IMC interview questions

Trader · Online assessment · Random walks · reported by a candidate

Player A has $1 and player B has $2. They play rounds for $1 each; A wins any round with probability 2/3. They play until one player is broke.

1. What is the probability that A wins everything?
   Answer and worked solution

   Answer: 4/7 ≈ 0.57143

   Key idea. Set up equations for A's chance of winning from each amount of money.

   1. Let P₁, P₂ be A's chances of winning everything from $1 and $2.
   2. From $1: A must win the round, P₁ = (2/3)P₂.
   3. From $2: win now, or drop to $1: P₂ = 2/3 + (1/3)P₁.
   4. Substitute: P₁ = (2/3)(2/3 + P₁/3) = 4/9 + 2P₁/9, so (7/9)P₁ = 4/9.
   5. P₁ = 4/7 ≈ 0.571.

   Check. Gambler's ruin formula with r = q/p = 1/2, start 1, target 3: (1 − r)/(1 − r³) = (1/2)/(7/8) = 4/7.

Interviewers may push further: Expected number of rounds?

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Trader · Interview · Expected value · reported by a candidate

Two dice are rolled.

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

   Answer: 161/36 ≈ 4.4722

   Key idea. For a maximum, P(max ≤ m) is easy: every die must be ≤ m.

   1. P(max ≤ m) = (m/6)².
   2. P(max = m) = (m² − (m − 1)²)/36 = (2m − 1)/36.
   3. E = Σ m(2m − 1)/36 = (1 + 6 + 15 + 28 + 45 + 66)/36 = 161/36.
   4. = ≈ 4.47.

   Check. It must be above 3.5 (the max beats a single die) and below 6.

Interviewers may push further: And the lower? (By symmetry, 7 − 4.47 = 2.53.)

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

You roll a die repeatedly.

1. What is the expected number of rolls until you see an odd number?
   Answer and worked solution

   Answer: 2

   Key idea. Waiting for an event with fixed probability p each try takes 1/p tries on average.

   1. Each roll is odd with probability 1/2.
   2. E = 1/(1/2) = 2.

   Check. Or: E = 1 + (1/2)E, so E = 2.

Interviewers may push further: Expected rolls until you have seen all three odd faces? (Each new one takes 6/3, then 6/2, then 6/1 rolls on average: 2 + 3 + 6 = 11.)

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Graduate quant researcher · Interview · Conditional probability · reported by a candidate

A and B take turns flipping a fair coin, A first. Whoever flips the first head wins.

1. (a) What is the probability that B wins?
   Answer and worked solution

   Answer: 1/3 ≈ 0.33333

   Key idea. If A's first flip is tails, B is in exactly A's starting position.

   1. P(B) = ½ × P(A).
   2. P(A) + P(B) = 1, so P(A) = 2/3.
   3. P(B) = 1/3.

   Check. Going first is an advantage, so B < ½.

2. (b) Given that B won, what is the probability B won on their first flip?
   Answer and worked solution

   Answer: 0.75

   Key idea. Conditional probability: P(B wins on the first flip | B wins).

   1. B wins on their first flip if A flips T then B flips H: ¼.
   2. Divide by P(B wins) = 1/3.
   3. (1/4)/(1/3) = 3/4.

Interviewers may push further: What if the coin lands heads with probability p?

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Quantitative analyst · Interview · Random walks · reported by a candidate

You start at 0. Each fair coin flip moves you up 1 (heads) or down 1 (tails).

1. (a) What is the probability you reach −100 before +50?
   Answer and worked solution

   Answer: 1/3 ≈ 0.33333

   Key idea. A fair walk is a martingale: its expected position at the stopping time is still 0.

   1. Let P = P(hit −100 first).
   2. −100P + 50(1 − P) = 0.
   3. 150P = 50, so P = 1/3.

   Check. The nearer barrier (+50) is twice as likely to be hit first.

2. (b) Follow-up (ours): how many flips does it take on average?
   Answer and worked solution

   Answer: 5000

   Key idea. X² − n is also a martingale for a fair ±1 walk, so E[time] = E[X²] at the end.

   1. E[X²] = (1/3)(100²) + (2/3)(50²) = 10,000/3 + 5,000/3 = 5,000.
   2. E[time] = 5,000.

   Check. In general it's a × b for barriers at distances a and b.

Interviewers may push further: What if the coin is 51% heads?

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Graduate Trader · Interview · Probability · reported by a candidate

Traders A, B and C each make an independent profit uniform between €0 and €100k.

1. What is the probability that A makes more than B, and B more than C?
   Answer and worked solution

   Answer: 1/6 ≈ 0.16667

   Key idea. Symmetry: all orderings of three continuous independent values are equally likely.

