The Monty Hall Problem: Why Switching Doors Wins Two-Thirds of the Time
October 29, 2026
In the Monty Hall problem, a game-show host shows you three doors. Behind one is a car; behind the other two are goats. You pick a door, say door 1. The host, who knows what is behind every door, opens another door that has a goat, say door 3, and asks if you want to switch to door 2. Should you? Many people think it makes no difference, with a 50-50 chance. In fact, switching wins the car two thirds of the time, while staying wins only one third of the time.
The problem became famous in 1990 when Marilyn vos Savant published the correct answer, and thousands of readers, including mathematicians, wrote in to say she was wrong. Computer simulations and careful reasoning proved her right.
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Suppose you always pick door 1. There are three equally likely cases for where the car is. Case 1: the car is behind door 1 (probability 1/3). The host opens door 2 or 3, and if you switch, you lose. Case 2: the car is behind door 2 (1/3). The host must open door 3, so switching to door 2 wins. Case 3: the car is behind door 3 (1/3). The host must open door 2, so switching to door 3 wins. Switching wins in two of the three cases: 2/3. Staying wins only in Case 1: 1/3.
The host's action is not random: he always avoids the car and your pick. That is what makes his action informative.
| Car location | Probability | Host opens | Stay wins? | Switch wins? |
|---|---|---|---|---|
| Door 1 (your pick) | 1/3 | Door 2 or 3 | Yes | No |
| Door 2 | 1/3 | Door 3 (forced) | No | Yes |
| Door 3 | 1/3 | Door 2 (forced) | No | Yes |
| Total win probability | 1/3 | 2/3 |
The 100-Door Intuition
Imagine 100 doors with one car. You pick one door (1% chance). The host, knowing where the car is, opens 98 of the other doors, all with goats, leaving your door and one other closed. Should you switch? Your first pick still has only a 1% chance of being right, as nothing the host does changes that. The other remaining door carries the combined 99% chance that the car was among the doors you did not pick. The host's action concentrates that probability on the one door he did not open. Switching wins 99% of the time. The three-door version is the same logic at smaller scale: your door keeps its 1/3, and the unopened door inherits the 2/3.
Assumptions That Matter
The 2/3 result depends on the host's rules: he knows where the car is, always opens a door with a goat that you did not choose, and always offers the switch. If the host opened a door at random and it happened to show a goat, the odds would be 50-50. If the host offers the switch only when you pick the car, switching would always lose. The problem is a lesson in conditional probability: the same observation (a goat behind door 3) has different meaning depending on how it came about. Bayes' theorem handles it formally: the probability the car is behind door 2 given the host opened door 3 is 2/3.
Testing It Yourself
Simulation is the most convincing evidence. With three cups and a coin under one, play 30 rounds with a friend as the host; the strategy of switching should win about 20 times and staying about 10. Or write a short program that plays 100,000 games: switching wins around 66,667 times. The paradox is a favorite for teaching probability because intuition fails and calculation succeeds. It also appears in real decisions, such as updating beliefs when new information arrives from a source that knows more than you do.
Frequently Asked Questions
Should I switch doors in the Monty Hall problem?
Yes. Switching wins 2/3 of the time, staying 1/3.
Why isn't it 50-50?
Because the host's choice is informed and not random, so it gives you information.
What if the host doesn't know where the car is?
If he opens a door at random and reveals a goat, switching and staying are equally likely to win.
How does the 100-door version help?
It makes clear that your first pick keeps its small chance while the remaining door gets the rest.
Who made this problem famous?
Marilyn vos Savant, in her 1990 column.