— GUIDE —

PROBABILITY INTERVIEW QUESTIONS

Nine questions with worked solutions, grouped by the idea each one is really testing. Try them before reading the answers.

UPDATED 2026-09-08 · 7 MIN READ · BY THE VARIANCE TEAM

Work these with a timer and a pen. The gap worth measuring is not whether you know the method, but how long it takes you to notice which method applies.

These are original representative questions, not real or leaked firm questions. If you want the underlying methods first, the probability shortcuts guide covers the five techniques these questions lean on.

Counting and complements

  1. You roll a fair die four times. What is the chance you see at least one six?

    Count the opposite. No six on a single roll has probability 5/6, so 1 − (5/6)⁴ = 1 − 625/1296 = 671/1296, a little over 51%. Anyone who starts listing cases here has already lost a minute.

  2. Two cards are drawn from a standard deck without replacement. What is the chance both are hearts?

    (13/52) × (12/51) = 1/17, close to 5.9%. The second factor is where people slip; the deck has changed.

  3. In a room of 23 people, is it more likely than not that two share a birthday?

    Yes, marginally: about 50.7%. The complement is a product of 364/365 × 363/365 × …, and it crosses one half at 23. Worth memorising as a reference point, since interviewers use it to see whether your intuition updates.

Conditional probability

  1. A disease affects 1% of people. A test catches 90% of true cases and returns a false positive 5% of the time. You test positive. What is the chance you have it?

    (0.9 × 0.01) / (0.9 × 0.01 + 0.05 × 0.99) = 0.009 / 0.0585 ≈ 15.4%. The base rate dominates. Interviewers like this one because the intuitive answer is wildly wrong and the correction is one line of arithmetic.

  2. A family has two children. At least one is a boy. What is the chance both are boys?

    As usually stated, 1/3: the equally likely cases are BB, BG, GB, and the condition removes only GG. The better answer flags the ambiguity, because if you learned this by meeting a specific child the answer becomes 1/2. Saying that out loud scores higher than either number alone.

  3. Three doors, one prize. You pick one, the host opens a different door showing no prize, and offers a switch. Do you switch?

    Switch. Your first pick wins 1/3 of the time, so switching wins 2/3. The host's knowledge is what makes the extra information real; if the door had been opened at random, switching would gain nothing.

Expectation

  1. How many rolls of a fair die do you expect before the first six?

    Six. For a repeated trial with probability p, the expected wait is 1/p. If you want the one-line derivation: E = 1 + (5/6)E, so E = 6.

  2. Shuffle n numbered cards into n numbered slots. How many land in their own slot on average?

    Exactly one, for any n. Each card has probability 1/n of matching, and expectations add regardless of dependence, so the total is n × 1/n = 1. The independence trap is the whole point of the question.

Symmetry

  1. You flip 21 fair coins and I flip 20. What is the chance you get strictly more heads than I do?

    One half. With one extra coin, exactly one of two things happens: you get strictly more heads, or you get strictly more tails, never both and never neither. Those two outcomes are symmetric, so each has probability 1/2. No summation required.

How to use these

Time yourself, and separate two failure modes. Not knowing the method is a study problem. Knowing it but freezing is a practice problem, and only repetition under a clock fixes it. For a broader mix across mental math, sequences, and logic, see the full practice question set.