   1. There are 3! = 6 orderings.
   2. P(A > B > C) = 1/6.

   Check. The distribution and the €100k are distractions.

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Trader · Interview · Options · reported by a candidate

Sketch the profit at expiry of (a) a long call and (b) a short put. What is the most you can lose, and the most you can make, in each? Why would anyone sell a put, given the limited upside?

Interviewers may push further: Which of the two positions is long volatility? How would you hedge the short put?

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Trader · Interview · Estimation · reported by a candidate

Estimate how long it would take to swim from Amsterdam to New York.

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Trader · Interview · Conditional probability · reported by a candidate

A family has two children, and you learn that one of them is a boy.

How you learn about the boy changes the answer. Ask which situation the interviewer means.

1. (a) If you were told "at least one of them is a boy", what is the probability both are boys?
   Answer and worked solution

   Answer: 1/3 ≈ 0.33333

   Key idea. List the equally likely families and condition on what you're told.

   1. BB, BG, GB, GG.
   2. "At least one boy" removes GG.
   3. 1 of the 3 remaining is BB: 1/3.

   Trap. Answering ½ without asking how the information arose.

2. (b) If instead you meet one of the children at random and it's a boy?
   Answer and worked solution

   Answer: 0.5

   Key idea. Now the information is about a specific child.

   1. In BB you meet a boy for sure; in BG and GB only half the time.
   2. P(BB | met a boy) = (¼ × 1)/(¼ × 1 + ¼ × ½ + ¼ × ½) = ½.

   Check. The other child is independent of the one you met.

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Trader · Interview · Combinatorics · reported by a candidate

Three balls are placed in three different random squares of a 4 × 4 grid.

We assume the three balls go in three different squares, chosen at random.

1. What is the probability that at least two share a row or a column?
   Answer and worked solution

   Answer: 29/35 ≈ 0.82857

   Key idea. Complement: no two balls share a row or column.

   1. Ball 1: anywhere.
   2. Ball 2 must avoid one row and one column: 9 of 15 remaining squares.
   3. Ball 3 must avoid two rows and two columns: 4 of 14.
   4. P(none shared) = 9/15 × 4/14 = 6/35, so P = 29/35 ≈ 0.83.

   Check. Three balls on a small grid usually line up somewhere, so a high answer makes sense.

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Trader Intern · Interview · Probability · reported by a candidate

Three people are chosen at random; each is equally likely to have been born on any day of the week.

Translated from a French report. Also reported at DRW (2010).

1. What is the probability that at least two were born on the same weekday?
   Answer and worked solution

   Answer: 19/49 ≈ 0.38776

   Key idea. Complement: all three born on different days.

   1. P(all different) = 1 × 6/7 × 5/7 = 30/49.
   2. P(at least two share) = 19/49 ≈ 0.39.

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Trader · Interview · Behavioural · reported by a candidate

Behavioural questions candidates report:

• What is your appetite for risk, and how do you manage it?
• Talk me through your personality-test results. Do you agree with them?
• Why IMC, and why quantitative trading?
• Name two qualities that would make you a good trader.

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

Alex and Blake each flip a fair coin until their first head.

1. Given that Alex needed fewer flips than Blake, what is Alex's expected number of flips?
   Answer and worked solution

   Answer: 4/3 ≈ 1.3333

   Key idea. Write P(A = k and B > k) and normalise.

   1. P(A = k) = (½)^k and P(B > k) = (½)^k, so P(A = k, B > k) = (¼)^k.
   2. P(A < B) = Σ (¼)^k = 1/3.
   3. E[A; A < B] = Σ k(¼)^k = (¼)/(¾)² = 4/9.
   4. E[A | A < B] = (4/9)/(1/3) = 4/3.

   Check. Check the shape: P(A = k | A < B) = 3(¼)^k = ¾(¼)^k−1, a geometric distribution with success probability ¾, whose mean is 4/3.

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

A bowl holds 100 strands of spaghetti. You repeatedly pick two free ends at random and tie them together, until no free ends remain.

Also reported by a D. E. Shaw quantitative analyst candidate (as noodles).

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

   Answer: 3.2843

   Key idea. Tie one end at a time and use linearity.

   1. With n strands left there are 2n free ends. Pick any end and tie it to a random other end.
   2. It closes a loop only if it's tied to its own other end: probability 1/(2n − 1).
   3. E[loops] = Σ_n=1¹⁰⁰ 1/(2n − 1) = 1 + 1/3 + 1/5 + … + 1/199.
   4. ≈ 3.28.

   Check. Grows like ½ ln n: very slowly.

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