Thirty-two classic interview-style brain teasers, grouped by skill. 10 fully solved here — all 32 solved step-by-step in the app. These are public puzzle structures rewritten in our own words — not leaked exam items from any firm.
Probability
10 puzzles
The false-positive alpha
Exactly 1% of tickers in a universe are true frauds. A screening model is advertised as "99% accurate": it flags 99% of real frauds and also correctly clears 99% of legitimate names. The model flags a new ticker as fraud. What's the probability it actually is a fraud?
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Exactly 1/2 (50%) — not 99%.
Base-rate fallacy. Imagine 10,000 tickers: 100 frauds, 9,900 clean. The model catches 99 of 100 frauds and falsely flags 1% of 9,900 ≈ 99 clean names. Among ~198 alarms, half are real. Formally: P(F|alarm) = (0.99·0.01) / (0.99·0.01 + 0.01·0.99) = 1/2. Always ask for prevalence before trusting a "high accuracy" claim.
The three dark pools
A large block is hidden uniformly at random in one of three dark pools A, B, or C. You send a ping to A. Before it fills, an oracle that knows the true location (and never "kills" the true pool) publicly shuts pool B, proving the block is not in B, and offers you a free redirect to C. Stick with A or switch to C?
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Switch to C. Staying with A wins with 1/3; switching wins with 2/3.
Monty Hall. Your first pick locks in 1/3. The complementary event "B or C" has 2/3. The oracle never shuts the true location, so all of that 2/3 collapses onto C once B is shut. Treating the remaining two pools as 50/50 would be correct only if the oracle had chosen independently of your first pick.
Double sixes on the open
A fair six-sided die is rolled until you see two sixes in a row. Expected number of rolls?
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42.
States: E0 (no current six), E1 (last roll a six), E2 = 0. E0 = 1 + (1/6)E1 + (5/6)E0, E1 = 1 + (5/6)E0. Solving gives E0 = 42. In general, expected waits for a probability-p symbol twice in a row is (1+p)/p².
Matching birthdays on the desk
A 23-person trading desk. Birthdays independent, uniform over 365 days. Probability that at least two share a birthday — and why is that not 23/365?
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About 50.7%. Compute the complement: all distinct, then subtract from 1.
P(collision) = 1 − 365/365 × 364/365 × … × 343/365 ≈ 0.507. The naive 23/365 treats a different question (someone matches a fixed person). Here every pair can collide: C(23,2) = 253 pairs, and 1 − exp(−253/365) ≈ 0.50.
The shuffled login tokens
n traders each have a unique login token, shuffled uniformly. Probability that nobody receives their own token? Limit as n grows?
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!n / n! → 1/e ≈ 36.79%, already close by n = 6.
Derangements via inclusion-exclusion: Σ_{k=0}^{n} (−1)^k / k!, the Taylor partial sum for e^{−1}. Related surprise: expected number of fixed points is exactly 1 for every n ≥ 1, by linearity, even while P(zero fixed points) sits near 1/e.
Two children, at least one boy
A colleague has two children. You learn at least one is a boy. Probability both are boys? What if you instead learn the older one is a boy?
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Write the sample space of equally likely children-pairs before you condition on the extra information.
85% of cabs in a city are Blue, 15% Green. A witness says a cab in a hit-and-run was Green; witnesses are right 80% of the time (either colour). Posterior that it was Green?
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Base rates first, then the witness's accuracy — same move as the fraud-screener.
Five-card hand from a 52-card deck. Probability of at least one ace? Don't compute all the ways to have two, three, four aces separately if you can avoid it.
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Complement: probability of zero aces is easier than summing exactly-k.
You interview 10 candidates in random order, one at a time. After each you must hire or permanently pass. You want the single best of the 10. Optimal policy, and roughly what success probability?
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Reject the first 3 as a benchmark, then hire the first later candidate better than all of those 3. Success ≈ 39.9% (→ 1/e ≈ 37% as n → ∞).
Secretary / optimal stopping. For large n, auto-reject the first ~n/e, then take the first candidate better than the sample. For n = 10 the best pure cutoff is r = 4 (reject first 3). If the overall best sits at i ≥ r, you hire them iff the best among the first i−1 sat inside the sample.
The market-making random walk
Inventory starts at +2 lots. Each trade is +1 or −1 with equal probability. Stopped at 0 or +5. Probability you hit +5 before 0?
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2/5. Symmetric gambler's ruin is linear: i / N.
P(i) = ½ P(i+1) + ½ P(i−1), P(0)=0, P(5)=1, so P(i)=i/5. At i=2 that is 2/5. Biased case: (1−r^i)/(1−r^N) with r = q/p. Expected duration in the symmetric case is i(N−i) = 6 steps here.
Flips until first heads
A fair coin is flipped until the first heads. Expected number of flips? Expected payout if you win $2^k when heads first appears on flip k?
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A geometric wait has a closed-form expectation; write the recurrence or the series.
A bet pays even money and you win with probability 60%. You have a bankroll of $100 you cannot refill. What fraction should you stake if you can repeat the bet many times and care about long-run growth?
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Kelly criterion: fraction of bankroll, not 'always go big'.
A firm locks 100 unique bonus envelopes labeled 1–100 in a room; a random permutation maps each ID to one envelope. Each of 100 traders may open at most 50 envelopes. They must all find their own ID for anyone to get paid. They may plan a strategy in advance, but once the room is open they cannot communicate. What's a strategy that gives the group roughly a 30% chance of collective success — and why is random guessing hopeless?
