SIG (Susquehanna) interview questions

Trader · Interview · Expected value · reported by a candidate

Roll three dice. All three the same: you win $10. Exactly two the same: you win $5. All different: you lose $2.

1. What is your expected profit per game?
   Answer and worked solution

   Answer: 1.25

   Key idea. Find the probability of each outcome type, then weight the payouts.

   1. 216 equally likely rolls.
   2. Triples: 6, so P = 6/216 = 1/36.
   3. All different: 6 × 5 × 4 = 120, so P = 20/36.
   4. Exactly a pair: the rest, 1 − 1/36 − 20/36 = 15/36.
   5. EV = (10·1 + 5·15 − 2·20)/36 = 45/36 = $1.25.

   Check. Check the pair count directly: choose the paired value (6) × the odd value (5) × its position (3) = 90 = 15/36 of 216.

   Trap. Counting pairs as 6 × 5 = 30 and forgetting the 3 positions of the odd die.

Interviewers may push further: How much would you pay to play 100 times? What is the standard deviation of your total?

Source: Glassdoor · Try it in the app →

Trading intern · Interview · Decision making · reported by a candidate

A painting has an 80% chance of being a fake. Real, it's worth $500k; fake, $10k. The seller wants $120k.

1. (a) What is the painting's expected value (in dollars)?
   Answer and worked solution

   Answer: 108000

   Key idea. Expected value: weight each outcome by its probability.

   1. Real: 0.2 × $500k = $100k.
   2. Fake: 0.8 × $10k = $8k.
   3. EV = $108,000, below the $120k asking price.

   Check. The answer must be between $10k and $500k, and much nearer $10k since fakes are likely.

2. (b) Follow-up (ours): above what probability of it being real would you buy at $120k?
   Answer and worked solution

   Answer: 11/49 ≈ 0.22449

   Key idea. Find the probability that makes the expected value equal the price.

   1. EV(p) = 500p + 10(1 − p) = 10 + 490p (in $k).
   2. Set 10 + 490p = 120: p = 110/490.
   3. p ≈ 0.224.

   Check. Your estimate (0.2) is only slightly below breakeven, so cheap information about authenticity is worth paying for.

Interviewers may push further: Would you buy it if you could resell it tomorrow for $115k with certainty to someone who believes it is real?

Source: The Wall Street Quants (reported SIG question) · Try it in the app →

Assistant Trader · Interview · Combinatorics · reported by a candidate

A deck has 97 distinct numbered cards. Four are drawn. If they come out in ascending order (not necessarily consecutive) you're paid $10; otherwise you pay $1.

1. (a) What is the probability the four cards are in ascending order?
   Answer and worked solution

   Answer: 1/24 ≈ 0.041667

   Key idea. Symmetry: any 4 distinct cards are equally likely to appear in each of their orders.

   1. 4 cards have 4! = 24 orders.
   2. Exactly one is ascending.
   3. P = 1/24 ≈ 0.0417.

   Check. The 97 doesn't matter. With 3 cards it would be 1/6; the pattern is 1/k!.

   Trap. Trying to count increasing 4-tuples from 97 cards. It works, but it's slow and error-prone: C(97, 4)/(97·96·95·94) is still 1/24.

2. (b) What is your expected profit per game?
   Answer and worked solution

   Answer: -13/24 ≈ -0.54167

   Key idea. Weight the win and the loss by their probabilities.

   1. Win $10 with probability 1/24, lose $1 with probability 23/24.
   2. EV = 10/24 − 23/24 = −13/24 ≈ −$0.54.

   Check. The fair payout would be $23: then 23/24 − 23/24 = 0.

Interviewers may push further: What payout makes this fair? What if the deck had repeated numbers?

Source: Glassdoor · Try it in the app →

Quant Trader · Interview · Probability · reported by a candidate

A card is drawn from a shuffled deck. You bet on whether the next card will be strictly higher in rank.

As reported, the question continues into bet sizing. The key insight is that the right bet depends on the card you see.

1. (a) Before seeing the first card, what is P(next card is strictly higher)?
   Answer and worked solution

   Answer: 8/17 ≈ 0.47059

   Key idea. "Higher" and "lower" are symmetric before you see anything, so only ties break the symmetry.

   1. After the first card, 51 remain and 3 share its rank: P(tie) = 3/51.
   2. P(higher) = (1 − 3/51)/2 = (48/51)/2 = 24/51.
   3. = 8/17 ≈ 0.471.

   Check. Just under ½, because ties count as a loss for "higher".

2. (b) Follow-up (ours): the first card is a 5 (ranks 2…A). What is P(next is higher)?
   Answer and worked solution

   Answer: 12/17 ≈ 0.70588

   Key idea. Count the cards strictly above a 5 among the 51 left.

