Monty Hall Problem Calculator

Adjust the number of doors and how many doors the host reveals. See why switching beats staying, even when the game scales beyond three doors.

Minimum of 3 doors to make the puzzle meaningful.

Host reveals goat doors but always leaves your initial choice and one mystery door closed.

Stay with Original Door

33.33%

Probability of winning if you refuse to switch

Switch to Remaining Door

66.67%

Probability of winning if you switch

Random Choice Among Remaining

50.00%

If you ignore strategy and choose randomly

After the host opens 1 goat door(s), 2 doors remain closed. Switching concentrates the initial 33.33% chance of picking a goat into 66.67% success when you move to the other door.

How to Use This Calculator

  1. Set the total number of doors in the game show.
  2. Choose how many goat doors the host reveals after your first pick.
  3. Compare the probabilities of staying versus switching.
  4. Use the result to explain why switching remains optimal as the puzzle scales.

Formula

P(stay) = 1 / N

P(switch) = (N − 1) / N

N = number of doors, assuming the host always reveals goats and leaves one alternative closed.

Your initial choice has a 1/N chance of being correct. Switching captures the remaining probability mass after the host reveals goats, making it the superior strategy.

Full Description

The Monty Hall problem famously defies intuition. Most contestants prefer to stay, believing the host's reveal makes the odds 50/50. In reality, the host's knowledge concentrates the winning probability in the unopened door you did not pick. As the number of doors grows, switching becomes overwhelmingly advantageous.

This calculator generalizes the puzzle. By allowing the host to reveal multiple goat doors, it mirrors versions of the game played with larger sets of prizes and demonstrates how conditional probability governs optimal play.

Frequently Asked Questions

Does switching still help with more doors?

Yes. The more doors there are, the stronger the benefit of switching. With 10 doors, switching wins 90% of the time.

What if the host sometimes reveals the prize by accident?

The classic puzzle assumes the host knows where the prize is and never reveals it. If the host acts randomly, the probabilities change.

Why isn't the random choice probability 50%?

After the host reveals goats, more than two doors may remain. Choosing randomly among them is worse than switching to the single door the host left closed.

Can I simulate the game?

Run repeated experiments with friends or a simple script. The empirical win rates will converge to the probabilities shown here.