Monty Hall, for quant interviews

The host rule is the whole puzzle

Three doors: one hides a car, two hide goats. You pick a door. The host — who knows what is behind every door — opens a different door, always revealing a goat, and offers to let you switch to the remaining door. Should you?

Before any arithmetic, state the three assumptions the answer depends on, because this is what a good interviewer is actually probing: the host always opens a losing door, the host never opens your door, and the host always offers the switch. Change any one of these and the number changes. The puzzle is not really about doors; it is about the fact that the host’s choice is constrained by information you don’t have, so the door he leaves closed is not a random door.

Just enumerate the three cases

The cleanest interview answer skips paradox-talk and counts. Fix your first pick as door 1 without loss of generality. The car is equally likely to be behind each door, giving three cases of probability $1/3$ each:

Staying wins only in the single case where your first guess was already right. Switching wins in both cases where it was wrong. Your first guess is wrong $2/3$ of the time, so switching wins $2/3$ of the time. That is the entire proof, and it fits on a whiteboard.

The same thing as a tree

If you prefer to see the strategy rather than the cases: switching is a machine that inverts your first pick. Pick the car and switch, you lose; pick a goat and switch, you are forced onto the car and win. So switching wins exactly when your first pick was a goat.

Deriving it with Bayes

When the interviewer says “do it formally,” they want the conditional probability. You picked door 1 and the host opened door 3. Compare the two survivors. The priors are $\Prob(C_i) = 1/3$. The only term that differs is the likelihood of the host opening door 3:

If the car is behind your door, the host picks freely between the two goats ($1/2$). If it is behind door 2, he is forced to open door 3 ($1$). He never opens the car ($0$). Bayes then gives the door you’d switch to:

Your original door keeps its prior $1/3$; the reveal poured the rest onto door 2. The forced $1$ versus the free $1/2$ is where the asymmetry lives.

Why 100 doors makes it obvious

The intuition pump interviewers love: play with $n$ doors, let the host throw open every goat door except one, then offer the switch. Your first pick is right with probability $1/n$, so switching wins $(n-1)/n$.

At $n = 100$ nobody argues 50/50 anymore: you almost certainly missed, and the host spent his knowledge sweeping away 98 goats to leave the car sitting there. The three-door case is the same effect, just quiet enough that people talk themselves out of it.

Variants that change the answer

This is where interviews are won or lost. The 2/3 result is not a fact about two closed doors; it is a fact about how the surviving door was chosen. Break the host rule and recompute:

• Ignorant host (“Monty Fall”). The host opens a random other door and it happens to show a goat. Now switching is a true $1/2$ — no edge — because the reveal was not steered away from the car.
• Malicious host (“Monty from Hell”). The host only offers a switch when you picked the car. Then switching loses with probability $1$, and the correct move is to stay.
• Host may reveal the car. If the game ends when a car is revealed, then seeing a goat is itself evidence, and the update is different again. Always ask what happens in the branch you were not shown.

Common mistakes

• “Two doors left, so it’s 50/50.” Counting doors instead of counting how the information was produced. The doors are not exchangeable.
• Answering 2/3 by reflex. Reciting the memorized number without ever stating the host rule reads as pattern-matching. The assumptions are the answer the interviewer wants.
• Forgetting the host’s free choice. When your first pick hides the car, the host chooses between two goats. That $1/2$ is exactly the term the Bayes denominator needs; drop it and you will “prove” the wrong number.