   1. Ranks above 5: 6, 7, 8, 9, 10, J, Q, K, A, which is 9 ranks.
   2. 9 × 4 = 36 cards.
   3. P = 36/51 ≈ 0.706.

   Check. A 5 is low, so "higher" should be well above ½.

Interviewers may push further: How much of a $100 bankroll would you bet after seeing a 5 at even odds? (Kelly: 2p − 1.)

Source: Glassdoor · Try it in the app →

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

You may roll a die up to twice and are paid the value of your last roll.

1. What is the expected value?
   Answer and worked solution

   Answer: 4.25

   Key idea. Compare each first roll with the value of rolling again (3.5).

   1. Keep 4, 5, 6; reroll 1, 2, 3.
   2. Kept: (4 + 5 + 6)/6 = 2.5. Rerolled: (1/2) × 3.5 = 1.75.
   3. EV = 4.25.

   Check. Same game as Jane Street's reroll question. It's common, so know it cold.

Interviewers may push further: How much would you pay for a third roll?

Source: Glassdoor · Try it in the app →

Assistant Trader · Interview · Optimal stopping · reported by a candidate

Roll a fair die: an even face pays twice its value, an odd face pays its value.

1. (a) What is a fair price to play?
   Answer and worked solution

   Answer: 5.5

   Key idea. List the payouts and average them.

   1. Faces 1–6 pay 1, 4, 3, 8, 5, 12.
   2. Sum = 33.
   3. Fair price = 33/6 = 5.5.
2. (b) Now you may reroll once, keeping the second result. What is it worth?
   Answer and worked solution

   Answer: 7

   Key idea. A reroll is worth 5.5. Keep only payouts above that.

   1. Keep 4 (pays 8) and 6 (pays 12); reroll 1, 2, 3, 5.
   2. Kept: (8 + 12)/6 = 20/6.
   3. Rerolled: (4/6) × 5.5 = 22/6.
   4. Value = 42/6 = 7.

   Check. Note that a 5 (paying 5) is rerolled, though 5 is "high": what matters is the payout, not the face.

Source: Glassdoor · Try it in the app →

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

You roll a fair die until you get a 6.

1. What is the expected number of rolls, including the 6?
   Answer and worked solution

   Answer: 6

   Key idea. Geometric waiting time: mean 1/p.

   1. p = 1/6, so E = 6.

   Check. Or: E = 1 + (5/6)E.

Interviewers may push further: Expected rolls until you see two 6s in a row? (42.)

Source: Glassdoor · Try it in the app →

Quant Trader Intern · Interview · Probability · reported by a candidate

You roll a fair die twice.

1. What is the probability that the first roll is strictly smaller than the second?
   Answer and worked solution

   Answer: 5/12 ≈ 0.41667

   Key idea. Symmetry, plus ties.

   1. P(tie) = 1/6.
   2. "First smaller" and "first larger" are equally likely: (1 − 1/6)/2 = 5/12.

   Check. Count: 15 of the 36 pairs have the first smaller.

Source: Glassdoor · Try it in the app →

Assistant Trader · Interview · Conditional probability · reported by a candidate

A six-chamber revolver has two bullets in adjacent chambers. The cylinder is spun, the trigger pulled, and it clicks: empty. You're next.

1. (a) If the cylinder just advances one chamber, what is your chance of an empty chamber?
   Answer and worked solution

   Answer: 0.75

   Key idea. Condition on the click: you're at one of the 4 empty chambers, equally likely.

   1. Chambers in order: B, B, E, E, E, E (in a circle).
   2. The next chamber after each empty one: E, E, E, B.
   3. P(safe if it advances) = 3/4.
2. (b) If it's spun again first?
   Answer and worked solution

   Answer: 2/3 ≈ 0.66667

   Key idea. Spinning forgets everything you learned.

   1. 4 empty chambers out of 6: 2/3.

   Check. 3/4 > 2/3: don't spin.

   Trap. Spinning because "a fresh spin is safer". It's the opposite here.

Interviewers may push further: What if the two bullets were placed in random, not necessarily adjacent, chambers?

Source: Glassdoor · Try it in the app →

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

You and a friend each toss three fair coins. If you get the same number of heads you win 2 points; otherwise you lose 1.

1. (a) What is the probability your head counts match?
   Answer and worked solution

   Answer: 0.3125

   Key idea. P(match) = Σ P(both get k heads).

   1. Three coins: 0, 1, 2, 3 heads with probabilities 1, 3, 3, 1 over 8.
   2. P(match) = (1² + 3² + 3² + 1²)/64 = 20/64 = 5/16.
2. (b) What is your expected score per game?
   Answer and worked solution

   Answer: -0.0625

   Key idea. Weight +2 and −1 by their probabilities.

   1. 2 × 5/16 − 1 × 11/16 = −1/16.

   Check. Slightly unfavourable. Breakeven would need P(match) = 1/3.

Source: Glassdoor: SIG quantitative trader · Try it in the app →

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

You flip a fair coin until the pattern HTH first appears.

