Monty Hall Variations: Game-Show Probability for Quant Interviews

The Monty Hall problem is the most famous probability puzzle ever posed in a popular column, and for that reason it shows up at quant-trading interviews with surprising regularity. The base puzzle is well-known by now, so interviewers at Jane Street, SIG, Optiver, Akuna, Citadel Securities, IMC, and others rarely stop at the original. Instead they ask variations — sometimes subtle, sometimes radical — that probe whether you understand why the original answer is what it is, not just that the answer is “switch.”

This guide walks through the original problem, its standard solutions, and the variations that interviewers actually ask. Mastering the variations is what separates “I memorized the answer” from “I understand conditional probability and information.”

The Original Problem

You’re on a game show. There are three doors. Behind one door is a car; behind the other two are goats. You pick a door (say door 1). The host, Monty Hall, who knows what’s behind every door, opens one of the other doors (say door 3) to reveal a goat. He then asks: “Do you want to switch to door 2?”

Should you switch? Yes. P(car behind door 2 | observed evidence) = 2/3. P(car behind door 1 | observed evidence) = 1/3.

Why Switching Wins

The intuition that throws candidates: “There are two doors left, so it’s 50/50, right?” No. The doors are not symmetric. Your initial choice was made before any information was revealed; the door Monty opened was chosen conditional on not revealing the car.

Three frames help build intuition:

Frame 1: Enumerate the cases

Suppose you initially pick door 1.

1/3 of the time, the car is behind door 1. Monty opens door 2 or door 3 (random). Switching loses.
1/3 of the time, the car is behind door 2. Monty must open door 3 (only goat available). Switching wins.
1/3 of the time, the car is behind door 3. Monty must open door 2. Switching wins.

Switching wins in 2 out of 3 cases.

Frame 2: 100 doors

Imagine 100 doors. You pick door 1. Monty opens 98 doors, all goats, leaving door 1 and one other door. Should you switch? Yes, obviously — Monty’s choice was constrained by avoiding the car, and the remaining door was selected from 99 doors with prior 99/100 of containing the car. Same logic at scale.

Frame 3: Bayes’ theorem

P(car behind 1 | Monty opens 3) = P(Monty opens 3 | car behind 1) × P(car behind 1) / P(Monty opens 3)

= (1/2 × 1/3) / (1/2) = 1/3.

Where P(Monty opens 3) = 1/2 from symmetry: across all car positions and Monty’s responses, door 3 gets opened half the time.

Variation 1: Monty Forgets

“Same setup, but Monty doesn’t know where the car is. He randomly picks one of the two doors you didn’t choose. He happens to open a goat. Should you switch?”

Now it’s 50/50. The asymmetry that gave switching its 2/3 advantage came from Monty’s knowledge — he intentionally avoided the car. If Monty is random, conditioning on “Monty revealed a goat” updates both remaining doors equally.

Bayes: P(car behind 1 | Monty random + revealed goat) = P(revealed goat | car behind 1) × P(car behind 1) / P(revealed goat)

= 1 × 1/3 / (2/3) = 1/2.

The denominator P(revealed goat) = 2/3 because Monty randomly opens a door and one-third of the time would reveal the car.

This variation tests whether you understood that Monty’s knowledge matters, not just the surface action.

Variation 2: Monty Always Opens Door 3

“Same setup, but Monty’s policy is: always open door 3 if possible (i.e., if the car isn’t behind door 3); otherwise open door 2. You pick door 1, and he opens door 3 to reveal a goat. Should you switch?”

Now switching gives the same probability as staying: 1/2. Wait — that doesn’t sound right at first. Let’s check.

Cases:

Car behind 1: Monty opens 3 (per policy). Observed.
Car behind 2: Monty opens 3 (per policy). Observed.
Car behind 3: Monty opens 2 (forced). Not observed.

Conditional on observing “Monty opens 3 with goat,” we eliminate case 3. Cases 1 and 2 are both consistent and originally equally probable. So P(car behind 1 | Monty opens 3) = P(car behind 2 | Monty opens 3) = 1/2.

Monty’s policy can change the answer. The original problem assumed Monty randomly chose between the two non-car doors when both were goats; if his choice rule is asymmetric, your inference changes.

Variation 3: Two Players

“Two contestants, three doors. Player A picks door 1. Player B picks door 2. Monty opens door 3 to reveal a goat. Both players are offered the chance to switch (to the other player’s door). What should each do?”

