Monty Hall Problem Calculator

Your details

The total number of doors in the game. The classic problem uses 3. Raising this makes the switching advantage even larger.
How many goat doors the host opens before offering the switch. Must be at least 1 and at most (total doors - 2) so at least one unopened door remains for you to switch to.
Choose the strategy you want to evaluate. You can compare both strategies using the outputs below.
Hypothetical number of rounds played with this strategy. Used to show expected wins and a convergence chart, not a Monte Carlo simulation.
Win probability (your strategy)Switching is better
0.67%

Exact probability of winning a car with your chosen strategy

Win probability - switch0.67%
Win probability - stay0.33%
Switch advantage (ratio)2
Expected wins in simulation rounds666.7
Expected losses in simulation rounds333.3
Switch0.67%
Stay0.33%
Switch strategy667.33%
Stay strategy333.67%

Switch advantage (ratio): 2

  • Win probability
  • Expected wins

Switching gives you a 66.67% chance - the optimal strategy.

  • Switching wins 66.67% of the time; staying wins only 33.33%. That is a 2.00x advantage for switching.
  • With 1000 rounds, switching earns about 667 wins vs 333 wins if you always stay.
  • In the classic 3-door game, your initial pick has a 1/3 chance. The host must reveal a goat, so the other closed door absorbs all of the remaining 2/3 probability.
  • The key insight: the host never reveals the car. That constraint makes the host's action informative, shifting probability toward the door you did not pick.

Next stepYou are already using the mathematically correct strategy. In a real game show, always switch when given the opportunity.

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