Quant Brain Teasers: 25 Real Interview Puzzles With Solutions 2026

Why Brain Teasers Still Matter

Despite years of finance reformers arguing that brain teasers don't predict on-the-job performance, every major prop trading firm and most hedge funds still use them. The reason isn't that the puzzles directly measure trading ability - it's that they measure two specific traits the firms care about: how you think out loud under pressure, and how you handle being stuck. The candidate who freezes at an unfamiliar problem behaves the same way at an unfamiliar trade.

This guide collects 25 of the most-asked brain teasers across SIG, Jane Street, Optiver, IMC, Citadel and Hudson River Trading interviews. For probability-specific questions, see our quant probability interview questions. For broader interview content, see our quant interview questions hub.

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Section 1: Classic Lateral Thinking (Questions 1-7)

1. The 100 doors

You have 100 doors, all closed. 100 people walk past in sequence. Person 1 toggles every door. Person 2 toggles every second door. Person 3 toggles every third door. And so on. After all 100 people have walked past, which doors are open?

Answer: Doors 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 - the perfect squares. A door is toggled once per divisor. Most numbers have an even number of divisors (paired: 12 = 112 = 26 = 34). Perfect squares have an odd number because the square root is its own pair (16 = 116 = 28 = 44).

2. The 100 light switches

A 100-foot rope hangs from the ceiling. You light it from one end. The rope burns at non-uniform speed but takes exactly 60 minutes to burn end-to-end. How do you measure 45 minutes using only this rope and matches?

Answer: Use two ropes. Light rope A from both ends and rope B from one end simultaneously. Rope A burns out in 30 minutes (regardless of non-uniformity). At that moment, light the other end of rope B. Rope B then burns out in 15 more minutes. Total: 45 minutes.

3. Two doors, two guards

You face two doors, one to heaven, one to hell. Each is guarded by someone. One always lies, one always tells the truth. You can ask one question to one guard. What do you ask to find the door to heaven?

Answer: Ask either guard: "If I asked the other guard which door leads to heaven, which would they point to?" Then take the opposite door. The truth-teller correctly reports the liar's lie; the liar lies about the truth-teller's truth. Either way you get the wrong door, so go the other way.

4. The 12 coins

You have 12 coins. One is counterfeit and weighs slightly different (you don't know if heavier or lighter). Using a balance scale only 3 times, find the counterfeit and determine if it's heavy or light.

Answer: Yes, it's solvable in 3 weighings. Split 4-4-4. Weigh first two groups. If balanced, the counterfeit is in the third group of 4 - one more weighing isolates it within 2 candidates, then a final weighing identifies it. If unbalanced, swap and rotate to identify both the coin and direction. The full solution is intricate but the construction exists - work through it on paper before the interview.

5. Two trains

Two trains 100 miles apart approach each other at 50 mph each. A bird starts at one train, flies to the other at 75 mph, immediately turns around when it reaches the other train, and continues bouncing back and forth until the trains meet. How far does the bird fly?

Answer: 75 miles. Don't fall into summing the bouncing distances. The trains meet in 1 hour (closing at 100 mph). The bird flies for 1 hour at 75 mph. Distance = 75 miles.

6. The blue-eyed islanders

100 islanders all have blue eyes. They live by a rule: if you ever know your own eye color is blue, you must leave the next morning. There are no mirrors or reflections, and they cannot communicate eye color directly. A visitor announces, "I see at least one blue-eyed person here." When do they leave?

Answer: Day 100. Common knowledge induction. With 1 blue-eyed person, they leave day 1. With 2, they each see one other and reason "if it were just one, that person would have left day 1; they didn't, so I must also be blue." They both leave day 2. By induction, n blue-eyed people leave on day n. With 100 blue-eyed islanders, all 100 leave on day 100.

7. The 5 hats

Five logicians stand in a line, each wearing a hat. There are 3 red hats and 2 blue hats; everyone knows this. Each can see only the hats in front of them, not their own or those behind. Starting from the back, each is asked their hat color in turn (and everyone hears every answer). What strategy ensures the most correct answers?

Answer: At least 4 correct, possibly 5. The first logician (at the back) sees 4 hats. They can encode parity: say "red" if they see an even number of red hats, "blue" if odd. The second deduces their own hat from the first's parity announcement and what they see. By induction, logicians 2 through 5 all answer correctly. Logician 1 may be wrong.

