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Monty Hall Problem Calculator

Enter the number of doors and prize doors to compare switching vs staying probabilities in the generalised Monty Hall problem.
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Luis GonzalezCreated by Luis GonzalezLast updated:

How to Use This Calculator

  1. 1

    Enter the Number of Doors

    Input the total count of doors available in the game. The classic problem uses 3 doors.

  2. 2

    Specify Doors With Prize

    Enter how many of the doors conceal a prize. This must be less than the total number of doors.

  3. 3

    Review Probabilities

    The calculator will display the win probabilities for both switching and staying, along with the switching advantage.

Example Calculation

A contestant faces the classic Monty Hall problem with 3 doors, one of which hides a prize, and wants to calculate the optimal strategy.

Number of Doors

3

Doors With Prize

1

Results

66.67%

Tips

Always Switch for Higher Odds

In the classic Monty Hall problem (3 doors, 1 prize, host knows where the prize is and always opens an empty door), switching your choice after a door is revealed doubles your odds of winning from 33.3% to 66.7%.

Understand the Information Gain

The host's action of opening a goat door provides crucial information. It concentrates the initial 2/3 probability of the prize being behind one of the *other* doors onto the *single remaining unopened door* you didn't initially pick.

Visualize with More Doors

If you have 100 doors and pick one, the probability of being correct is 1/100. If the host then opens 98 empty doors from the remaining 99, the probability of the prize being behind the *one* remaining unopened door is 99/100, making the switch overwhelmingly advantageous.

The Monty Hall Problem Calculator explores one of the most famous and counter-intuitive puzzles in probability theory. This tool allows users to simulate the game show scenario, calculating the win probabilities for both switching and staying with their initial choice, even with varying numbers of doors and prizes. For the classic setup of 3 doors and 1 prize, it reveals that switching your choice after a goat door is opened doubles your chance of winning from 33.33% to 66.67%, a result that challenges common intuition.

Decoding the Probabilities in the Monty Hall Problem

The core of the Monty Hall Problem lies in conditional probability and the information conveyed by the host's action. When you initially pick a door, you have a 1 out of N chance of being correct, where N is the number of doors. This means there's an (N-1)/N chance the prize is behind one of the other doors. When the host, who knows where the prize is, opens K empty doors from the unchosen set, the remaining (N-1-K) unchosen doors now collectively hold the (N-1)/N probability. If the host always opens N-2 empty doors (leaving only your door and one other), the (N-1)/N probability concentrates on that single remaining unchosen door.

probability if stay = doors with prize / number of doors
probability if switch = (number of doors - doors with prize) / number of doors

This simplified formula applies to the classic scenario where there is only one prize and the host opens all but one of the unchosen goat doors.

💡 For exploring other mathematical concepts involving multiple possibilities, our Infinite Solutions Identifier Calculator can help you understand systems with endless outcomes.

Example: The Classic 3-Door Scenario

Let's walk through the classic Monty Hall Problem setup: 3 doors, 1 prize. You pick Door #1.

  1. Initial Probability: Your chosen Door #1 has a 1/3 (33.33%) chance of having the prize. The other two doors (Door #2 and Door #3) collectively have a 2/3 (66.67%) chance.
  2. Host's Action: The host then opens one of the unchosen doors (e.g., Door #3), always revealing a goat. This is crucial: the host knows where the prize is and must open an empty door.
  3. New Information: The 2/3 probability that the prize was behind Door #2 or Door #3 now collapses entirely onto Door #2, because Door #3 is known to be empty.
  4. Switching Decision: If you switch from Door #1 (1/3 chance) to Door #2 (now 2/3 chance), you double your odds of winning.

In this example, the Probability If Switch is 66.67%, and Probability If Stay is 33.33%.

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Understanding Conditional Probability and Game Theory

The Monty Hall problem is a pedagogical cornerstone in the study of conditional probability and game theory, frequently used in introductory statistics courses to illustrate how new information can fundamentally alter probabilities. It showcases Bayesian reasoning, where initial beliefs (prior probabilities) are updated based on new evidence (the host opening a door). In game theory, it's a simple example of a non-zero-sum game with an intelligent opponent (the host) whose actions are not random. The problem's counter-intuitive solution often challenges human cognitive biases, such as the tendency to stick with an initial choice, making it a compelling case study for understanding decision-making under uncertainty and the power of mathematical logic over gut feeling.

Misapplications of the Monty Hall Logic

The logic of the Monty Hall problem is powerful, but its application is highly specific, and misapplications can lead to incorrect conclusions. The core reasoning relies on several strict conditions: the host must know where the prize is, must always open a goat door from the unchosen set, and must always offer the switch. If the host opens a door randomly, or if they sometimes open a prize door (which would end the game), or if they only offer a switch when you initially picked correctly, the probabilities change dramatically. For example, if the host doesn't know where the prize is and happens to open an empty door, then switching offers no advantage; the odds remain 1/2 for both remaining doors. Therefore, applying the "always switch" strategy to scenarios that deviate from these precise rules will yield misleading or incorrect results, undermining the statistical advantage.

Frequently Asked Questions

What is the Monty Hall Problem?

The Monty Hall Problem is a famous probability puzzle based on a game show scenario where a contestant chooses one of three doors, with a prize behind one and goats behind the others. After the contestant picks a door, the host (who knows where the prize is) opens one of the *other* doors, always revealing a goat. The contestant is then offered the chance to switch their original choice to the remaining unopened door. The counter-intuitive solution is that switching doors doubles the probability of winning the prize.

Why does switching doors increase the probability of winning?

Switching doors increases the probability of winning because the host's action of deliberately opening an empty door provides new information. Initially, your chosen door has a 1/3 chance of being correct. This means there's a 2/3 chance the prize is behind one of the *other two* doors. When the host reveals one of those two other doors as empty, the entire 2/3 probability from the unchosen set consolidates onto the *single remaining unchosen door*, making it the better choice. Your initial 1/3 probability never changes.

What are the conditions for the Monty Hall Problem to work?

The Monty Hall Problem's solution relies on three specific conditions: 1) The host must always know where the prize is. 2) The host must always open a door that the contestant did not pick. 3) The host must always open a door that does *not* contain the prize (i.e., a goat door). If any of these conditions are violated, the probabilities change, and switching may not be the optimal strategy.

Is the Monty Hall Problem relevant to real-life decision-making?

While a game show scenario, the Monty Hall Problem highlights the importance of understanding conditional probability and how new information can alter initial assessments, making it relevant to real-life decision-making. It teaches that sometimes, counter-intuitive choices are statistically superior when all variables are understood. It's a classic example used in statistics and cognitive psychology to illustrate human biases in probability assessment, such as the 'endowment effect' or 'status quo bias' where people prefer not to change their initial choice.