Monty Hall Problem Calculator

Explore the Monty Hall problem with analytical probabilities and Monte Carlo simulation for 3+ doors, variable reveals, and strategy comparison visuals.

About the Monty Hall Problem Calculator

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.

Why Use This Monty Hall Problem Calculator?

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.

How to Use This Calculator

  1. Set the number of doors (minimum 3) and doors revealed by the host.
  2. Choose your strategy: always switch or always stay.
  3. Click "Run Simulation" to run a Monte Carlo simulation with specified rounds.
  4. Compare analytical probabilities with simulation results.
  5. Study the multi-door table to see how the advantage scales.
  6. Try presets for 3, 4, 5, 10, and 100 doors.

Formula

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.

Example Calculation

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.

Tips & Best Practices

The Psychology of Probability

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.

Generalizations and Variants

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.

Game Theory and Decision Making

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).

Sources & Methodology

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Frequently Asked Questions

Why is switching better?

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.

Does it matter which door you initially pick?

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.

What if the host opens a door randomly?

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.

How does this relate to Bayesian reasoning?

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.

Why do people get it wrong so often?

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.

What if there are multiple prizes?

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.

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