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The Monty Hall problem

This puzzle caused more arguments among mathematicians than almost any other. When Marilyn vos Savant published the answer in 1990, she received over 10,000 letters (many from PhD mathematicians) telling her she was wrong.

She wasn't.

What do you think?
You're on a game show with 3 doors. One hides a car; two hide goats. You pick a door. The host (who knows where the car is) opens another door, revealing a goat. Should you switch to the remaining door?

The intuition trap

Most people think: "Two doors left. 50-50 chance. Doesn't matter."

This is wrong. You should always switch. Switching wins 2/3 of the time.

Why switching works

Tracing through all possibilities, say you initially pick Door 1.

Case-by-Case Analysis
You pick Door 1
Case 1
Case 1: Car at Door 11/3
Case 2: Car at Door 21/3
Case 3: Car at Door 31/3
1/4

Case 1: Car is behind Door 1 (probability 1/3)

  • You picked correctly
  • Monty reveals a goat behind Door 2 or 3
  • If you switch, you lose

Case 2: Car is behind Door 2 (probability 1/3)

  • You picked a goat
  • Monty must reveal the goat behind Door 3
  • If you switch to Door 2, you win

Case 3: Car is behind Door 3 (probability 1/3)

  • You picked a goat
  • Monty must reveal the goat behind Door 2
  • If you switch to Door 3, you win

Switching wins in 2 out of 3 cases. Your initial pick has a 1/3 chance of being right. If you picked wrong (2/3 chance), switching guarantees you the car!

The conditional probability view

To be rigorous, let:

  • CiC_i = car is behind door ii
  • HjH_j = host opens door jj

You pick Door 1. The host opens Door 3, revealing a goat.

We want: P(C1∣H3)P(C_1 | H_3) vs P(C2∣H3)P(C_2 | H_3)

Computing P(Car at Door 1 | Host opens Door 3)
P(C1)=P(C2)=P(C3)=13P(C_1) = P(C_2) = P(C_3) = \frac{1}{3}
Each door equally likely to have the car initially.
Step 1 of 7
Why is P(H₃|C₂) = 1 but P(H₃|C₁) = 1/2? (one word)

Play the game

Experience it yourself. Play manually, or simulate thousands of games:

The Monty Hall Problem
Pick a door. One has a car, two have goats.
Simulate:

Why our intuition fails

Monty's action carries information.

When Monty opens a door, he's not choosing randomly. He:

  1. Knows where the car is
  2. Never opens the door you picked
  3. Never opens the door with the car

This constraint means his choice is influenced by where the car is. When you picked a goat (2/3 probability), Monty's reveal effectively points to the car.

The problem changes completely if Monty opens doors randomly and happens to reveal a goat. In that case, switching doesn't help. It really is 50-50.

Informed vs Random Host
Always-switch strategy. 1,000 games per click.

The 100-door version

Still not convinced? Try this version:

There are 100 doors. One has a car; 99 have goats. You pick Door 1. Monty then opens 98 doors, all revealing goats. Only your Door 1 and Door 57 remain.

What do you think?
Should you switch to Door 57?
100-Door Monty Hall
Pick any door (1-100)
N-Door Monty Hall
3100200
Stay (1/3)33.3%
Switch (2/3)66.7%
Doors
3
P(stay)
33.3%
P(switch)
66.7%
Switching is 2× better than staying

With 100 doors, the intuition becomes clearer. Monty's opening of 98 specific doors concentrated all that probability onto Door 57.

Lessons from Monty Hall

1.Conditioning changes everything: Probability depends on what you've learned, not just what's possible. This is conditional probability at work.

2.Information has structure: Monty knows where the car is and must avoid it, which creates an asymmetry you can exploit.

3.Simulation beats intuition: When probabilities feel paradoxical, simulate. The numbers don't lie.

4.Even experts get fooled: Paul Erdős, one of history's greatest mathematicians, reportedly didn't believe the answer until shown a simulation.

Summary

StrategyWin ProbabilityWhy
Stay1/3You win only if you picked correctly initially
Switch2/3You win whenever your initial pick was wrong

General principle: When someone who knows the answer gives you constrained information, that information can dramatically shift probabilities, even when it seems like it shouldn't.

Test your understanding

You initially pick a door. What's the probability you picked the car? (decimal, e.g. 0.42)
After Monty reveals a goat, what's P(car behind your door)? (decimal, e.g. 0.42)
After Monty reveals a goat, what's P(car behind the other closed door)? (decimal, e.g. 0.42)
With 1000 doors (1 car, 999 goats), you pick one, Monty opens 998 goat doors. P(switch wins)? (decimal, e.g. 0.42)
True or False: If Monty opens a door randomly (might reveal the car), then it IS 50-50. (true or false)

What's next

The Monty Hall problem shows how conditional probability can be counterintuitive. You've now practiced conditioning, Bayes' Rule, and independence.

Next, we'll explore random variables: extracting numbers from random experiments.