Optimal stopping asks when to commit under uncertainty — and since trading is a stream of commit-or-wait decisions, quant loops return to it constantly. Three setups cover the interview canon.
1. The secretary problem (the 37% rule)
See $n$ candidates in random order, accept or reject each irrevocably, maximize the chance of picking the best. Optimal: observe the first $n/e \approx 37\%$ without accepting, then take the first candidate better than everything seen. Success probability: $1/e \approx 37\%$, independent of $n$. Fully worked here. Interview follow-ups: what if you want to maximize the VALUE rather than P(best)? What if you can recall past candidates? Each variant changes the rule — knowing THAT is the point.
2. Finite-horizon backward induction
Dice games with rerolls, "three offers, take one" — solve from the end: the value of the last stage sets the threshold for the stage before. The three-roll dice game is the canonical drill (thresholds 5, then 4.25). If you can narrate "value of continuing vs value of stopping" cleanly, every finite problem in this family falls.
3. Threshold strategies on continuous draws
Uniform draws with a stop decision — "observe up to $n$ uniforms, keep the last one you accept" — where the optimal rule is a declining sequence of thresholds. Worked here. The structural lesson interviewers probe: optimal rules in this family are ALWAYS thresholds ("accept if above x"), never randomized — because the value of continuing is a fixed number you compare against.
Why firms care
Market-making games embed stopping constantly: when to fade a quote, when to cut a position, when to stop trading a signal that might have decayed. The trading games exercise the same muscle live. And in options terms, every stopping problem is an American-option exercise decision — a bridge senior interviewers like you to notice out loud.
Practice with solutions
- The secretary problem
- Three-roll dice game
- Optimal stopping for the max of uniforms
- The expectation bank — 75 free problems in the parent family.
More topic guides
- The Airplane Seat Problem: Why the Answer Is 1/2 (Three Proofs)
- Bayes' Theorem in Quant Interviews
- Behavioral Interview Questions at Trading Firms (With Answer Frameworks)
- Coin Flip Questions in Quant Interviews
- C++ Low-Latency Interview Questions at HFT Firms
- Dice Questions in Quant Interviews
- Fermi Estimation Interview Questions at Trading Firms
- Gambler's Ruin in Quant Interviews
- The Kelly Criterion in Quant Interviews
- Linear Regression Interview Questions: OLS Assumptions, R² Traps & Regression to the Mean
- Machine Learning Quant Interview Questions: Overfitting, Cross-Validation & Feature Leakage
- The Market Making Game Interview: How to Answer 'Make Me a Market'
- Market Microstructure Interview Questions: Order Books, Spreads & Adverse Selection
- Markov Chains in Quant Interviews
- Martingales in Quant Interviews
- Mental Math for Trading Interviews: Training Plan, Zetamac Benchmarks & Firm Tests
- The Monty Hall Problem — and the Variants Interviews Actually Ask
- Number Sequence Tests in Trading Interviews: The 8 Pattern Types & How to Practice
- Options Pricing Interview Questions: Black-Scholes, Greeks & Put-Call Parity
- Quant Interview Cheat Sheet: Probability, Markov Chains, Options & Linear Algebra (Free PDF)
- Random Walks in Quant Interviews
- Statistics Interview Questions for Quant Roles: Hypothesis Testing, MLE & p-Value Traps
- Stochastic Calculus Interview Questions: Ito's Lemma, SDEs & Brownian Motion
- Time Series Interview Questions: Stationarity, ARMA & Autocorrelation Traps
- Top 50 Quant Interview Questions (With Full Solutions)
- Quant Mental Math Practice: The Tests, the Skills, and How to Drill Them
- Zetamac: What a Good Score Is, and How to Practice Past It
- All guides & explainers
Frequently asked questions
What is the 37% rule?
In the secretary problem, reject the first n/e ≈ 37% of candidates, then accept the first one better than all seen so far. It picks the single best candidate with probability 1/e ≈ 37%, regardless of n.
How do you solve finite optimal stopping problems?
Backward induction: compute the value of the final stage, then at each earlier stage stop if the current offer beats the expected value of continuing. In the classic three-roll dice game the thresholds are: keep ≥5 on roll one, keep ≥5 on roll two (value 4.25 to continue), keep anything on roll three.
Why are optimal strategies thresholds?
Because the value of continuing is a fixed number: accept exactly when the current observation exceeds it. This 'threshold structure' answer is itself a common interview question.
How does optimal stopping relate to trading?
Position exits, quote fades, and signal-decay decisions are stopping problems, and every American option is an optimal-exercise problem — the interview versions are the same math with dice and uniforms.
Practice the real thing
QuantVault has 2,800+ quant interview problems with full solutions, intuition, and hints, firm-by-firm interview funnels, and an auto-graded coding judge. Start free.