Monty Hall Problem

Problem Statement

You are a contestant on a game show. There are three doors. Behind one door is a brand-new car; behind the other two doors are goats.

1. You pick a door — say, Door 1.
2. The host, who knows what is behind each door, opens another door — say, Door 3 — to reveal a goat.
3. The host then asks: Would you like to switch to Door 2, or stay with Door 1?

Should you switch?

Hint: Think about the probability that your initial pick was correct (1 out of 3) versus incorrect (2 out of 3). The host's action gives you information — but only about the doors you didn't pick.

Why It's Counterintuitive

Most people intuit that after the host reveals a goat, there are two remaining doors and the car must be behind one of them — so the odds are 50/50. This is wrong, and understanding why it's wrong is the core of the problem.

The key is that the host is not acting randomly. The host always opens a door with a goat. This asymmetry is what breaks the 50/50 intuition.

Solution

You should always switch. Switching wins with probability 2/3.

Solution

Step 1: Your initial pick (Door 1) is correct with probability 1/3, and wrong with probability 2/3.

Step 2: If your initial pick was wrong (probability 2/3), the car is behind one of the other two doors. The host must reveal the door with the goat — leaving the car behind the door you'd switch to. Switching wins.

Step 3: If your initial pick was correct (probability 1/3), both remaining doors have goats. The host reveals one. Switching loses.

Step 4: Conclusion: P(win | switch) = 2/3, P(win | stay) = 1/3.

Formal Proof Using Bayes' Theorem

Let C_i be the event "car is behind Door i", and H_3 be "host opens Door 3."

P(C_1 | H_3) = P(H_3 | C_1) × P(C_1) / P(H_3)

Computing each term:

• P(C_1) = 1/3 — prior probability your door is correct
• P(H_3 | C_1) = 1/2 — host can open Door 2 or Door 3 equally
• P(H_3 | C_2) = 1 — host must open Door 3 (Door 1 is your pick, Door 2 has the car)
• P(H_3 | C_3) = 0 — host cannot open Door 3 (car is there)

By the law of total probability: P(H_3) = (1/3) × (1/2) + (1/3) × 1 + (1/3) × 0 = 1/2

Therefore: P(C_1 | H_3) = (1/6) / (1/2) = 1/3

So P(C_2 | H_3) = 1 - 1/3 = 2/3. Switching wins.

Key Insight: The host's knowledge and constrained behavior (always revealing a goat) transfers probability from your initial pick to the other door. The host is giving you information — just not about your own door.

Variants

Variant 1: What if there are 100 doors, you pick one, and the host opens 98 doors with goats? Should you switch?

Solution:

1. Your initial pick is correct with probability 1/100.
2. The remaining closed door holds the car with probability 99/100.
3. You should absolutely switch. The extreme version makes the intuition clearer: it's almost certain you picked wrong initially.

Variant 2: What if the host opens a door at random (not necessarily showing a goat, and sometimes revealing the car)?

Solution:

1. If the host acts randomly and happens to reveal a goat, the problem is now symmetric.
2. Conditional on the host not revealing the car, P(win | switch) = 1/2.
3. The host's knowledge is what drives the 2/3 result. Without it, you're back to 50/50.

Related Problems

This problem is commonly asked at Jane Street, Citadel, SIG, Two Sigma, and Goldman Sachs interviews.