Let's make a deal

Three doors. Behind one is a car, behind the other two are goats. You pick a door. The host, Monty Hall, who knows what's behind each door, opens one of the other two doors to reveal a goat. Then he asks: do you want to switch to the remaining unopened door?

If you haven't seen this puzzle before, take a moment to think. Most people's first instinct is wrong.

Play the game

The best way to build intuition is to play. Pick a door, watch Monty reveal a goat, then decide whether to switch or stay:

Play at least 10 rounds. Do you notice a pattern in how often switching wins?

Running thousands of games

One round at a time is revealing, but noise makes it hard to see the true probabilities. Run a batch of thousands of games:

After enough trials, staying wins about 33% and switching wins about 67%. The evidence is overwhelming.

The probability tree

Let's map out every possibility. You pick Door 1. Where could the car be, and what happens in each case?

Expand all three branches. In two out of three scenarios, switching wins. That's the whole argument.

Why 50/50 is wrong

The fallacy: "After Monty opens a door, there are two doors left, so each has a 50% chance."

This reasoning ignores how the two remaining doors were selected. Monty didn't pick randomly — he deliberately avoided the car. His action is informative.

If you've already explored the Monty Hall lesson, you've seen the interactive simulator. Here we focus on the deeper competition-level reasoning: the formal proof, and the generalization.

The key argument

The logic is disarmingly simple once you see it the right way:

Formal proof using Bayes' theorem

Label the doors 1, 2, 3. Without loss of generality, suppose you pick door 1 and Monty opens door 3 (revealing a goat). We want $P(\text{car at door 2} \mid \text{Monty opens door 3})$.

By Bayes' rule:

$P(\text{car at 2} \mid \text{Monty opens 3}) = \frac{P(\text{Monty opens 3} \mid \text{car at 2}) \cdot P(\text{car at 2})}{P(\text{Monty opens 3})}$

Computing each piece:

• $P(\text{car at 2}) = 1/3$
• $P(\text{Monty opens 3} \mid \text{car at 2}) = 1$ (Monty must open door 3; door 1 is yours, door 2 has the car)
• $P(\text{Monty opens 3}) = P(\text{opens 3} \mid \text{car at 1}) \cdot 1/3 + P(\text{opens 3} \mid \text{car at 2}) \cdot 1/3 + P(\text{opens 3} \mid \text{car at 3}) \cdot 1/3 = 1/2 \cdot 1/3 + 1 \cdot 1/3 + 0 \cdot 1/3 = 1/2$

$P(\text{car at 2} \mid \text{Monty opens 3}) = \frac{1 \times 1/3}{1/2} = \frac{2}{3}$

Generalizing to N doors

With $N$ doors (1 car, $N-1$ goats), you pick one door. Monty opens $N-2$ goat doors, leaving yours and one other. Should you switch?

Why this problem is famous

The Monty Hall problem became a cultural phenomenon when Marilyn vos Savant published the correct answer in Parade magazine in 1990. Nearly 10,000 readers wrote in to disagree, including many with PhDs in mathematics. Even Paul Erdős, one of the most prolific mathematicians in history, initially got it wrong.

The core issue is that conditional probability is genuinely counterintuitive. Our brains struggle with the fact that how information is revealed matters as much as what is revealed.

Competition appearances

• Classic probability puzzle: Appears in virtually every probability textbook
• Quant interviews: Tests understanding of conditional probability and Bayes' rule
• Programming contests: "Simulate 10,000 Monty Hall games" is a standard introductory problem
• Variations: Monty Hall with N doors, Monty Hall where the host doesn't know (the "Monty Fall" problem), multiple rounds

Takeaway

The Monty Hall problem is the poster child for why conditional probability matters. Switching wins 2/3 of the time because your initial pick was wrong 2/3 of the time, and Monty's reveal converts "wrong initial pick" directly into "switch and win." The generalization to $N$ doors makes the intuition crystal clear: with 100 doors, staying with your 1-in-100 guess when Monty eliminates 98 goats is obviously suboptimal.