0.333333
0.666667
33.33
%
66.67
%
2
0.333333
0.666667
33.33
%
66.67
%
2
The Monty Hall Problem Calculator computes your probability of winning by staying versus switching in the famous Monty Hall scenario. Named after the host of the TV game show "Let's Make a Deal," this probability puzzle has confused mathematicians, professors, and even Paul Erdős since it was popularized by Marilyn vos Savant in 1990. The correct answer — that switching doors doubles your chances of winning — remains one of the most counter-intuitive results in probability theory.
This calculator extends the classic 3-door problem to any number of doors and revealed doors, letting you explore how the advantage of switching scales with the game's parameters.
In the standard Monty Hall problem with d doors and r doors revealed by the host:
$$P(\text{win by staying}) = \frac{1}{d}$$
$$P(\text{win by switching}) = \frac{d - 1}{d \cdot (d - 1 - r)}$$
For the classic 3-door, 1-revealed scenario:
$$P(\text{stay}) = \frac{1}{3} \approx 33.3\%$$
$$P(\text{switch}) = \frac{2}{3 \cdot 1} = \frac{2}{3} \approx 66.7\%$$
The key insight is that your initial choice has probability 1/d of being correct. When the host reveals r losing doors (the host always knows what's behind the doors and never reveals the prize), the probability (d-1)/d that the prize is behind one of the other doors doesn't disappear — it concentrates onto the remaining (d-1-r) unchosen doors.
With the classic 3 doors: initially, there's a 2/3 chance the prize is behind one of the two doors you didn't pick. When the host opens one of those and shows a goat, the entire 2/3 probability transfers to the one remaining door. Switching captures this concentrated probability.
The generalization shows that as more doors are revealed, the switching advantage increases dramatically. With 100 doors and 98 revealed, P(switch) = 99/100 = 99%, making the advantage nearly absolute.
The advantage ratio shows how many times better switching is compared to staying. In the classic problem, switching is exactly 2× better. With more doors and more reveals, this ratio can become extremely large. The key lesson: new information (revealed doors) should update your strategy, and the optimal Bayesian response is to switch.
This problem is a powerful illustration of conditional probability and why human intuition about probability often fails. Our brains tend to see two remaining doors and assume 50/50, ignoring the information gained from the host's deliberate reveal.
Inputs
Results
You pick door 1. The host opens door 3 (a goat). Should you switch to door 2? Yes! Staying wins 1/3 of the time; switching wins 2/3. The host's action doesn't change your initial 1/3 odds — it concentrates the remaining 2/3 onto the single unchosen door.
Inputs
Results
With 10 doors, your initial pick has a 10% chance. The host reveals 8 losing doors, leaving just your door and one other. Switching wins 90% of the time — a 9× advantage. This extreme case makes the logic clear: the host is funneling 90% of the probability into one door.
Because the host's reveal is not random. The host always opens a losing door, which is a deliberate action carrying information. If the host opened a door at random (potentially revealing the prize), then yes, the remaining doors would be 50/50. But since the host's choice is constrained (must show a goat, must not open your door), the reveal creates an asymmetric update. Your initial door's probability stays at 1/3; the unrevealed other door absorbs the full 2/3.
Yes, extensively. Computer simulations with millions of trials consistently confirm that switching wins approximately 2/3 of the time in the 3-door version. These simulations were crucial in convincing skeptics after Marilyn vos Savant published the correct answer in 1990 and received thousands of letters — including from PhD mathematicians — insisting she was wrong. The simulation approach is a powerful tool for verifying counter-intuitive probability results.
The standard problem assumes: (1) the host always opens a losing door after your choice; (2) the host knows what's behind each door; (3) the host gives you the option to switch; (4) the prize placement is random (uniform distribution). If any assumption changes, the answer changes. For example, if the host opens doors randomly and happens to show a goat, switching and staying both give 50%. The host's knowledge and deliberate action are what create the switching advantage.
With d doors and (d-2) reveals, the switching advantage ratio is (d-1). For 3 doors: 2× advantage. For 10 doors: 9×. For 100 doors: 99×. As d increases, staying becomes increasingly foolish. The 100-door version is often used to build intuition: you pick 1 door out of 100, the host opens 98 losing doors, leaving yours and one other. Would you switch? Almost everyone says yes — revealing the same logic that applies to 3 doors.
Yes, the Monty Hall problem is a classic application of Bayesian reasoning. Your prior probability for each door is 1/d. The host's action (revealing specific losing doors) provides evidence that updates these probabilities via Bayes' theorem. The posterior probability for your chosen door remains 1/d, while the posterior for the remaining unchosen door becomes (d-1)/(d-1-r) normalized by d. The Monty Hall problem elegantly demonstrates how new evidence should change beliefs.
With multiple prizes, the analysis changes. If k out of d doors have prizes, the probability of winning by staying is k/d. The switching probability depends on how many prizes are among the revealed doors (usually zero, since the host avoids revealing prizes). The general formula becomes more complex, but the switching advantage typically persists whenever the host reveals information. For the standard single-prize case, the mathematics is cleanest and the switching advantage is most dramatic.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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