Each player initially has 1/3 prior on their own door and 1/3 on each other door. When Monty (knowing) opens door 3, he’s avoiding the car. Conditioning on “Monty avoided opening the car door,” the posterior weight on door 3 goes to 0, distributed between door 1 and door 2 in proportion to how much “Monty would open door 3 given car position.”

If car behind 1: Monty must open 2 or 3, but neither contestant chose 3, so… wait, Monty must choose from non-contestant-chosen doors that aren’t the car. Only door 3 is available; he opens it.
If car behind 2: Same logic; he opens door 3.
If car behind 3: He can’t open door 3; he opens nothing or the rules break down.

Conditional on “Monty opens door 3 (a goat),” cases 1 and 2 are equiprobable, case 3 ruled out. P(car behind 1) = P(car behind 2) = 1/2. Switching is irrelevant; both players have the same probability either way.

The two-player variation kills the original asymmetry because both remaining doors were “chosen by a player” rather than “remaining after Monty’s filter.”

Variation 4: N Doors, K Reveals

“100 doors, 1 car, 99 goats. You pick a door. Monty opens 50 doors, all goats. Should you switch to one of the remaining 49 unopened doors?”

Yes. P(car behind your door | observed) = 1/100. P(car behind any specific other unopened door | observed) = (99/100) / 49 ≈ 0.0202. Switching gives roughly 2× the probability of staying. The general rule: if you started with N doors and Monty reveals K goats from the remaining N-1, your door has prior 1/N, and each other unopened door has (N-1)/N divided by (N-1-K) others.

Variation 5: Monty Reveals After You Commit

“You pick door 1. Monty asks: ‘Are you locked in?’ You say yes. Then he opens door 3 to reveal a goat. He says, ‘Too bad — you committed.’ Did you make the right choice?”

The probability conditional on what’s now known is identical: P(car behind 1) = 1/3 still. But you can’t switch, so your realized probability is 1/3. The variation is testing whether you understand that information flow precedes commitment, and committing pre-information is suboptimal in this game.

What Interviewers Look For

Recognizing the role of Monty’s knowledge. The “Monty forgets” variation distinguishes candidates who memorized the answer from those who understand why.
Computing probabilities by conditioning on observation, not action. Bayes’ theorem mechanics, applied cleanly.
Verbalizing the asymmetry. Strong candidates explicitly say: “Door 1 was chosen by me without information. Door 3 was opened by a knowing party who avoided the car. The two doors are not symmetric.”
Handling variations gracefully. When the interviewer changes the rules, recompute from scratch rather than pattern-matching to the original.

Frequently Asked Questions

Is the Monty Hall problem really still asked? It’s so famous.

The original isn’t asked much anymore because it’s too well-known. But variations are asked frequently, often because the interviewer wants to see whether you can adapt. “Same as Monty Hall, but Monty doesn’t know where the car is” is a common follow-up that catches candidates who memorized the answer. If an interviewer asks you the original Monty Hall, expect a follow-up immediately afterward; don’t relax.

What’s the cleanest one-sentence explanation of why switching wins?

“Your door has 1/3 probability because you chose it from three options without information; the remaining unopened door has 2/3 because the opened door was specifically chosen to not be the car.” If you can deliver that clearly under interview pressure, you’ve demonstrated understanding. Avoid getting tangled in formulas; the verbal intuition is what interviewers want to hear.

How do I handle a variation I’ve never seen?

Set up Bayes’ theorem from scratch. List the prior probabilities, compute the likelihood of the observation under each hypothesis, divide by the marginal. Don’t try to pattern-match to the original. Every Monty Hall variation is a clean Bayesian problem; the only question is what’s prior, what’s likelihood, and what’s being asked.

What’s the trickiest variation?

“Monty Hall with three doors, but Monty has a 50% chance of being honest (opens a goat) and 50% chance of being random (opens any other door, possibly the car). You pick door 1, he opens door 3 to reveal a goat. Should you switch?” The answer involves weighting over Monty’s two policies. Setup: P(Monty honest) = 0.5, P(Monty random) = 0.5. Compute P(car behind each door | Monty opens 3 with goat) by mixing the two cases. The math is doable but multi-step; this is the kind of variation a senior interviewer might ask.

Why do quant firms care about this specifically?

The Monty Hall problem maps directly to trading: information is revealed by other market participants whose actions are not random — their choice of when to trade, what to quote, which side to take, all reflect their private information. A market maker who treats other participants’ actions as uninformative is leaving money on the table; one who correctly conditions on counterparty actions makes better trades. Monty Hall is a stylized version of this skill.