Jane Street interview questions

Trader · Phone · Expected value · reported by a candidate

You're paid in dollars the number showing on a fair die.

1. (a) What is the expected value of a single roll?
   Answer and worked solution

   Answer: 3.5

   Key idea. Expected value is the average payout, weighting each outcome by its probability.

   1. Each face 1–6 has probability 1/6 and pays its value.
   2. E = (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 = 3.5.

   Check. It's the midpoint of 1 and 6, as it should be for a symmetric distribution.

2. (b) After seeing your roll you may reroll once, but must keep the second result. What is the game worth under optimal play?
   Answer and worked solution

   Answer: 4.25

   Key idea. Decide using what the alternative is worth: a reroll is worth 3.5, because you must keep whatever comes up.

   1. Keep any roll worth more than 3.5 (4, 5, 6); reroll 1, 2, 3.
   2. Kept rolls contribute (4 + 5 + 6)/6 = 2.5.
   3. You reroll with probability 1/2, and a reroll is worth 3.5: (1/2) × 3.5 = 1.75.
   4. Total: 2.5 + 1.75 = 4.25.

   Check. The option to reroll can only help, so the value must be above 3.5; and it can't reach 6. 4.25 is in range.

   Trap. Rerolling a 4 because "it's below average": 4 beats the 3.5 a reroll is worth.

3. (c) Now you may reroll up to twice. What is it worth?
   Answer and worked solution

   Answer: 14/3 ≈ 4.6667

   Key idea. Work backwards. The value of the last decision tells you the threshold for the one before it.

   1. With one reroll left the game is worth 4.25 (part b). So on the first roll, a reroll is worth 4.25.
   2. Keep only rolls above 4.25: 5 and 6. Reroll 1–4.
   3. Value = (5 + 6)/6 + (4/6) × 4.25 = 11/6 + 17/6 = 28/6.
   4. = 14/3 ≈ 4.67.

   Check. Each extra reroll adds less: 3.5 → 4.25 → 4.67. The value creeps towards 6 but never reaches it.

   Trap. Using the threshold 3.5 on the first roll. Your threshold rises as you have more chances left.

Interviewers may push further: What if each reroll costs $0.25? How would you price it with n rerolls? What does the value converge to?

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Trader · Interview · Expected value · reported by a candidate

You roll a 30-sided die; your opponent rolls a 20-sided die. The higher roll wins, and ties go to your opponent. The loser pays the winner the value of the winning roll in dollars.

A figure of 10.2 circulates online for this one. Computing directly from the rules as stated gives 8.15, so check it with the simulation, and in an interview confirm the rules before you start.

1. What is the expected value of this game for you?
   Answer and worked solution

   Answer: 8.15

   Key idea. Split the game into what you win and what you pay, and condition on your roll.

   1. There are 30 × 20 = 600 equally likely (x, y) pairs.
   2. You win when x > y. For x ≤ 20 that's x − 1 of their faces (ties go to them); for x > 20 it's all 20. Each win pays you x.
   3. E[winnings] = Σ x·min(x − 1, 20)/600 = 7760/600 ≈ 12.93.
   4. You lose when y ≥ x. For a given y that happens for x = 1…y, i.e. y times, and you pay y: E[payments] = Σ y²/600 over y = 1…20 = 2870/600 ≈ 4.78.
   5. EV = 12.93 − 4.78 ≈ 8.15.

   Check. Your die is much bigger, so you should win most games and the EV should be clearly positive.

   Trap. A figure of 10.2 circulates online. It comes from misreading the rules (e.g. ties). Always confirm the rules before computing.

Interviewers may push further: What is your probability of winning a round? How much would you pay to swap your die for a 40-sided one?

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Trader · Interview · Optimal stopping · reported by a candidate

A 100-sided die shows 1 to 100. After each roll you can take that many dollars, or pay $1 to roll again, as many times as you like.

1. (a) What is the game worth under optimal play?
   Answer and worked solution

   Answer: 87.357

   Key idea. Use a threshold strategy and optimise the threshold: stop on the top n faces.

   1. Each roll stops with probability n/100, so the number of rolls is geometric with mean 100/n; you pay $1 for each of the 100/n − 1 rerolls.
   2. When you stop, your roll is uniform on the top n faces: average (100 + (101 − n))/2 = (201 − n)/2.
   3. V(n) = (201 − n)/2 − (100/n − 1) = 101.5 − n/2 − 100/n.
   4. Calculus: dV/dn = −1/2 + 100/n² = 0 at n = √200 ≈ 14.1. Check n = 14: 101.5 − 7 − 7.14 ≈ 87.36.

   Check. Cross-check with the stopping rule: reroll when your roll is below V − 1 ≈ 86.4. That means keeping 87–100, the top 14 faces. Consistent.

2. (b) What is the lowest roll you should accept?
   Answer and worked solution

   Answer: 87

   Key idea. Keep a roll exactly when it beats the value of continuing, which costs $1.

   1. Continuing is worth V − 1 ≈ 86.36 (pay $1, then the game restarts).
   2. Accept a roll x only if x ≥ 86.36, so the lowest acceptable roll is 87.

   Check. This matches n = 14 from part (a): the top 14 faces are 87–100.

Interviewers may push further: How does the answer scale if rerolls cost $c?

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Trader · Interview · Probability · reported by a candidate

To roll the higher number against an opponent, would you rather use one 20-sided die or the sum of three 6-sided dice (your opponent gets the other)?

Reported as a choice question. The numeric parts below are how you'd justify the answer.

1. (a) What is the probability the 20-sided die beats 3d6 (strictly higher)?
   Answer and worked solution

   Answer: 0.475

   Key idea. Both totals have the same mean (10.5) and are symmetric about it, so the difference is symmetric about 0.

