The puzzle
You face three closed doors. Behind one is a car; behind each of the other two is a goat. You pick a door, say door 1, but do not open it. The host, who knows where the car is, then opens one of the other two doors to reveal a goat, and offers you a choice: stay with your door, or switch to the remaining closed door.
Under the standard rules, switching wins the car with probability 2/3, and staying wins with probability 1/3. Many people expect 50-50, since two doors remain, and that is the source of the puzzle’s fame.
The name comes from Monty Hall, who hosted the television game show “Let’s Make a Deal”. The problem became widely discussed around 1990 after it appeared in a magazine column by Marilyn vos Savant, which prompted many readers to write in disagreeing with the answer. The disagreement is understandable, and the correct answer has since been confirmed by simple counting and by computer simulation.
The rules that matter
The answer depends on how the host behaves. The standard version assumes:
- The car is equally likely to be behind any of the three doors.
- The host always opens a door other than yours.
- The host always opens a door hiding a goat, which he can do because he knows where the car is.
- The host always makes the offer to switch.
If any of these is changed, the answer can change. That is worth remembering whenever the puzzle is retold from memory.
Counting every case
Suppose you always pick door 1. Because the car is equally likely to be anywhere, there are three equally likely situations.
| Car is behind | Host opens | If you stay | If you switch |
|---|---|---|---|
| Door 1 | Door 2 or 3 (either) | Win | Lose |
| Door 2 | Door 3 (forced) | Lose | Win |
| Door 3 | Door 2 (forced) | Lose | Win |
Staying wins in one of three situations. Switching wins in two of three. Your first pick had a 1 in 3 chance of being right, and nothing the host does changes that, because he always has a goat door available to open whatever you chose.
Picking door 1 is no restriction: by symmetry the same table applies to any first pick.
Where the intuition goes wrong
The “it is 50-50” argument treats the two remaining doors as equal. They are not, because the host’s choice was not random. When the car is behind one of the two doors you did not pick, the host is forced to open the other one, and so the closed door that remains is where the car is. When your first pick was right, which happens one time in three, the host has a free choice, and the remaining door hides a goat.
Think of it as two groups. Your door is group one, with a 1/3 chance. The other two doors together are group two, with a 2/3 chance. The host then opens a goat door from group two, which does not change the group’s combined chance, but it concentrates all of it on the single door left. Opening a door gave you information about group two without changing the chance that your original pick was right.
The hundred-door version
Suppose there are 100 doors, with one car. You pick one, with a 1 in 100 chance of being right. The host, who knows where the car is, opens 98 of the other doors, all goats, leaving your door and one other. Should you switch?
Your first pick is right only 1 time in 100. In the other 99 cases the car is behind one of the doors you did not pick, and the host has been careful to leave exactly that door closed. So switching wins 99 times in 100. Most people find the answer obvious here, and the three-door puzzle is the same logic with smaller numbers.
A probability tree view
You can also solve it with a probability tree. The first branch is where the car is, each with probability 1/3. The next branch is the host’s action. Staying wins along only one first-level branch, which has probability 1/3. Switching wins along the other two, which total 2/3. This is also a case of conditional probability, since the host’s reveal is information that the puzzle asks you to use.
When it is 50-50
Change the rules and the answer changes. Suppose the host does not know where the car is and opens a random other door. If, by luck, he reveals a goat, then the two remaining doors are equally likely to hide the car, and switching gives no advantage. The host’s knowledge, and the guarantee that he will not reveal the car, is what makes the standard answer 2/3.
A host who only offers the switch when you have picked correctly would make switching a bad idea. The puzzle quietly relies on the host following a fixed, known procedure.
Checking it yourself
A reliable way to build confidence is to simulate it. A Monte Carlo simulation of a few thousand rounds with cards or in a spreadsheet will show a staying win rate near one third and a switching win rate near two thirds. Because individual runs vary, the counts will not be exact, and a small number of rounds can easily look closer to half. A larger sample size smooths this out. You can also try it physically with three cards, one of them an ace, with a friend as the host.
What the puzzle teaches
The result is a reminder that probabilities depend on the process that produced the information, not merely on how many options remain. A fair-looking count of outcomes can mislead when the revealing step is not random. This is the same principle that applies in card games when a card is revealed by a person who has a choice about what to show.
The puzzle is a teaching example about conditional reasoning. It does not carry over to casino games, where the odds are fixed by the rules and games with a house edge lose on average. If gambling is causing problems, a local support service can help.