Monty Hall Problem Simulator
Print PageWhy Switching is always Better
The Monty Hall problem is based on conditional probabilities:
- When you first pick a door, you have a **1/3** chance of selecting the Car, and a **2/3** chance of choosing a Goat.
- If you choose a Goat (which happens 2/3 of the time), Monty *must* reveal the other Goat. The remaining closed door is therefore guaranteed to contain the Car. Thus, if you switch, you win the Car **2/3 of the time**!
- If you stay, you only win the Car if your initial choice was correct (which is **1/3 of the time**).
A Monty Hall Problem Calculator (also known as a Monty Hall Paradox Simulator, 3-Door Switching Odds Utility, Let’s Make a Deal Probability Analyzer, or Bayesian Door Simulator) computes and visualizes the exact winning probabilities of Switching versus Staying when playing the famous Monty Hall game show scenario. Whether you choose between 3 doors (containing 1 sports car and 2 goats) or an extended N-door game, a Monty Hall calculator proves why switching doors doubles your win probability from 33.33% (1/3) to 66.67% (2/3).
The paradox deceives the human brain because people falsely assume that after the host opens a goat door, the remaining two closed doors have equal 50/50 odds. However, because host Monty Hall knows what is behind every door and is forced to open a goat door, switching transfers the initial 66.67% chance of picking a goat directly onto the unopened door.
Our free online Monty Hall Problem Calculator provides instant odds calculations across custom door counts and game rules:
- Probability of Winning by Staying (P(Win | Stay)):
P(Stay) = 1 ÷ N(For 3 doors:1 ÷ 3 = 33.33%). - Probability of Winning by Switching (P(Win | Switch)):
P(Switch) = (N - 1) ÷ N(For 3 doors:2 ÷ 3 = 66.67%). - General N-Door Formula (Host opens p goat doors):
P(Switch) = (N - 1) ÷ [ N · (N - p - 1) ]. - 100-Door Extreme Scenario:
P(Stay) = 1.00%vsP(Switch) = 99.00%.
Master Monty Hall Door Count & Strategy Win Odds Table
The table below displays the exact mathematical probabilities for Staying versus Switching across different total door counts (assuming the host opens all remaining goat doors except one):
| Total Doors in Game (N) | Goat Doors Opened by Host (p) | Strategy: STAY Win Probability | Strategy: SWITCH Win Probability | Advantage Ratio of Switching |
|---|---|---|---|---|
| 3 Doors (Classic Monty Hall) | 1 Goat Door Opened | 1 in 3 (33.33%) | 2 in 3 (66.67%) | EXACTLY DOUBLE (2.00x) |
| 4 Doors | 2 Goat Doors Opened | 1 in 4 (25.00%) | 3 in 4 (75.00%) | TRIPLE ODDS (3.00x) |
| 5 Doors | 3 Goat Doors Opened | 1 in 5 (20.00%) | 4 in 5 (80.00%) | QUADRUPLE ODDS (4.00x) |
| 10 Doors | 8 Goat Doors Opened | 1 in 10 (10.00%) | 9 in 10 (90.00%) | 9.00x Greater Chance |
| 100 Doors (Intuition Visualizer) | 98 Goat Doors Opened | 1 in 100 (1.00%) | 99 in 100 (99.00%) | 99.00x Greater Chance |
Step-by-Step Bayes’ Theorem Proof for the 3-Door Monty Hall Problem
To mathematically prove why switching yields 66.67% win probability using Bayes’ Theorem (assume contestant picks Door 1, and host opens Door 3 revealing a goat):
Step 1 (Define Prior Probabilities): P(Car behind Door 1) = P(C1) = 1/3, P(C2) = 1/3, P(C3) = 1/3
Step 2 (Host Action Conditional Probabilities P(Host opens Door 3 = H3)):
If Car is behind Door 1 (C1): Host can open Door 2 or Door 3 randomly &implies; P(H3 | C1) = 1/2
If Car is behind Door 2 (C2): Host is forced to open Door 3 &implies; P(H3 | C2) = 1.0
If Car is behind Door 3 (C3): Host cannot open Door 3 &implies; P(H3 | C3) = 0
Step 3 (Calculate Total Host Action Probability P(H3)):
P(H3) = P(H3|C1)P(C1) + P(H3|C2)P(C2) + P(H3|C3)P(C3) = (1/2 × 1/3) + (1 × 1/3) + (0 × 1/3) = 1/6 + 1/3 = 1/2
Step 4 (Apply Bayes' Theorem for Switching to Door 2 P(C2 | H3)):
P(C2 | H3) = [ P(H3 | C2) · P(C2) ] ÷ P(H3) = [ 1.0 × (1/3) ] ÷ (1/2) = (1/3) ÷ (1/2) = 2/3 = 66.67%
Thus, Bayes’ Theorem proves that the probability the car is behind Door 2 is exactly 2/3 (66.67%), while staying with Door 1 remains 1/3 (33.33%).
Why the Human Brain Fails: 50/50 Illusion vs. 100-Door Intuition
Below is a comparative reference chart explaining why human intuition struggles with conditional host knowledge:
| Perspective Concept | Common Fallacious Assumption | Mathematical Reality |
|---|---|---|
| The 50/50 Fallacy | “There are 2 closed doors left, so each door must have a 50% chance.” | Ignores that the host’s choice was non-random and informed by hidden car location. |
| The 100-Door Intuition Tool | Hard to visualize with 3 doors. | If you pick 1 door out of 100 (1% chance), and Monty opens 98 goat doors, the unopened door has a 99% chance! |
History & Controversy: 1975 Steve Selvin to 1990 Marilyn vos Savant
1975 Steve Selvin & The American Statistician
The problem was first posed by statistician Steve Selvin in a letter to The American Statistician in 1975, naming it after Monty Hall, the host of the TV game show Let’s Make a Deal.
1990 Marilyn vos Savant & The Parade Magazine Controversy
In 1990, Marilyn vos Savant (listed in the Guinness Book of World Records for highest IQ) answered the Monty Hall problem in her Parade magazine column “Ask Marilyn”, stating that contestants should always switch. She received over 10,000 letters—including nearly 1,000 from PhD mathematicians and university professors—falsely insisting she was wrong. Famous mathematician Paul Erdős refused to accept the 2/3 solution until presented with a Monte Carlo computer simulation.
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Frequently Asked Questions (FAQ)
Should you switch doors in the Monty Hall Problem?
Yes! You should always switch doors. Switching doubles your probability of winning the car from 33.33% (1 in 3) to 66.67% (2 in 3).
Why is the probability not 50/50?
Because the host does not open a door at random. Monty Hall knows where the car is and MUST open a goat door. When you first choose, there is a 66.67% chance you picked a goat. When Monty eliminates the other goat, switching wins whenever your initial pick was a goat.
What are the odds of winning if you switch in a 100-door Monty Hall game?
If there are 100 doors and the host opens 98 goat doors, switching gives you a 99.0% chance of winning the car.