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Section 2: Numerical Puzzles (Questions 8-14)

8. The 25 horses

You have 25 horses and a 5-track racetrack. Each race tells you the order of the 5 horses but not their times. What's the minimum number of races to determine the top 3 fastest?

Answer: 7. Race 5 groups of 5 (5 races). Then race the 5 winners (race 6) - the winner is the fastest horse overall. To find 2nd and 3rd: candidates are 2nd and 3rd from race 6, plus 2nd from race 6's 1st-place group, plus 2nd and 3rd from race 6's 2nd-place group, plus 2nd from race 6's 3rd-place group. Wait - that's too many. Carefully: only 5 candidates need racing in race 7 (the runner-up of the winner's heat, the 2nd and 3rd of the winner of race 6 etc.) Race 7 determines positions 2 and 3.

9. The 100 prisoners

100 prisoners are each given a unique number 1-100. The warden randomly places papers numbered 1-100 in 100 boxes. Each prisoner enters the room one at a time, can open at most 50 boxes, and must find the paper with their own number. They cannot communicate after starting. If all succeed, all are freed; if any fail, all die. The prisoners can strategize beforehand. What's their optimal strategy and survival probability?

Answer: ~31% survival. Each prisoner opens the box matching their own number, then opens the box matching the number on the paper they find, etc. - following a "cycle" through the boxes. They survive if and only if no cycle has length > 50. The probability that a random permutation of 100 has no cycle longer than 50 is (1 - \sum_{k=51}^{100} 1/k \approx 0.31).

10. Average weight problem

A scale measures up to 100 kg. You have 8 weights of identical appearance, but one is slightly different from the others. You can use the scale a maximum of 2 times. Can you find the odd weight?

Answer: Not always. With 8 unknowns and 2 weighings producing 3 outcomes each (left heavier, balanced, right heavier), you have 9 possible information states - just enough for 8 candidates if you also need to know direction (heavy or light). Without knowing direction in advance, the answer is no. With direction known, yes - split 3-3 first, then 1-1 on the relevant subset.

11. The Birthday Problem

In a room of 23 people, what's the probability at least two share a birthday?

Answer: ~50.7%. (P(\text{none}) = 365!/((365-23)! \cdot 365^{23}) \approx 0.493). So (P(\text{at least 2}) \approx 0.507). Worth memorising the ~50% threshold at 23 people.

12. The 1000 wines

You have 1000 wine bottles, exactly one of which is poisoned. The poison takes 24 hours to show effect. You have 10 prisoners and 24 hours. How do you identify the poisoned bottle?

Answer: Number bottles 0 to 999 in binary (10 bits). Prisoner i drinks from every bottle whose i-th binary digit is 1. After 24 hours, the binary representation of which prisoners died identifies the poisoned bottle exactly.

13. The 100-floor egg drop

You have a 100-floor building and 2 identical eggs. There exists a critical floor F such that eggs break when dropped from floor F or above. Find F using the minimum number of drops in the worst case.

Answer: 14 drops. Strategy: drop the first egg from floor 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99 (decreasing increments of 14, 13, 12, ...). When it breaks, use the second egg linearly within the previous bracket.

14. The 1000 doors that open and close

1000 doors numbered 1-1000 are all closed. You walk through and toggle them in passes. Pass 1 you toggle every door. Pass 2 you toggle every second door. ...Pass 1000 you toggle every 1000th door. After pass 1000, how many doors are open?

Answer: 31. Same as Question 1 with N=1000: the open doors are the perfect squares 1, 4, 9, ..., 961 = 31². So 31 open doors.

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Section 3: Logic and Sets (Questions 15-20)

15. The 10 stacks of coins

You have 10 stacks of 10 coins. 9 stacks contain real coins (10g each); one stack contains fake coins (11g each). Using a digital scale only ONCE, identify the fake stack.

Answer: Take 1 coin from stack 1, 2 from stack 2, 3 from stack 3, ..., 10 from stack 10. Total weight if all real: (550 \times 10 = 5500)g. The actual weight will be 5500 + n where n is the fake stack number.

16. The 5 pirates

5 pirates have 100 gold coins. Pirate 1 (most senior) proposes a distribution. All pirates vote (Pirate 1 included); if majority approves, the distribution is final. If not, Pirate 1 walks the plank and Pirate 2 proposes, and so on. Pirates are perfectly rational and prefer (in order): self-preservation, more gold, and watching others walk the plank. What does Pirate 1 propose?

