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 explores the classic door-switching puzzle using both exact probability and Monte Carlo simulation. In the standard setup, you pick one door, the host reveals a losing door, and you decide whether to stay or switch.
The point of the puzzle is that the host's informed reveal changes the odds. What looks like a 50/50 choice after one door opens is actually asymmetric, and switching is better in the classic three-door case.
This page lets you vary the number of doors and reveals to see how the advantage changes when the puzzle is generalized beyond the television version.
Monty Hall is a compact way to show how information changes probability. The host's reveal is not neutral, so the remaining doors are not equally informative even though they look symmetrical at first glance.
Seeing the exact odds and the simulation together makes the switching advantage easier to trust than a verbal explanation alone.
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).
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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.