Monty Hall Problem Calculator

The Monty Hall problem is a famous probability puzzle based on a game show scenario. You choose one of three doors — behind one is a car, behind the others are goats. After your choice, the host (who knows what's behind each door) opens a different door to reveal a goat. You are then asked: should you switch to the remaining door or stick with your original choice? The counterintuitive answer is that switching doubles your chances of winning.

It seems like 50/50 because two doors remain, but this ignores the information gained from the host's action. When you first picked, you had a 1/3 chance of being right. The host always reveals a goat — which means the other unopened door now carries the full 2/3 probability that the car is behind it. The host's knowledge changes the odds, so it's never a true 50/50 split.

Switching wins 2/3 of the time (approximately 66.7%) while staying wins only 1/3 of the time (approximately 33.3%). This is because your initial pick has a 1/3 chance of being correct. When the host removes a losing door, the remaining door absorbs the entire 2/3 probability that your original choice was wrong. Switching is always the mathematically superior strategy.

In the classic 3-door version, switching gives you a 2/3 (≈66.7%) chance of winning the car. Staying with your original door gives only a 1/3 (≈33.3%) chance. The more doors there are (with the host revealing all but one), the greater the advantage of switching becomes.

Marilyn vos Savant is a columnist who famously explained the correct solution to the Monty Hall problem in her 1990 Parade magazine column. She correctly stated that contestants should switch doors to maximize their odds. Her answer sparked enormous controversy — thousands of readers, including many mathematicians, wrote in to disagree — but she was ultimately proven correct.

With more doors (and the host always revealing all goats except one other door), the advantage of switching becomes even more dramatic. For example, with 10 doors, staying wins only 10% of the time while switching wins 90% of the time. The general formula is: switching wins (n−1)/n × 1/(n−2) × (n−2) = (n−1)/n of the time, which simplifies to a much higher probability than staying.

This simulator runs the Monty Hall game hundreds or thousands of times using random numbers. Each trial randomly places the car behind a door, randomly picks your initial door, then the host reveals a goat door. It tallies how often staying vs. switching would have won. Running more simulations (e.g. 1000) brings the results very close to the theoretical probabilities of 66.7% and 33.3%.

It originated from the real TV game show 'Let's Make a Deal', hosted by Monty Hall. While simplified slightly for the mathematical puzzle, the core mechanics reflect an actual game show format. The problem became a landmark example in probability and decision theory, and it is now routinely used to teach Bayesian reasoning and conditional probability in statistics courses worldwide.