Akuna Capital interview questions

Quant trader · Interview · Probability · reported by a candidate

You have a coin that lands heads 70% of the time. How can you use it to make a fair 50/50 decision?

1. Using the standard method (flip twice; HT means A, TH means B, otherwise flip twice again), how many flips do you need on average?
   Answer and worked solution

   Answer: 100/21 ≈ 4.7619

   Key idea. HT and TH are equally likely (0.7 × 0.3 each), so they make a fair coin.

   1. Each pair of flips decides with probability 0.21 + 0.21 = 0.42.
   2. The number of pairs is geometric: mean 1/0.42.
   3. Expected flips = 2/0.42 ≈ 4.76.

   Check. With a fair coin it would be 2/0.5 = 4: a bias costs extra flips.

Interviewers may push further: Can you do better on average by reusing the HH and TT outcomes?

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

Three fair coins are flipped and you're told that one of them is heads.

Reported twice with the same ambiguity: is it "at least one is heads", or "this particular coin is heads"? Ask, then answer both.

1. (a) If that means "at least one is heads", what is the probability all three are heads?
   Answer and worked solution

   Answer: 1/7 ≈ 0.14286

   Key idea. List the 8 outcomes and keep those with at least one head.

   1. 7 remain, equally likely.
   2. 1 is HHH: 1/7.

   Trap. Answering 1/4 without asking how you were told.

2. (b) Same reading: probability of exactly two heads?
   Answer and worked solution

   Answer: 3/7 ≈ 0.42857

   Key idea. Same 7 outcomes.

   1. Exactly two heads: HHT, HTH, THH. 3/7.
3. (c) If instead you're shown one particular coin and it's heads, what is the probability all three are heads?
   Answer and worked solution

   Answer: 0.25

   Key idea. If a specific coin is shown, the other two are untouched.

   1. (½)² = 1/4.

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

A game pays the highest of several fair dice. Price it with two dice, then three.

1. (a) Two dice: expected maximum?
   Answer and worked solution

   Answer: 161/36 ≈ 4.4722

   Key idea. P(max ≤ m) = (m/6)², so P(max = m) = (2m − 1)/36.

   1. E = Σ m(2m − 1)/36 = 161/36 ≈ 4.47.
2. (b) Three dice?
   Answer and worked solution

   Answer: 119/24 ≈ 4.9583

   Key idea. Tail-sum formula: E[X] = Σ P(X ≥ m) for m = 1…6.

   1. P(max ≥ m) = 1 − ((m − 1)/6)³.
   2. E = 6 − (0 + 1 + 8 + 27 + 64 + 125)/216 = 6 − 225/216.
   3. = 119/24 ≈ 4.96.

   Check. More dice push the maximum towards 6.

Interviewers may push further: With n dice? (6 − Σ_k=1⁵ (k/6)ⁿ, which tends to 6.)

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Quant Trader Intern · Online test · Sequences · reported by a candidate

What comes next: 12, 1112, 3112, 132112, …?

1. Next term:
   Answer and worked solution

   Answer: 1113100000

   Key idea. Each term describes the previous one aloud.

   1. 12 → "one 1, one 2" → 1112.
   2. 1112 → "three 1s, one 2" → 3112. 3112 → 132112.
   3. 132112 → "one 1, one 3, one 2, two 1s, one 2" → 1113122112.

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Junior Quantitative Trader · Interview · Sequences · reported by a candidate

Continue the sequence M, V, E, M, J, … and explain it.

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

A fair coin is tossed until it lands heads. If that takes k tosses you win $2^k.

Reported by name ("what would you pay for the St Petersburg lottery"); these are the standard rules.

1. (a) If the payout is capped at $2²⁰ (about $1 million, all the bank can pay), what is the game worth?
   Answer and worked solution

   Answer: 21

   Key idea. Write the expectation as a sum over the number of tosses k.

   1. P(first head on toss k) = 2^−k; the prize is 2^k.
   2. For k ≤ 20, each contributes 2^k × 2^−k = $1: $20 in total.
   3. For k > 20 the prize is capped at 2²⁰, with probability 2⁻²⁰: $1 more.
   4. Value = $21.

   Check. Uncapped, the sum is 1 + 1 + 1 + … = ∞: the paradox. A real counterparty can't pay infinite amounts.

2. (b) Follow-up (ours): with the cap at $2³⁰ (about $1 billion)?
   Answer and worked solution

   Answer: 31

   Key idea. Same sum with the cap at 2³⁰.

   1. 30 terms of $1, plus $1 for the tail: $31.

   Check. A thousand times bigger bankroll adds only $10: the "infinite" value lives in absurdly unlikely outcomes.

Interviewers may push further: Without a cap, how would risk aversion (e.g. log utility) price it? (About $4 for log utility of the prize alone.)

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

We roll two dice. If the total is 10, you win $100; otherwise you lose your stake.

1. What is the most you should risk (the stake that makes the bet fair)?
   Answer and worked solution

   Answer: 100/11 ≈ 9.0909

   Key idea. A fair bet balances expected gain and expected loss.

   1. P(total 10) = 3/36 = 1/12.
   2. Fair stake s: (1/12) × 100 = (11/12) × s.
   3. s = 100/11 ≈ $9.09.

   Check. Odds against are 11 to 1, so you risk 1/11 of the prize.

Interviewers may push further: The same interview asked: make a market on the Chicago metro population, then how much you would risk to win $100 if the truth is inside your market.

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