Monty Hall Problem Calculator

Understanding the Monty Hall Setup

The puzzle begins with three closed doors. Unknown to you, one door conceals a prize (traditionally a car), while the other two hide non-prizes (traditionally goats). You select one door, but it remains closed.

The critical moment arrives when the host—who knows what lies behind each door—deliberately opens one of the two remaining doors, always revealing a non-prize. This leaves two unopened doors: your original choice and one other. The host then asks if you wish to switch.

What makes this problem intellectually challenging is that it defies the naive assumption that two closed doors mean equal odds. Your initial 1/3 probability does not become 1/2 simply because one door opens. The host's informed action creates asymmetry: the unopened door you didn't choose inherits the probability from the door the host eliminated.

Why Switching Delivers Better Odds

Imagine labelling your first choice Door A and the two unchosen doors Doors B and C. On your initial pick, you had a 1/3 chance of selecting the car—and a 2/3 chance the car sits behind B or C combined.

When the host opens one of B or C (always revealing a non-prize), the 2/3 probability does not vanish. Instead, it concentrates entirely on whichever of B or C remains closed. Therefore:

• If you stay with Door A: your winning probability remains 1/3.
• If you switch to the remaining unopened door: your winning probability becomes 2/3.

The host's knowledge and deliberate action ensure that switching is statistically superior. Over many repetitions, switching wins roughly twice as often as staying.

Monty Hall Probability Framework

The underlying mechanics involve random assignment of the prize location, your door choice, and the host's mandatory reveal of a non-prize door. The simulation tracks these variables across multiple trials to establish empirical probabilities.

• Prize location — Door number (1, 2, or 3) randomly assigned to contain the car
• Your initial choice — The door you pick before any door is opened
• Host opens — A door the host reveals, always showing a non-prize and never your original choice
• Remaining unopened door — The single door left closed that you did not pick
• Trials — Number of game iterations in a simulation run

Common Pitfalls and Misconceptions

Understanding the Monty Hall problem requires letting go of several intuitive but incorrect assumptions about probability.

1. The Symmetry Fallacy — After the host opens a door, many people assume both remaining doors have equal 50/50 odds. This ignores that the host acted with knowledge and never opened the door you chose or the one hiding the prize. Your original choice's probability does not update; the unopened door you didn't pick inherits all remaining probability mass.
2. Conflating Sample Space with Outcomes — It's tempting to reason that three doors initially mean 1/3 odds, so two doors must mean 1/2 odds per door. But probability in the Monty Hall setup depends on your information and the host's deliberate action, not simply the count of closed doors. The host's reveal provides information that shifts probability without affecting your first choice.
3. Forgetting the Host's Constraint — The host never randomly opens doors; they always avoid your choice and always avoid the prize. This non-random behaviour is essential. If the host opened doors blindly and happened to reveal a non-prize, the problem would be different. The host's knowledge and constraint are what create the 2/3 advantage for switching.
4. Misinterpreting Simulation Results — If you run a small simulation (say, 10 trials) and switching loses more than staying, don't conclude the mathematics is wrong. Probability relies on large sample sizes. Run hundreds or thousands of trials; the switching win rate will converge to approximately 66.7%, while staying converges to 33.3%.

Testing the Problem Through Simulation

The easiest way to overcome scepticism about the Monty Hall problem is to test it empirically. This calculator allows you to simulate hundreds or thousands of games, employing either a 'always stay' strategy or an 'always switch' strategy.

Over a large sample—say, 1,000 games—the switching strategy should yield roughly 667 wins, and the staying strategy should yield roughly 333 wins. This empirical verification has convinced many sceptics, including mathematicians who initially doubted the theoretical analysis.

You can also play a single round interactively, making your own strategic choice. Whatever you decide, the mathematics remains consistent: switching wins 2/3 of the time in the long run.