Optiver interview questions

Quantitative Trader · Technical · Markov chains · reported by a candidate

In a tennis game you're tied at 30–30. You win each point with probability 0.6.

1. (a) What is your probability of winning the game?
   Answer and worked solution

   Answer: 9/13 ≈ 0.69231

   Key idea. 30–30 is strategically the same as deuce: you need two points in a row, and a split brings you back.

   1. Look at the next two points. Win both: p² = 0.36, you win. Lose both: q² = 0.16, you lose. Split: 0.48, back to 30–30 (or deuce).
   2. The splits just restart, so only the deciding pairs matter.
   3. P(win) = p²/(p² + q²) = 0.36/0.52 = 9/13 ≈ 0.692.

   Check. 0.692 > 0.6: needing two in a row amplifies the better player's edge.

   Trap. Trying to enumerate every path to the end of the game. The pairs trick makes it one line.

2. (b) Now p = 0.9. What is your probability of winning?
   Answer and worked solution

   Answer: 0.9878

   Key idea. Same formula, p = 0.9.

   1. p² = 0.81, q² = 0.01.
   2. P = 0.81/0.82 ≈ 0.988.

   Check. The better player's edge is amplified further: a 90% point-winner almost never loses the game from 30–30.

Interviewers may push further: What is the expected number of points played from 30–30?

Source: Glassdoor (Nov 2018) · Try it in the app →

Quantitative Trading Intern · Interview · Combinatorics · reported by a candidate

You roll three dice at once and read them left to right.

1. What is the probability the three numbers are strictly increasing?
   Answer and worked solution

   Answer: 5/54 ≈ 0.092593

   Key idea. Only rolls with three different values can be strictly increasing, and each such set has exactly one increasing order.

   1. Choose 3 different values: C(6, 3) = 20 ways.
   2. Each gives exactly one increasing roll.
   3. P = 20/216 = 5/54 ≈ 0.093.

   Check. Compare with 1/6 if the three numbers had to be different: P(all different) = 120/216, and 1/6 of those are increasing: 20/216. Same answer.

   Trap. Answering 1/6, which ignores repeated values.

Interviewers may push further: Non-decreasing instead? (C(8,3) = 56 multisets → 56/216.)

Source: Glassdoor · Try it in the app →

Trading intern · Interview · Probability · reported by a candidate

You roll a fair die six times.

1. (a) What is the probability that every number 1–6 appears exactly once?
   Answer and worked solution

   Answer: 0.015432

   Key idea. Count favourable sequences over all sequences.

   1. All sequences: 6⁶ = 46,656.
   2. Each face exactly once: the orderings of 1–6, 6! = 720.
   3. P = 720/46,656 = 5/324 ≈ 0.0154.

   Check. Very unlikely: about 1 in 65.

2. (b) Follow-up (ours): on average, how many rolls until you have seen every face?
   Answer and worked solution

   Answer: 14.7

   Key idea. Coupon collector: break the wait into stages, one per new face.

   1. With k faces seen, a roll is new with probability (6 − k)/6: wait 6/(6 − k).
   2. Sum over k = 0…5: 6(1/6 + 1/5 + 1/4 + 1/3 + 1/2 + 1) = 6 × 2.45.
   3. = 14.7.

   Check. The last face takes 6 rolls on average on its own, so the answer must be well above 6.

Interviewers may push further: How likely is it that you need more than 20 rolls?

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Trader · Interview · Optimal stopping · reported by a candidate

Three face-down cards show n, n + 1 and n + 2, in random order, and you don't know n. Turn one over: keep it, or turn another. Keep that, or take the last card. You win the value of the card you keep.

1. Under the best strategy, how much more than n do you win on average?
   Answer and worked solution

   Answer: 4/3 ≈ 1.3333

   Key idea. The first card tells you nothing, since n is unknown, but the second can be compared with it.

   1. Always look at a second card.
   2. If it's higher than the first, it's the middle or top card, equally likely: keep it (worth n + 1.5 on average, versus n + 1 for the last card).
   3. If it's lower, the last card is better on average: take it.
   4. Over the 6 equally likely orders this strategy wins 1, 2, 2, 2, 1 and 0 above n: 8/6, so 4/3 ≈ 1.33 above n on average.

   Check. Doing nothing clever gives n + 1 on average; the comparison adds a third.

Interviewers may push further: With four cards n … n + 3? Now you are told n is 0, 10 or 20, equally likely. Does the strategy change?

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Trader · Online test · Mental maths · reported by a candidate

Examples from a report of Optiver's 80-in-8 arithmetic test (no calculator).

1. (a) 46²
   Answer and worked solution

   Answer: 2116

   Key idea. Square a round number, then correct.

   1. 46 = 50 − 4.
   2. 2500 − 400 + 16 = 2116.

   Check. Ends in 6 (6² = 36).

2. (b) 2/17 to four decimal places
   Answer and worked solution

   Answer: 0.1176

   Key idea. Know 1/17 = 0.0588235… and double it.

