Monty Hall Problem Calculator

Explore the famous probability puzzle and calculate winning chances

About the Monty Hall Problem Calculator

The Monty Hall Problem Calculator is a tool designed to demonstrate one of the most famous and controversial puzzles in probability theory. Named after the host of the television game show Let's Make a Deal, the problem presents a scenario where a contestant chooses one of three closed doors. Behind one door is a car (the prize), and behind the others are goats (losing options). After a choice is made, the host—who knows what is behind each door—opens one of the remaining doors to reveal a goat and offers the contestant a chance to switch their choice to the last remaining door.

Many people intuitively believe that once one door is removed, the chances of winning are 50/50 between the two remaining doors. However, this calculator uses Bayesian probability to show that switching actually doubles your chances of winning from 33.3% to 66.7%. This tool is frequently used by statistics students, teachers, and logic enthusiasts to visualize how adding doors or changing the number of revealed doors impacts the final odds. By adjusting the parameters, users can see how the logic scales beyond the standard three-door setup into more complex variations of the puzzle.

Formula

P(Win | Switch) = (N - 1) / [N * (N - K - 1)]

In the standard version, N is the total number of doors (3), and K is the number of doors the host opens (1). The formula calculates the probability of winning the prize if you decide to switch your choice after the host reveals one or more losing options. For the classic 3-door game, this simplifies to 2/3.

P(Win | Stay) is simply 1/N. This represents the probability that your very first selection was correct, which does not change regardless of what the host reveals, provided the host's actions are pre-determined by the rules of the game.

Worked examples

Example 1: A contestant plays the classic version of the game with 3 doors and the host opens 1 door.

N = 3, K = 1
P(Stay) = 1 / 3 = 0.3333
P(Switch) = (3 - 1) / [3 * (3 - 1 - 1)]
P(Switch) = 2 / [3 * 1] = 2/3 = 0.6667

Result: The probability of winning by switching is 66.67%, while staying is 33.33%.

Example 2: A version of the game with 100 doors where the host opens 98 doors to reveal goats.

N = 100, K = 98
P(Stay) = 1 / 100 = 0.01
P(Switch) = (100 - 1) / [100 * (100 - 98 - 1)]
P(Switch) = 99 / [100 * 1] = 99/100 = 0.99

Result: The probability of winning by switching is 99%, while staying is 1%.

Example 3: A game with 5 doors where the host reveals 2 goats and leaves 2 other doors closed besides your own.

N = 5, K = 2
P(Stay) = 1 / 5 = 0.20
P(Switch) = (5 - 1) / [5 * (5 - 2 - 1)]
P(Switch) = 4 / [5 * 2] = 4/10 = 0.40

Result: The probability of winning by switching is 40%, while staying is 20%.

Common use cases

Teaching introductory statistics students about conditional probability and Bayes' Theorem.
Verifying the results of a classroom experiment involving large-scale trials of the simulation.
Explaining the 'Law of Large Numbers' by comparing calculated theoretical probability to real-world outcomes.

Pitfalls and limitations

The calculator assumes the host must always open a door and must always reveal a goat.
Probabilities shift if the host only offers the choice to switch when they know the contestant has already picked the winning door.
Calculations are invalid if the host opens a door at random without knowing where the prize is located.

Frequently asked questions

should i always switch in the monty hall problem

Yes, switching doors always gives you a 2/3 (66.67%) chance of winning, while staying with your original choice only leaves you with a 1/3 (33.33%) chance. This is because your initial choice was likely wrong (2/3 of the time), and the host has eliminated an incorrect option for you.

does the monty hall problem work if the host doesn't know where the car is

The Monty Hall problem assumes the host already knows where the prize is and must deliberately reveal a goat. If the host opens a door at random and happens to find a goat by accident, the probabilities change to 50/50.

why is monty hall so counterintuitive

It is hard to believe because humans tend to view the final two doors as a new, independent 50/50 choice. However, the probability is locked at the moment of your first choice; switching capture the 66% chance that you were wrong at the start.

how does monty hall work with 100 doors

You have a 1/N chance of being right initially and an (N-1)/N chance of being wrong. When the host opens all but one other door, that (N-1)/N probability collapses onto the single remaining closed door, making it the statistically superior choice.

is the monty hall problem actually true

Standard probability theory (Bayes' Theorem) confirms that switching is the optimal strategy. Despite the fame of the Marilyn vos Savant column, mathematicians have mathematically proven the 2/3 advantage through simulations and formal proofs.