Monty Hall Probability Calculator

Enter the number of doors to compare staying versus switching.

How it is calculated

The model has N doors, one prize, and a host who knows where it is. You pick a door, then the host opens exactly one losing door that you did not pick, and you may switch to any of the other N − 2 closed doors at random. Staying wins only if your first pick was right: P(stay) = 1 ÷ N. Switching wins when your first pick was wrong, which has probability (N − 1) ÷ N, and you then land on the prize with chance 1 ÷ (N − 2): P(switch) = (N − 1) ÷ (N × (N − 2)). The advantage is P(switch) ÷ P(stay) = (N − 1) ÷ (N − 2), shown with two decimals. N must be a whole number of 3 or more.

Worked example

Take 5 doors. If you stay, you win only when your first pick was the prize: 1 ÷ 5 = 20.00%. If you switch, your first pick must have been wrong, probability 4 ÷ 5, and then you choose one of the 3 remaining closed doors, so the chance is 4 ÷ (5 × 3) = 4 ÷ 15 = 26.67%. The advantage is 26.67 ÷ 20.00 = 4 ÷ 3 = 1.33×. The tool shows switch 26.67%, stay 20.00%, advantage 1.33×.

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Common mistakes

  • Assuming the host opens doors at random. The result depends on the host knowing where the prize is and always opening a losing door; if the opened door could be the prize, the odds change.
  • Expecting the host to open all but one door. This tool models the host opening a single door, so with 10 doors you still pick among 8 closed ones, and switching is only 1.13× better.
  • Believing it is 50/50 after the reveal. The two remaining doors are not equally likely; your first pick keeps its original 1 in N chance.
  • Typing 2 or a decimal such as 3.5. The game needs at least 3 whole doors, so the tool returns nothing for those values.
  • Reading 100% as the sum of switch and stay. They do not add up to 100% for N above 3 because the other doors still hold probability that you could not choose.

About this calculator

The Monty Hall problem, named after the host of the game show Let's Make a Deal, is a famous probability puzzle: a contestant picks one of several doors, the host — who knows what's behind each door — opens a losing door, and the contestant is offered the chance to switch. Counterintuitively, switching is almost always the better strategy: with three doors, staying wins with probability 1/3 while switching wins with probability 2/3. This calculator generalizes the puzzle to any number of doors, showing exactly how the switching advantage shrinks (but never disappears) as more doors are added.

The problem is a mainstay of probability theory and decision-theory education because it exposes how strongly conditional information — the host's deliberate, non-random choice of which door to open — changes the odds compared to naive intuition. It routinely appears in statistics courses, game theory discussions, and even in classic debates over conditional probability that once split professional mathematicians.

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