Monty Hall Problem Simulator
Run thousands of Monte Carlo simulations of the Monty Hall problem and watch the “always switch” strategy win about two-thirds of the time.
Each trial randomly places the car, makes a first pick, has the host reveal a goat, then records whether switching or staying would have won. Theory predicts switching wins 2/3 of the time.
About the Monty Hall problem
The Monty Hall problem, named after the host of the game show Let’s Make a Deal, is one of the most famous counterintuitive results in probability. You choose one of three doors; a car hides behind one and goats behind the other two. The host — who knows where the car is — opens one of the doors you did not pick to reveal a goat, then asks whether you want to switch to the other unopened door. Although it feels like a 50/50 choice, switching wins two-thirds of the time. Your original door keeps its initial 1-in-3 chance, so the remaining door inherits the full 2-in-3 probability once a goat is revealed. This simulator plays the game thousands of times at random so you can watch the win rates converge on 66.7% for switching and 33.3% for staying — a hands-on confirmation of the math.
How to use it
- Enter the number of simulations to run (more trials converge closer to the theoretical values).
- Click Run simulation.
- Compare the empirical win rates for always switching versus always staying.
- Run it again or with more trials to see the results stabilize near 2/3 and 1/3.
Frequently asked questions
- What is the Monty Hall problem?
- A probability puzzle based on a game show. You pick one of three doors; behind one is a car, behind the others goats. The host, who knows what is behind each door, opens a different door revealing a goat, then offers you the chance to switch to the remaining door. The counterintuitive answer: you should switch.
- Why should you switch doors?
- Your first pick has a 1/3 chance of being the car, so the other two doors together hold 2/3. When the host reveals a goat behind one of those two, that entire 2/3 probability concentrates on the single remaining door. Switching therefore wins 2/3 of the time, versus 1/3 for staying.
- Does this simulator prove it?
- Empirically, yes. Running thousands of random trials, the “always switch” win rate converges on about 66.7% and “always stay” on about 33.3%. More trials give a tighter convergence to the theoretical values — a demonstration of the law of large numbers.
- Why is the answer so counterintuitive?
- People tend to think that once one door is removed, the two remaining doors are equally likely (50/50). But the host’s choice is not random — he always reveals a goat — which transfers information and breaks the symmetry, giving the switch door the advantage.
- What if the host opened a door at random?
- Then the puzzle changes. If the host might reveal the car (and the game only continues when a goat happens to appear), switching and staying become equally good (50/50). The 2/3 advantage depends specifically on the host knowingly always showing a goat.
- Does switching guarantee a win?
- No. Switching wins about two times out of three, not every time. In any single game you might lose by switching; the advantage only shows up over many plays, which is exactly what this simulator demonstrates.