1. What is the expected number of flips?
   Answer and worked solution

   Answer: 10

   Key idea. Track how much of HTH you've built: nothing (E₀), "H" (E₁), "HT" (E₂).

   1. E₀ = 1 + ½E₁ + ½E₀.
   2. E₁ = 1 + ½E₁ + ½E₂ (another H keeps "H").
   3. E₂ = 1 + ½ × 0 + ½E₀ (T sends you back to the start).
   4. Solving gives E₀ = 10.

   Check. Shortcut: 2³ = 8, plus 2 because HTH ends with its own first letter (H), giving 10. HHH: 8 + 4 + 2 = 14.

   Trap. Expecting 8 for every 3-flip pattern.

Interviewers may push further: Why does HHH take 14 flips but HTT only 8?

Source: Glassdoor search: quantitative trader (2026) · Try it in the app →

Quantitative Trader · Interview · Logic · reported by a candidate

You have 25 swimmers and a pool with five lanes but no stopwatch: a race only tells you the order of the five in it.

Also reported at Squarepoint with horses. We assume speeds are consistent and all different.

1. What is the fewest races needed to find the three fastest?
   Answer and worked solution

   Answer: 7

   Key idea. Use each race's information: who can still be in the top three?

   1. Races 1–5: heats of five.
   2. Race 6: the five heat winners. Its winner is the fastest overall.
   3. Candidates for 2nd and 3rd: 2nd and 3rd from the winner's heat; 1st and 2nd from the heat of the race-6 runner-up; 1st from the heat of the race-6 third place. Five swimmers.
   4. Race 7: those five. Total: 7 races.

   Check. Anyone else has at least three swimmers known to be faster.

Source: Glassdoor search: quantitative trader (2026) · Try it in the app →

Quantitative Trader · Interview · Probability · reported by a candidate

The probability of seeing at least one hawk in an hour is 0.8.

You need a model for when hawks arrive. The natural one: arrivals at a constant rate, independent across time (a Poisson process). State that assumption out loud.

1. What is the probability of seeing at least one in half an hour?
   Answer and worked solution

   Answer: 0.55279

   Key idea. Assume hawks arrive at a steady rate, independently over time (a Poisson process).

   1. P(none in an hour) = 0.2 = P(none in ½ hour)².
   2. P(none in ½ hour) = √0.2 ≈ 0.447.
   3. P(at least one in ½ hour) ≈ 0.553.

   Check. More than half of 0.8, because two halves can both contain hawks.

   Trap. Answering 0.4. Probabilities don't halve with time; the rate does.

Source: Glassdoor search: quantitative trader (2026) · Try it in the app →

Quantitative Trader · Interview · Bayes · reported by a candidate

You pick drawer X or Y with equal probability. 40% of the socks in X are red, 70% in Y. You draw two socks and both are red.

We assume socks are drawn with replacement (or the drawers are large), which the report doesn't say.

1. What is the probability you picked drawer X?
   Answer and worked solution

   Answer: 0.24615

   Key idea. Bayes with equal priors: compare the likelihood of two reds from each drawer.

   1. P(RR | X) = 0.4² = 0.16; P(RR | Y) = 0.7² = 0.49.
   2. P(X | RR) = 0.16/(0.16 + 0.49) = 16/65 ≈ 0.246.

   Check. Two reds point towards the redder drawer, Y.

Source: Glassdoor search: quantitative trader · Try it in the app →

Assistant Trader Intern · Interview · Game theory · reported by a candidate

Two players take turns adding a whole number from 1 to 10 to a running total that starts at 0. Whoever makes the total exactly 50 wins (you can't go over).

1. You go first. What number should you say first?
   Answer and worked solution

   Answer: 6

   Key idea. Work backwards from 50 to find the totals you want to say.

   1. If you say 39, every reply (40–49) lets you reach 50.
   2. The same holds 11 below each winning total: 50, 39, 28, 17, 6.
   3. Say 6 first, then always bring the total to the next one in the list.

   Check. 11 is (largest move + 1): whatever they add, you add the rest.

Interviewers may push further: What if the target were 55? (Then you'd want to go second.)

Source: Glassdoor search: trader assistant intern · Try it in the app →

Assistant Trader Intern · Interview · Combinatorics · reported by a candidate

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

1. What is the probability their ages increase going round in one direction or the other?
   Answer and worked solution

   Answer: 1/12 ≈ 0.083333

   Key idea. Fix the youngest person's seat.

   1. The other four sit 4! = 24 ways.
   2. 2 are increasing (one each way).
   3. P = 2/24 = 1/12.

Source: Glassdoor search: trader assistant intern · Try it in the app →

Trader · Interview · Behavioural · reported by a candidate

Behavioural questions candidates report:

• Why do you want to trade, and why SIG?

Source: Glassdoor · Try it in the app →