How this instrument works
In the classic Monty Hall problem, there are 3 doors, a prize behind exactly one of them, and you pick one door without opening it. The host — who knows what's behind every door — then opens one of the two remaining doors, always revealing a door without the prize, and offers you the choice to stay with your original pick or switch to the last unopened door. The result that famously surprises people: switching wins 2/3 of the time, while staying wins only 1/3.
The reason is that the host's action isn't random — the host always avoids opening the door with the prize, which injects information into the situation that a naive 'two doors left, so it must be 50-50' reading misses entirely. Your original pick had a 1/3 chance of being right when you made it, and nothing that happens afterward changes that original 1/3 — but the remaining 2/3 chance that the prize is behind one of the other two doors doesn't vanish when the host opens one of them; it concentrates entirely onto the single door the host left closed.
This instrument also generalizes the problem to more than 3 doors, keeping the host's rule the same — pick one door, the host opens exactly one other door known to be empty, regardless of how many doors there are in total. With more doors, staying still wins with probability 1/(number of doors), but the switching advantage per remaining door shrinks, because the host has only removed one losing door out of many rather than eliminating almost all the alternatives, as happens in the tight 3-door case.
- Enter the total number of doors into Number of doors — at least 3, matching the classic setup or a generalized version with more doors.
- Read Probability of winning if you stay for the odds if you keep your original pick after the host's reveal.
- Read Probability of winning if you switch for the odds if you instead switch to one specific remaining unopened door.
- In every case switching beats staying, but by how much depends on the door count — the advantage is largest in the classic 3-door version and narrows as the number of doors grows.
Worked example — the classic 3-door version
Enter 3 into Number of doors. Probability of winning if you stay reads 0.333333 (1/3) — your original pick's odds, fixed the moment you chose it. Probability of winning if you switch reads 0.666666 (2/3) — exactly double.
Here's why: your first pick has a 1/3 chance of hiding the prize and a 2/3 chance of being wrong. If you're wrong (2/3 of the time), the host is forced to open the only other empty door, which means the single door you didn't pick and the host didn't open must be the prize door — so switching converts that whole 2/3 chance into a win. If you're right to begin with (1/3 of the time), switching loses. Averaged out, switching wins 2/3 of the time and staying wins only 1/3.
Questions
Why does switching double your odds instead of it being 50-50?
Because the host's reveal isn't random information — the host always knows where the prize is and always opens a door without it. Your original pick locks in a 1/3 chance of being right; the remaining 2/3 chance that the prize is behind one of the other two doors doesn't split evenly once the host opens one of them, it collapses entirely onto whichever single door is left unopened. That's why switching wins 2/3 of the time, not half.
Does the advantage still exist with more than 3 doors?
Yes, switching still beats staying with any number of doors — but the advantage per door shrinks as the door count grows, since the host in this generalization only opens one door out of many, not almost all the alternatives. With 10 doors, staying wins 10% of the time and switching to one specific remaining door wins 11.25% of the time — still better, but nowhere near the dramatic 1/3-versus-2/3 gap of the classic 3-door case.
What if the host opens doors randomly instead of always avoiding the prize?
Then the whole analysis changes, because the host's action would no longer be reliably informative. If the host might accidentally reveal the prize, and simply didn't this time, that changes what the reveal tells you about the unopened doors — this calculator assumes the standard rule that the host always knows the prize's location and always opens an empty door on purpose, which is the setup that produces the classic switching advantage.
Why did so many people, including mathematicians, initially get this wrong?
The problem is a well-documented case of strong, misleading intuition: seeing two closed doors left tempts people to assume 50-50 odds, ignoring that one of those doors was deliberately, non-randomly kept closed by someone who knew where the prize was. When columnist Marilyn vos Savant published the correct 2/3 answer in Parade magazine in 1990, she received thousands of letters, including from PhDs, insisting she was wrong — until computer simulations settled the question decisively in her favor.
Where does this problem come from?
It's named after Monty Hall, the longtime host of the television game show 'Let's Make a Deal,' where a similar choice-and-reveal scenario played out. The mathematical version was formalized by Steve Selvin in a 1975 letter to the journal The American Statistician, and it became widely known after vos Savant's 1990 column sparked national debate over the correct answer.
Is there a way to build intuition for why 2/3 is right?
One helpful trick is to imagine far more doors — say, 100 — with the host opening 98 empty ones, leaving just your original pick and one other door. It becomes obvious your first guess was very unlikely to be right among 100 doors, so the single door the host deliberately left closed is very likely the winner. The 3-door version follows the exact same logic, just with smaller, less dramatically obvious numbers.