SIG Quant Research Phone Interview Experience (2026) - Probability Drill with Penney's Game, Bayes & Three Points on a Circle, Pending

Problem 3: Waiting Time Paradox

Problem: Three bus routes arrive at a stop independently. Route A's arrival time is uniform on (0, 10). Route B is uniform on (0, 20). Compute the average waiting time until the first bus of any kind.

Expected minimum of two non-identical uniforms. Integrate the product of survival functions: E[min(X, Y)] = integral from 0 to 10 of (1 - t/10)(1 - t/20) dt , since after t=10 bus A has arrived with certainty. Evaluating gives roughly 4.17 minutes. The trap is letting the outer bound run to 20 or confusing "first bus" with "bus A."

This question is coming soon to HackTheRounds.

Problem 4: Three Random Points on a Circle

Problem: Three points are chosen independently and uniformly on a unit circle. Compute the probability they form an obtuse triangle.

Three points form an obtuse triangle if and only if all three lie on some semicircle. The probability that all three lie on a common semicircle is 3 / 4 . So P(obtuse) = 3/4 , and P(right triangle) = 0 (measure zero), P(acute) = 1/4 .

This is the one I had pre-memorized. The interviewer followed up with: "Now compute the expected length of the chord between two randomly chosen points on the unit circle." This is integrating 2 sin(theta / 2) over theta uniform on (0, 2 pi), which gives expected chord length 4 / pi .

Problem 5: Penney's Game

Problem: Player A picks HHT. Player B picks HTH. A fair coin is flipped repeatedly, and the first player whose pattern appears first wins. Who has the advantage?

B has the advantage, winning with probability 2/3. The reason is non-transitivity: after a flip, the pattern that "recovers" faster from a miss wins more often. For HHT versus HTH, if the first two flips are HT and the third is H, A resets entirely but B is already partway matched. If you have not seen Penney's game before, this is very hard to derive under time pressure. I had the answer cached and spent the time on the why rather than rediscovering it.

Problem 6: Poker Raise Minimum Win Rate

Problem: You and a friend each put $10 into the pot. Your friend raises by another $20. If you fold, you forfeit the $10. What is the minimum win rate you need to call?

Pot odds. Call costs $20, potential win is the $40 already in the pot. Required win rate 20 / (20 + 40) = 1/3 , so 33.3%. The trap is treating the $10 blind as part of the call cost; it is already sunk. I walked through the sunk-cost logic out loud and the interviewer confirmed the framing.

This question is coming soon to HackTheRounds.

Problem 7: Biased Coin from 99 Fair Coins

Problem: A bag contains 99 fair coins and one double-headed coin. You pick a coin uniformly at random and flip it 7 times. All 7 flips are heads. What is the probability the coin is double-headed?

Bayes. Numerator 1 (1/100) , denominator (1/100) + (99/100) (1/128) . Final: 128 / 227 ≈ 0.564 . The interviewer pushed on intuition: prior 1%, posterior 56% after 7 heads, because 7 heads is 128 times more likely under the double-headed hypothesis.

Problem 8: Intelligence Quiz Break-Even

Problem: Staying gets you $1,000. A correct answer gets you $4,000. A wrong answer leaves $250. What is the minimum accuracy at which risking is worth it?

Break-even: p 4000 + (1 - p) 250 = 1000 , so p = 750 / 3750 = 20% . I solved it in under twenty seconds and the interviewer moved past quickly.

This question is coming soon to HackTheRounds.

Behavioral Tail (15 min)

Five quick questions: why QR, what other interviews am I in the middle of, describe my research, why SIG, and what my post-graduation goals are (technical / academic / financial). I kept each answer under 90 seconds and tied every answer to a specific detail rather than an abstraction. The recruiter-style screen at SIG grades on specificity.

Result

Pending. Recruiter said results within two weeks.

Tips

1. Cache the five canonical answers: three points on a circle (3/4), Penney's game (HHT vs HTH goes 2/3 to HTH), waiting time paradox integral form, Bayes with low prior and massive update, and expected-value break-even. SIG's QR pool reuses these every cycle. Memorize the answers plus a two-sentence justification.
2. Ask for clarification on ambiguous phrasings before computing. The "two trips, one international in December" problem is deliberately ambiguous. Candidates who charge ahead with an unspoken assumption about which trip is fixed get the wrong framing. Ask, then solve.
3. Know how to set up expected minimum of two non-identical uniforms. `E[min(X, Y)] = integral of P(X > t) * P(Y > t) dt` with the integration bound capped at the smaller range's upper end. This shows up on waiting-time problems across every quant shop.
4. For the "game ends on A's turn" problem, answer probability 1 quickly and move on. Candidates who try to compute the expected geometric stopping time in the first pass waste thirty seconds on a trivial sub-question. The expected rolls follow-up is where the time should go.
5. In the poker problem, name the sunk-cost reasoning explicitly. SIG's graders listen for "the $10 blind is sunk, only the $20 call is relevant" out loud. Solving it silently and stating only the final percentage makes you look like you lucked into the answer.
6. Practice with the Green Book. Mark Joshi's Quant Job Interview Questions and Answers plus A Practical Guide to Quantitative Finance Interviews cover 80% of the SIG QR pool. Grind the probability and combinatorics sections specifically. Brainteasers section is lower yield.

SIG's QR phone interview is a speed check on classical probability. If you can produce a clean derivation under 90 seconds for every problem in the canonical pool, you will advance. If you are improvising from first principles, you will run out of time.