Monty Hall Problem Calculator

Play the game yourself, then see exactly why switching wins twice as often as staying. Try it with more doors, or with a host who doesn't know where the car is, to see what actually makes switching the better move.

Last updated 23 September 2026. How we calculate

Set up the game

One car, goats behind the rest. After you pick, the host opens all but one of the other doors.

Results update as you type. Press Calculate or Enter to save one to your recent list.

Before you see the answer

Three doors, one car. You pick a door, the host opens another to show a goat, and offers you a switch. What's your gut instinct?

Practice questions

Five fresh questions each time, with worked answers. Teachers can print them as a worksheet with an answer key, or save it as a PDF from the print window.

    The problem

    A car is behind one of three doors, and a goat is behind each of the other two. You pick a door. The host, who knows where the car is, opens one of the other two doors to reveal a goat, then asks: do you want to switch to the remaining door, or stay with your first pick?

    Most people's instinct is that it doesn't matter, since two doors are left and the car is behind one of them. The correct answer is that switching wins 2 times out of 3, twice as often as staying.

    Why switching wins

    Your first pick has a 1 in 3 chance of being the car, and nothing that happens afterwards changes that. The host always opens a door with a goat, whichever door you picked, so his action gives you no new information about your own door.

    That means the other 2 in 3 chance, the chance the car is somewhere among the two doors you didn't pick, doesn't go away. The host has simply removed one of those two doors from consideration, a goat door, and concentrated that entire 2 in 3 chance onto the one door he left closed. Switching to it wins whenever your first pick was wrong, which is 2 times out of 3.

    With more doors

    P(switch wins) = (n − 1) / n

    With n doors, your first pick is right 1 time in n. The host opens n − 2 goat doors, leaving one other door closed. Switching wins whenever your first pick was wrong: (n − 1)/n of the time. With 10 doors, that's 90%.

    The one detail that changes everything

    The result depends entirely on the host knowing where the car is and always revealing a goat on purpose. If he opens a door at random and simply happens to reveal a goat, the two remaining doors become equally likely, and switching only wins half the time. Try the "host guesses" option above to see this yourself: some rounds will now show the car by accident and can't be played at all.

    A history of getting it wrong

    The problem was posed by Steve Selvin in a 1975 letter to The American Statistician. It became famous in 1990, when Marilyn vos Savant answered a reader's letter about it in her Parade magazine column. She gave the correct answer, and received an estimated 10,000 letters disagreeing with her, including many from mathematicians and statisticians. Paul Erdős, one of the most prolific mathematicians of the 20th century, reportedly remained unconvinced until he was shown a computer simulation.

    Why our brains get this wrong

    Getting the Monty Hall problem wrong isn't a sign of poor maths ability. Studies find that university students, statistics teachers and working mathematicians fall for it at similar rates to everyone else, which is part of why it became so notorious in the first place.

    Psychologists call the main culprit the equiprobability bias: the automatic assumption that if there are two possible outcomes, they must be equally likely. A 2014 study by Lore Saenen and colleagues, comparing primary school pupils, secondary school pupils and university students, found this bias was the most common reasoning error across every age group, and that people who reasoned this way consistently answered "50/50" regardless of how much maths they knew.

    The bias survives because of how the problem is presented. You see two closed doors at the end and your mind treats that snapshot as the whole story, ignoring the process that produced it. The host's decision to open a specific door, forced by his knowledge of where the car is, carries real information. Two options that look symmetric on the surface can hide very different odds underneath, and our intuition for probability is built around what things look like right now, not around how they got that way.

    This is also why the "before you see the answer" prompt at the top of this page, and the door game itself, tend to work better than reading an explanation alone. Several studies have found that guided practice with real feedback shifts people's intuitions more effectively than a written proof does, even though the proof is what actually justifies the answer.

    Common mistakes

    Treating it as a fresh 50/50 choice. Once a door is opened, it's tempting to forget how it was chosen. The host's choice was not random, and that's exactly why the odds aren't even.

    Assuming your own door's odds improve. They don't. Your first pick is stuck at 1/n forever, because nothing about the door you chose ever changes. All the improvement happens to the other closed door.

    Ignoring the host's knowledge. The "host opens doors at random" version above gives switching no advantage at all. The advantage comes entirely from the host being forced to avoid the car.

    How we calculate

    The exact answer follows directly from conditional probability: given that the host revealed goats (which he always can, if he knows where the car is), the chance your original pick was right stays at 1/n. The simulation runs 10,000 random games in your browser using the same rule, and counts how often each strategy wins. When the host opens doors at random, games where he reveals the car by chance are excluded, since those games are never actually played out. Every number here was checked against an independent calculation in Python.

    References

    • Lore Saenen, Mieke Heyvaert, Ilke Grosemans and colleagues, "The Equiprobability Bias in the Monty Hall Dilemma: A Comparison of Primary School, Secondary School, and University Students", Proceedings of the Annual Meeting of the Cognitive Science Society, 36(36), 2014.
    • Steve Selvin, "A Problem in Probability", letter to the editor, The American Statistician, 29(1), 1975.
    • Marilyn vos Savant, "Ask Marilyn", Parade magazine, September 1990.
    • Charles Grinstead and J. Laurie Snell, Introduction to Probability, 2nd edition, American Mathematical Society, 1997.

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    Questions people ask

    Should you always switch in the Monty Hall problem?

    Yes, if the host knows where the car is and always reveals a goat. Switching wins 2 times out of 3, against 1 in 3 for staying.

    Why doesn't it become 50/50 once a door is opened?

    Because the host's choice wasn't random. He can always open a goat door, whichever one you picked, so his action carries information about where the car probably is, not just about the door he opened.

    Does the answer change with more doors?

    Switching wins (n − 1)/n of the time with n doors, and the host opening all but one of the rest. With 10 doors that's 90%, an even bigger advantage than the classic 3-door version.

    What if the host doesn't know where the car is?

    Then switching and staying both win half the time, among the games where he happens to reveal a goat. The host's knowledge is what creates the advantage, not the act of opening a door itself.

    Who first posed the Monty Hall problem?

    Steve Selvin, in a 1975 letter to The American Statistician. It became widely known in 1990 after Marilyn vos Savant answered it in her Parade magazine column and received thousands of letters disputing the correct answer.