Quant Trading Interview Probability Questions with Solutions

AbstractEight questions in the style we use in interviews, each with a full solution and, where useful, a Monte Carlo check. They test the same skills a prediction-market desk uses daily: expected value, conditioning, recursion, consistency between related prices and sizing. All numbers are computed in research/interview.py.

1. HH versus HT#

Question. You flip a fair coin until you see two heads in a row (HH). What is the expected number of flips? And until you see heads followed by tails (HT)?

Solution. For HH, let $E$ be the expected flips from scratch and $E_{H}$ from a state where the last flip was H. Then $E = 1 + \frac{1}{2} E_{H} + \frac{1}{2} E$ and $E_{H} = 1 + \frac{1}{2} \cdot 0 + \frac{1}{2} E$ . Solving gives $E = 6$ . For HT, once you have an H you only wait for a T (an H keeps you in the same state), so $E = 2 + 2 = 4$ . Monte Carlo over 100,000 runs: 5.994 and 3.998.

What it tests. Whether you can set up states, and whether you notice that patterns which overlap themselves (HH) take longer to appear.

2. The honest coin (our monthly puzzle)#

Question. A contract pays $1 if at least 60 of 100 fair flips are heads. Price it before any flips, after 34 heads in the first 50, and after one more flip each way.

Solution. $Pr (Bin (100, \frac{1}{2}) \geq 60)$ = 0.0284, so about 2.8¢. After 34 of 50 you need at least 26 of the remaining 50: $Pr (Bin (50, \frac{1}{2}) \geq 26)$ = 0.4439, or 44.4¢. A tail next leaves 26 needed from 49: 0.3877 (38.8¢). A head leaves 25 from 49: exactly 0.5 by symmetry. One flip moves the price by −5.6¢ or +5.6¢, which is the information the market maker must survive (the puzzle page asks about the quote).

3. One re-roll#

Question. Roll a die and be paid its face in dollars, or pay nothing to re-roll once and be paid the second roll. What is the game worth, played optimally?

Solution. A re-roll is worth 3.5, so keep 4, 5 or 6 and re-roll 1–3. Value $= \frac{1}{6} (4 + 5 + 6) + \frac{1}{2} \cdot 3.5 = 4.25$ . The general lesson: the value of an option is computed backwards from its exercise rule.

4. A signal and a market price#

Question. A contract trades at 30¢. You have a signal that is correct 80% of the time, independent of the market's information, and it says Yes. Where would you value the contract?

Solution. Treating the market price as the prior, $Pr (Yes ∣ s) = \frac{0.3 \times 0.8}{0.3 \times 0.8 + 0.7 \times 0.2} = 0.632$ . Strong candidates then question the independence assumption: if the market already reflects the signal, the posterior is 30¢.

5. Best of seven#

Question. A team wins each game independently with probability 0.55. What is the probability it wins a best-of-seven series?

Solution. Sum over the number of losses $k$ before the fourth win: $\sum_{k = 0}^{3} (k 3 + k) (0.55)^{4} (0.45)^{k} = 0.6083$ . A 55% per-game edge becomes a 61% series edge: longer series amplify small edges, a fact worth knowing when pricing a series contract against single-game contracts.

6. Consistent joint prices#

Question. Contract A trades at 60¢ and contract B at 50¢. What range of prices for "A and B" admits no arbitrage?

Solution. The Fréchet bounds: $max (0, 0.6 + 0.5 - 1) = 0.1$ to $min (0.6, 0.5) = 0.5$ , so 10¢ to 50¢. Independence gives 30¢. Inside the band the joint price is a view on correlation; see no-arbitrage bounds.

7. How much to bet#

Question. You believe an outcome has probability 0.6 and can buy it at 50¢. What fraction of your bankroll maximises long-run growth? What if you believe 0.5 and the price is 40¢?

Solution. Kelly for a $1 contract at cost $c$ : $f^{*} = (p - c) / (1 - c)$ , giving 0.20 and 0.167. The follow-up we always ask is "would you actually bet that much?", and the right answer involves how sure you are of 0.6 (Kelly with estimation error).

8. Gambler's ruin with a small edge against you#

Question. You have $3 and bet $1 at a time until you reach $10 or $0. With a fair coin, what is the probability of reaching $10? With a 48% coin?

Solution. Fair: $3/10 = 0.3$ . With win probability $w$ and $ρ = (1 - w) / w$ , $Pr = \frac{1 - ρ ^{3}}{1 - ρ ^{10}} = 0.2213$ for $w = 0.48$ . A 2-point edge against you cuts your chance by about a quarter over a short horizon, which is the same arithmetic that makes small negative edges ruinous for a high-frequency trader.

How we use questions like these#

We care more about the setup than the answer: states, conditioning, what is independent of what, and whether the candidate checks their answer with a quick sanity bound. If these were fun, look at open roles and the monthly puzzle.

Questions#

What probability questions are asked in quant trading interviews?

Expected value and conditional probability, Markov chains and recursions (such as waiting times for coin patterns), dynamic programming for optimal stopping, Bayes' rule with market priors, combinatorics, and position sizing such as the Kelly criterion.

How should I prepare for a quant trading interview?

Practise setting up problems out loud, derive rather than recall, and check answers with quick bounds or simulation. Interviewers care most about clear reasoning under time pressure.

DisclosureThis note is research and education. It is not investment advice or an offer to trade, and nothing in it describes the firm's own results.