prior chance of 1/3, the probability that the car is behind door 1 is (1/3 × 1/2) divided by (1/3 × 1/2 + 1/3 × 1), which is 1/3. The probability that it is behind door 2 is therefore 2/3.
Why does intuition fail?
The mistaken thought is that two doors remain, so each must be 50 percent. But two outcomes are not necessarily equally likely, and the host's knowledge is what makes the leftover door special. Psychology adds to the confusion: people dislike giving up a choice they made, and fear the regret of switching away from a winner, a pattern related to loss aversion. In a 2010 study, pigeons learned the switching strategy through repeated trials, while human participants mostly kept their first choice. The host can also be viewed as a player with a strategy, which links the puzzle to game theory.
Try it yourself
A short simulation settles the matter:
import random
trials = 100000
wins_by_switching = 0
for _ in range(trials):
car = random.randrange(3)
pick = random.randrange(3)
opened = next(d for d in range(3) if d != pick and d != car)
switched = next(d for d in range(3) if d != pick and d != opened)
wins_by_switching += (switched == car)
print(wins_by_switching / trials) # about 0.667
Run it and the result settles near 0.667, the two-thirds that the argument predicted.
Tags
bayes conditional probability decision making probability puzzles
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