QUANT INTERVIEW PROBLEM·conditional-probability·difficulty 3

The Monty Hall Problem

Always switch. The counter-intuitive 2/3 answer and why the host's knowledge matters.

also known as: three doors problem · goat problem

PROBLEM

You're shown three doors. Behind one is a car; behind the other two, goats. You pick door 1. The host — who knows where the car is and will always open a goat door — opens door 3 to reveal a goat. You're offered the chance to switch to door 2 or stay with door 1. Should you switch, and what's the probability of winning if you do?

ANSWER

Yes, switch. P(win | switch) = 2/3.

CANONICAL SOLUTION

Before any door opens, your chosen door has probability 1/3 of hiding the car; the other two doors collectively have probability 2/3. When the host opens a goat door from the other two, the full 2/3 probability doesn't vanish — it concentrates on the remaining unopened door. So switching gives you P = 2/3 vs. staying at 1/3.

Formally, by Bayes' theorem: let C = event 'car is behind door 2', H = event 'host opened door 3'. Then

$P(C \mid H) = \frac{P(H \mid C) \cdot P(C)}{P(H)} = \frac{1 \cdot \tfrac{1}{3}}{\tfrac{1}{2}} = \tfrac{2}{3}$

The key is that $P(H \mid C) = 1$ — if the car is behind door 2, the host is forced to open door 3 (since door 1 is yours and door 2 has the car).

AI ALTERNATIVE FRAMING

Scale up to 100 doors. You pick door 1. The host opens 98 goat doors, leaving only door 1 (your pick) and, say, door 57. Should you switch to door 57? Almost everyone intuitively says yes — the host has basically pointed at the car. That same intuition applies to the 3-door version, just less dramatically. The crucial ingredient in both cases is the host's deliberate, informed choice: he never opens the car door, so his action carries genuine information.

On QuantPrep, every practice problem gets an AI-generated alternative explanation like this — tailored to the canonical solution, streaming live.

COMMON VARIATIONS

Host with no knowledge

The host opens a door at random — sometimes revealing the car (in which case the game ends and you lose). Conditional on the host revealing a goat, should you still switch?

No — P becomes 1/2 for both doors, because the host's action carries no information (he got 'lucky' with equal probability).

Monty Fall variant

The host slips and accidentally opens a door, revealing a goat by chance. Should you switch?

Same as the no-knowledge case — 1/2 each. The switching advantage comes entirely from the host's knowledge, not from any door-opening mechanism.

Your friend picks

Instead of you, a friend who doesn't know the rules picked door 1 at random. The host opens door 3 showing a goat. Should YOU now switch to door 2?

Yes, same as standard — the original pick's lack of information doesn't matter; what matters is the host's informed reveal.

FIRMS THAT ASK THIS

Jane Street SIG Optiver IMC Citadel

TECHNIQUE

Conditional Probability

"Given X happened, what's the probability of Y?" — the second-most-common framing in quant interviews.

FREE QUANT BRAINTEASERS WEEKLY

Not ready to sign up? Get 5 hand-picked quant interview problems every week. No spam, unsubscribe any time.

TRY A PROBLEM NOW · NO SIGNUP

#042 · optimal_stoppingdifficulty 4

In the classical secretary problem, as n → ∞, what fraction should you reject outright before accepting the next best-so-far candidate?

FAQ

Why does intuition scream 1/2?

Because after the host opens a door, you see two doors — and the naive heuristic is 'two options, so 50/50'. But that ignores the fact that the host's choice was constrained (he can't open the car door or your door). The information from his forced choice breaks the apparent symmetry.

Does this come up in real quant interviews?

Not as the literal 3-doors problem — it's too famous. But its structural cousins appear constantly: any problem where a second actor has information and makes a constrained choice, and candidates must update correctly on the action taken.

What's the cleanest way to explain this in an interview?

The 100-door version. Say: 'You pick 1 of 100. The host opens 98 goat doors. Switching feels obviously right because the host has effectively pointed at the car. The 3-door version is the same mechanism, just with less dramatic numbers.' This reliably convinces skeptical interviewers.

How is this connected to Bayes' theorem?

The formal solution is a direct application of Bayes: $P(C \mid H) = P(H \mid C) P(C) / P(H)$ . The key term is $P(H \mid C) = 1$ , which encodes the host's constraint. Without that (if the host chose randomly), the update vanishes.

Practice variants of this problem

400+ problems on QuantPrep, including variants of the monty hall problem and every technique it draws on.

Start practicing free