Monty Hall Problem Calculator

The classic puzzle involves three doors: behind one is a brand-new car, and behind the other two are goats. When you pick an initial door, say Door 1, the host (Monty) who knows what is behind the doors opens another door, say Door 3, revealing a goat. You are then given the option to stick with your original choice or switch to the remaining unopened door (Door 2). Mathematically, your initial choice has a 1/3 probability of holding the car, while the remaining unopened door absorbs the remaining 2/3 probability. When you run simulations, the algorithm randomly places the car (car = floor(data/100)), has the host open a goat door, and tracks your win rate based on whether you choose to switch (final_door = if_else(choice_change, choice_door, change)).

Imagine running a simulation of 1,000 game rounds to test the strategy of always switching doors. Suppose you pick Door 1 in every round. In approximately 333 rounds, the car happens to be behind Door 1, meaning switching would lead to a loss. However, in the remaining 667 rounds, the car is behind either Door 2 or Door 3. Because Monty always reveals a goat behind the unchosen door that you didn't pick, switching forces your final choice onto the only remaining door, which successfully captures the car 667 times out of 1,000. Thus, your simulated win rate for switching settles near 66.7%, confirming the 2/3 theoretical probability.

When experimenting with probability calculators, always run a high number of simulations (such as 1,000 or more) to see the Law of Large Numbers in action; small sample sizes like 5 or 10 rounds can produce misleading anomalies. Remember that the host's action is not random—Monty actively avoids opening the door with the car and avoids opening the door you initially selected, which is the crucial mechanism that shifts the mathematical weight to the unopened door.