   1. 1/17 = 0.05882352941…
   2. × 2 = 0.1176470588…
   3. To four places: 0.1176.

   Check. 2/17 is a little under 2/16 = 0.125.

Interviewers may push further: Keep going: how many digits of 2/17 can you give in a minute?

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Trader intern · Interview · Mental maths · reported by a candidate

Quick estimates, no calculator.

From a 2014 report; formats may have changed.

1. (a) 44/157 in basis points, to the nearest 5
   Answer and worked solution

   Answer: 2802.5

   Key idea. Basis points: multiply the fraction by 10,000. Approximate the denominator, then correct.

   1. 44/160 = 0.275.
   2. 157 is about 1.9% less than 160, so 44/157 ≈ 0.275 × 1.019 ≈ 0.2803.
   3. ≈ 2,803 bp.

   Check. 44/157 is a bit more than 0.28, since 0.28 × 157 = 43.96.

2. (b) 7^2.5 to the nearest integer
   Answer and worked solution

   Answer: 130

   Key idea. 7^2.5 = 7² × √7.

   1. √7 ≈ 2.6458 (2.65² = 7.02).
   2. 49 × 2.6458 ≈ 129.6.
   3. Nearest integer: 130.

   Check. Between 7² = 49 and 7³ = 343, much closer to 49 × 2.6.

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Quantitative Trader · Interview · Combinatorics · reported by a candidate

Six people of different ages sit at random around a round table.

1. What is the probability that, starting from the youngest, ages increase going either clockwise or anticlockwise?
   Answer and worked solution

   Answer: 1/60 ≈ 0.016667

   Key idea. Rotations don't matter: fix the youngest person's seat.

   1. The other five can be arranged 5! = 120 ways.
   2. Exactly 2 are increasing (one clockwise, one anticlockwise).
   3. P = 2/120 = 1/60.

   Check. With 5 people it's 2/4! = 1/12, so this should be smaller.

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Trader Trainee · Interview · Expected value · reported by a candidate

You flip five fair coins, winning $1 for each head and losing $1 for each tail.

1. (a) What is your expected profit?
   Answer and worked solution

   Answer: 0

   Key idea. Linearity: the total is the sum of five coins.

   1. Each coin is +1 or −1 with equal probability: mean 0.
   2. Five coins: 0.
2. (b) Follow-up (ours): what is the standard deviation of your profit?
   Answer and worked solution

   Answer: 2.2361

   Key idea. Independent variances add.

   1. One coin: (±1 − 0)² = 1, so variance 1.
   2. Five coins: variance 5, SD = √5 ≈ 2.24.

Interviewers may push further: What would you pay for the right to stop after any flip?

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Graduate Trader · Interview · Statistics · reported by a candidate

You flip 100 fair coins.

1. (a) What is the standard deviation of the number of heads?
   Answer and worked solution

   Answer: 5

   Key idea. Heads ~ Binomial(100, ½).

   1. Var = np(1 − p) = 100 × ¼ = 25.
   2. SD = 5.
2. (b) Roughly what is the probability of exactly 50 heads?
   Answer and worked solution

   Answer: 0.0796

   Key idea. Approximate the binomial by a normal with σ = 5. The chance of one exact value is about the density there.

   1. Peak density of a normal: 1/(σ√(2π)) = 1/(5 × 2.507) ≈ 0.080.
   2. Exact: C(100, 50)/2¹⁰⁰ ≈ 0.0796.

   Check. Even the most likely single outcome happens less than 1 time in 12.

   Trap. Answering ½ because 50 is "expected". Expected doesn't mean likely.

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Junior Quant Trader · Interview · Expected value · reported by a candidate

You roll two dice and are paid their total, except that a double is cancelled.

The report says doubles are "cancelled" without saying how, so both readings are here.

1. (a) If a cancelled double pays nothing, what is the game worth?
   Answer and worked solution

   Answer: 35/6 ≈ 5.8333

   Key idea. Start from the expected total of two dice and remove what the doubles contribute.

   1. E[total] = 7.
   2. Doubles (1,1) … (6,6) each have probability 1/36 and totals 2, 4, …, 12: contribution 42/36 = 7/6.
   3. E = 7 − 7/6 = 35/6 ≈ 5.83.
2. (b) If you reroll after a double instead, what is it worth?
   Answer and worked solution

   Answer: 7

   Key idea. Rerolling doubles means conditioning on "not a double".

   1. E[total | not double] = (35/6)/(30/36) = 7.

   Check. Doubles average 7 too, so removing them doesn't change the mean.

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Trader · Interview · Behavioural · reported by a candidate

Behavioural questions candidates report:

• Tell me about an accomplishment you are particularly proud of.
• How have you worked in a team, and how did you handle conflict?
• Why Optiver, and why do you want to be a trader?

Source: Glassdoor: Optiver interviews (QT intern, Oct 2026) · Try it in the app →