Answer: 98, 0, 1, 2, 0 to Pirates 1-5 respectively. Backward induction. With just Pirate 5 left, they keep all 100. With Pirate 4 alive, Pirate 4 keeps everything (votes self only; majority since 50% of 1 vote). With Pirates 3-5: Pirate 3 needs 1 ally - gives Pirate 5 one coin (since Pirate 5 would otherwise get nothing). Continue.

17. The 100 flights

At a 100-passenger flight, the first passenger is drunk and sits in a random seat. Each subsequent passenger sits in their assigned seat if available, or randomly if not. What's the probability the 100th passenger sits in their assigned seat?

Answer: 0.5. The state collapses to "the right seat or the drunk's original seat will be filled." By symmetry, the 100th passenger has 50% probability of sitting in their assigned seat regardless of N.

18. The 5 children and 5 ages

5 children's ages multiply to 72 and sum to a number which I'll tell you. After hearing the sum, you say "I cannot determine the ages." After hearing this, I say "the oldest plays the violin." Now you can. What are the ages?

Answer: 3, 3, 2, 2, 6. Sum is 14. There are exactly two factorisations of 72 with sum 14: {3, 3, 2, 2, 6} and {6, 6, 1, 1, 2}. The "oldest" being unique distinguishes them - the former has a single oldest (6); the latter has two 6s tied for oldest.

19. The 12 marble puzzle

You have 12 marbles, one of which is heavier or lighter than the others. Using a balance scale 3 times, identify the marble and determine if it's heavy or light.

Answer: Yes, this is solvable. Split into 4-4-4 groups; first weighing identifies which third the odd marble is in. The full strategy is intricate; work through it on paper. (See Question 4 for a related setup.)

20. The 3 light bulbs

Three light bulbs are in a sealed room. Three switches are in a hallway, each controlling exactly one bulb. You can flip switches as much as you want, but you can enter the room only ONCE. How do you determine which switch controls which bulb?

Answer: Turn switch 1 on for 5 minutes. Turn it off. Turn switch 2 on. Enter the room. The bulb that's on is switch 2. The bulb that's warm is switch 1. The bulb that's cold is switch 3.

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Section 4: Real Trading Interview Brain Teasers (Questions 21-25)

21. SIG hat puzzle

3 people each wear a hat that's either red or blue with equal probability. Each can see the others' hats but not their own. Simultaneously, each guesses their own color or passes. They win collectively if at least one guesses correctly and no one guesses incorrectly. What strategy maximises win probability?

Answer: 75% win rate. Strategy: pass if you see two different colors; if you see two of the same color, guess the opposite. The team wins on 6 of 8 configurations.

22. Jane Street equilateral triangle

Three points are chosen uniformly at random on a circle. What's the probability they form an obtuse triangle?

Answer: 3/4. Place the first point anywhere (WLOG). Parameterize the other two by their angular positions. The triangle is acute iff the largest arc between consecutive points is less than π. Compute this probability and complement.

23. Optiver coin variant

A coin is tossed repeatedly. What's the probability that 3 consecutive heads appear before 5 consecutive tails?

Answer: ~94%. Markov chain with 8 states (current "head streak" 0-2, current "tail streak" 0-4, plus absorbing states for win/lose). Solve the linear system. The asymmetry is severe: 3 heads is much more likely first because the absorbing barrier is closer.

24. Citadel dice

You roll a fair die. If you roll 1, 2, or 3, you win nothing. If you roll 4, 5, or 6, you win the face value in dollars. What's the fair price to play?

Answer: $2.50. (E[X] = (1/6)(0+0+0+4+5+6) = 15/6 = 2.50).

25. HRT coin flipping race

Two players each flip a fair coin until they get a head. The winner is whoever gets a head first; if simultaneous, tie. What's the probability of a tie?

Answer: 1/3. Each player's flips form a geometric distribution. P(both first heads on flip k) = ((1/2)^{2k}). Sum: (\sum (1/4)^k = 1/3).

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How to Use This Guide

Practice the questions in order. Aim to recognise the technique within 30 seconds: is this a recursion-on-states problem, a parity problem, a backward-induction problem, a base-rate Bayes problem? Pattern recognition matters more than computational speed.

For probability-specific drilling, see our quant probability interview questions. For firm-specific interview content, see our SIG interview, Jane Street interview and Optiver interview guides.

For the broader hub of interview prep content, see our quant interview questions.