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Follow the cycle: each trader starts at the envelope labeled with their own ID, then follows the ID found inside to the next envelope, up to 50 opens. Success probability ≈ 1 − ln(2) ≈ 30.7%. Independent guessing is about 2⁻¹⁰⁰.
The envelopes define a permutation of 100 IDs, which decomposes into disjoint cycles. Trader k finds their ID iff the cycle containing k has length ≤ 50. The group fails iff there is a cycle longer than 50. There can be at most one such cycle, and P(a random permutation of n has a cycle of length j) = 1/j. For n = 100, P(fail) = sum_{j=51}^{100} 1/j ≈ ln(2), so P(success) ≈ 1 − ln(2) ≈ 0.307. Independent search fails because each person has only ~½ chance and the attempts are essentially independent.
Dividing the bonus pool by seniority
Five perfectly rational, purely self-interested traders are ranked 1 (most senior) through 5. They must split a $100m bonus pool. The most senior proposes an integer-dollar allocation; if at least 50% of remaining traders (including the proposer) vote yes, it passes. If not, the proposer is fired with $0 and the next most senior proposes under the same rule. What does trader 1 propose?
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$98 to trader 1, $0 to 2, $1 to 3, $0 to 4, $1 to 5 (or any $1 bribes to the two who would get $0 in the 4-person subgame).
Backward induction. With 2 left: 4 takes $100, 5 gets $0. With 3: 3 needs 2 of 3 votes, buys 5 for $1, keeps $99. With 4: 2 needs one extra vote, buys the person who would get $0 next (trader 4). With 5: 1 needs two extra votes; in the 4-person subgame, 3 and 5 get $0, so they are the lowest-cost bribes at $1 each. Seniority still captures almost everything once you price continuation values.
Nine chips, one light
You have 9 identical-looking chips. Exactly one is lighter; the other eight are equal. You have a two-pan balance. Minimum weighings that guarantee you find the light chip?
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2 weighings — ternary split each time.
A balance has three outcomes, so it trisects the search space: up to 3^k candidates in k weighings. 3² = 9, so two suffice. Divide into three groups of three, weigh 1 vs 2, then repeat inside the lighter group (or the untouched group if they balance).
Gold, empty, and the lying labels
Three boxes: one gold, one empty, one mixed. Every label is wrong. You draw one item from the box labeled "mixed" and it is gold. Which box is pure gold, and what is in the other two?
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A statement that cannot be true under a label is your first cut.
Four people need to cross a bridge at night with one torch. Crossing times 1, 2, 5, 10 minutes. At most two at a time, and they walk at the slower person's pace. The torch must be carried each way. Fastest total?
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Two slow people cannot both walk with the flashlight at the end without a cheap return.
1,000 bottles, at most one poisoned. A tester who sips poison shows a result after a delay, and you have one test round. 10 testers. How do you find the poisoned bottle?
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Each tester is a bit. Think binary encoding, not one-bottle-per-tester.
n people have blue eyes, the rest brown. Each can see others' eyes but not their own. They leave on the nth midnight iff they deduce their own colour. A visitor says "I see someone with blue eyes." For n=2, what happens?
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Common knowledge plus induction from the 1-person case.
Two traders, red or blue hats independent fair coins. Each sees the other's hat, not their own. They guess simultaneously or pass. They win if at least one guesses correctly and nobody guesses wrong. Give a strategy that wins with probability 1/2.
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Agree a coding of what you see before the hats go on.
An interviewer will roll a fair die and pay you the face value in dollars. Quote a two-way market (bid / ask) on that payoff. You will have to trade at least one side. Where do you quote, and why is the spread not zero?
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Your two-way quote has to survive being lifted on one side.
A widget's common value V is unknown; your estimate is unbiased. You and several others each submit a sealed bid. You win if you bid highest. Should you bid your estimate?
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Conditional on winning, your estimate was the most optimistic in the room.
Decompose, assign units, sanity-check the order of magnitude.
Want more?
Ten of these 32 include a full worked solution on this page so search engines can index them. The other 22 give a one-line hint here and the full solution in the app, which also has a larger library with difficulty tiers and tracking.
FAQ
Are these the exact questions these firms actually ask?
No — and no serious prep site can promise that. Each puzzle below is adapted from classic quant-interview material (and from TradeMind's own practice library), written in our own words. Firm labels reflect where similar puzzles are most often discussed in public prep sources (Glassdoor threads, WSO, Heard-on-the-Street-style collections), not a confirmed or exclusive current interview question. The same structure often shows up at many firms, and outside finance entirely.
Why these 32, and not only the airplane / two-guards classics?
Those are still fair warm-ups, but candidates see them everywhere. This set is grouped by the tools that show up in trading interviews: probability, expected value, logic, market-making intuition, and Fermi estimation — closer to how quants actually reason in a room.
Why do quant firms ask brain teasers at all?
Less to check whether you already know a trick, more to watch how you think out loud: do you model the problem, name assumptions, pick a structure (cycles, induction, Bayes table), and stay calm when the first intuition is wrong. The path is usually worth more than the final number.
What's the best way to practice these?
Cover the answer, set a 3–5 minute timer, and narrate out loud as if an interviewer were listening — silent staring is a different skill from talking through structure under mild pressure. Then re-solve once with notes to lock the framework (e.g. "always start pirate games from the end").
These puzzles are independent adaptations of classic interview brain teasers, rewritten in our own words. They are not confirmed, exclusive, or current exam questions from any company. The same structures appear across many firms and outside quantitative finance. TradeMind is not affiliated with, endorsed by, or sponsored by any firm named elsewhere on this site.
