Explore the Monty Hall problem with analytical probabilities and Monte Carlo simulation for 3+ doors, variable reveals, and strategy comparison visuals.
The Monty Hall problem calculator lets you explore one of probability's most famous puzzles — both analytically and through Monte Carlo simulation. In the classic setup, a contestant picks one of three doors, the host reveals a goat behind another door, and the contestant can switch or stay.
Counterintuitively, switching doubles your chances of winning from 1/3 to 2/3. This calculator extends the problem to any number of doors and any number of reveals, computing exact probabilities and running simulations to verify.
Adjust the door count and reveal count to see how the switching advantage changes. The generalized formula shows that with n doors and r reveals, P(win | switch) = (n−1) / (n(n−1−r)), which always exceeds P(win | stay) = 1/n. Check the example with realistic values before reporting. Use the steps shown to verify rounding and units. Cross-check this output using a known reference case. Use the example pattern when troubleshooting unexpected results.
The Monty Hall problem is a gateway to understanding conditional probability, Bayesian updating, and the psychology of probabilistic reasoning. This calculator makes the unintuitive answer concrete through both mathematics and simulation.
Whether you're a student encountering the problem for the first time, a teacher building intuition, or someone who wants to explore generalizations, the interactive simulation settles any doubt.
P(Win | Stay) = 1/n. P(Win | Switch) = (n−1)/(n(n−1−r)), where n = doors, r = doors revealed. For classic 3-door: P(Stay) = 1/3, P(Switch) = 2/3.
Result: P(Win | Switch) = 66.67%, P(Win | Stay) = 33.33%
With 3 doors, your initial choice has a 1/3 chance of being correct. The host reveals a goat door, so the remaining door has a 2/3 chance. Switching doubles your odds.
The Monty Hall problem famously stumped even mathematicians when Marilyn vos Savant published the correct answer in 1990. The error stems from ignoring conditional probability — our intuition treats the two remaining doors as equally likely, but the host's informed action breaks this symmetry.
The N-door generalization shows the effect scales dramatically. With 1,000,000 doors, switching gives 99.9999% win probability. The "Monty Fall" variant (host opens a random door and happens to reveal a goat) changes the answer to 50/50 — proof that the host's knowledge is the crucial ingredient.
The Monty Hall problem illustrates a broader principle: when someone with information takes an action, that action itself contains information. This principle appears in poker (reading opponents), negotiation (interpreting offers), and machine learning (feature selection based on outcomes).
Your initial pick has a 1/n chance of being right. The remaining n−1 doors collectively hold (n−1)/n probability. When the host eliminates losing doors, that (n−1)/n probability concentrates on the remaining unchosen door(s). Switching accesses this concentrated probability.
No. All initial choices are equally likely to be correct (1/n). The advantage comes from switching after information is revealed, not from the initial choice.
If the host might accidentally reveal the prize, the problem changes completely. When the host knowingly avoids the prize door, that knowledge creates the asymmetry. Random opening means no advantage to switching.
The Monty Hall problem is a perfect Bayesian updating example. Your prior (1/n for each door) gets updated by the evidence (host opening specific doors), yielding a posterior that favors switching.
Humans tend to assume the remaining two doors are equally likely (50/50 with 3 doors). This ignores the asymmetry: the host's choice depended on your initial pick, creating conditional probability that's not 50/50.
With multiple prizes and the host still only revealing non-prize doors, the analysis changes but switching still generally helps. The exact advantage depends on the specific rules.