   1. By symmetry, P(D > S) = P(S > D).
   2. P(tie): whatever 3d6 shows (3 to 18), the d20 matches it with probability 1/20.
   3. So P(D > S) = (1 − 1/20)/2 = 0.475.

   Check. It's an even contest apart from ties, so each side should win just under half the time.

   Trap. Assuming the d20 is better "because it can roll higher". Range doesn't matter for who's higher on average; symmetry does.

2. (b) Follow-up (ours): what is the variance of the d20, and of 3d6?
   Answer and worked solution

   Answer: 33.25

   Key idea. For a fair n-sided die, Var = (n² − 1)/12. Variances of independent dice add.

   1. d20: (400 − 1)/12 = 33.25.
   2. 3d6: 3 × (36 − 1)/12 = 3 × 35/12 = 8.75.

   Check. The d20 is far more spread out, even though the head-to-head is even.

   Trap. That's why the choice matters when you need a high target (say 17+): the wide d20 gets there far more often.

Interviewers may push further: What if ties are re-rolled? Which would you pick if you needed at least 17?

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Trader · Interview · Estimation · reported by a candidate

Two fair dice are rolled. You want to guess a number as close as possible to their sum.

1. (a) What number should you guess?
   Answer and worked solution

   Answer: 7

   Key idea. To minimise expected absolute error, guess the median.

   1. Two dice: totals 2–12, symmetric around 7.
   2. Mean, median and mode all equal 7.
   3. Guess 7.

   Check. 7 is also the single most likely total (6 ways out of 36).

2. (b) Follow-up (ours): what is your expected absolute error?
   Answer and worked solution

   Answer: 35/18 ≈ 1.9444

   Key idea. Average the distance |S − 7| over the distribution.

   1. Distances 1, 2, 3, 4, 5 each occur on both sides of 7, with 5, 4, 3, 2, 1 ways per side.
   2. E|S − 7| = 2 × (1·5 + 2·4 + 3·3 + 4·2 + 5·1)/36 = 2 × 35/36 = 70/36.
   3. = 35/18 ≈ 1.94.

   Check. The standard deviation of two dice is about 2.4; the mean absolute deviation is always a bit smaller.

Interviewers may push further: What if you are penalised by squared error instead? (Still 7: the mean minimises squared error.)

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Trader · Interview · Expected value · reported by a candidate

Flip four fair coins and get $1 for each head. After seeing them, you may re-flip one coin.

1. What is the most you would pay to play?
   Answer and worked solution

   Answer: 2.4688

   Key idea. Value the coins without the re-flip, then add what the re-flip option is worth.

   1. Four coins at $0.50 each: $2.
   2. Re-flip a tail whenever you have one: it becomes a head with probability ½, gaining $0.50 on average.
   3. You have at least one tail with probability 1 − 1/16 = 15/16.
   4. Value = 2 + (15/16)(0.5) = 2 + 15/32 = $2.47 (79/32).

   Check. The option adds a bit less than $0.50, since sometimes there's nothing to re-flip.

Interviewers may push further: What if you can re-flip every tail once? What if the re-flip costs $0.40?

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

A bag holds three coins: one with heads on both sides, one with tails on both sides, and one normal coin. You pick one at random, toss it, and see heads.

1. What is the probability you picked the double-headed coin?
   Answer and worked solution

   Answer: 2/3 ≈ 0.66667

   Key idea. Count the heads faces you could be looking at, not the coins.

   1. Heads faces: two on the double-headed coin, one on the normal coin, none on the double-tailed coin.
   2. You're equally likely to be looking at any of those 3 faces.
   3. 2 of them belong to the double-headed coin: P = 2/3.

   Check. Bayes gives the same: (1/3 × 1)/(1/3 × 1 + 1/3 × ½) = 2/3.

   Trap. Answering ½ ("it's one of two coins that can show heads"). The double-headed coin shows heads twice as often.

Interviewers may push further: You toss the same coin again and get heads again. Now what?

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Assistant Trader Intern · Interview · Expected value · reported by a candidate

A box is worth an amount V, uniform between $0 and $1000, that only the seller knows. You name a bid b. The seller sells if V is below your bid, and you pay b. You can then resell the box for 1.5V.

Another report gives the same game with values 0–100 and no distribution; uniform is the standard assumption.

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

   Answer: 0

   Key idea. Adverse selection: your bid is only accepted when the box is worth less than you paid.

   1. Bid b. The seller sells only if V < b, so given a trade, V is uniform on [0, b] with mean b/2.
   2. You pay b and resell for 1.5 × b/2 = 0.75b: a loss of 0.25b on every trade.
   3. Expected profit = (b/1000) × (−0.25b) < 0 for any b > 0.
   4. Best bid: 0. Don't bid.

   Check. The 1.5× resale looks generous, but you only ever buy the cheap half.

   Trap. Using the unconditional mean of V (500) instead of its mean given that the seller accepts.

2. (b) If you bid $1000, so you always buy, what is your expected profit?
   Answer and worked solution

   Answer: -250

   Key idea. Bidding the maximum removes the selection: you always buy.

   1. V is uniform on [0, 1000], mean 500.
   2. Resale = 1.5 × 500 = 750; you paid 1000.
   3. Expected profit = −$250.

   Check. Breakeven needs a resale multiplier of 2 at any bid: 2 × b/2 = b.

Interviewers may push further: What resale multiplier would make bidding worthwhile? (Above 2.) How does this relate to quoting a market to someone who knows more than you?

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Quantitative Trader · Interview · Combinatorics · reported by a candidate

Put N points on a circle and draw every chord between them, placed so that no three chords meet at one point inside. The chords cut the disc into regions.

1. (a) How many regions with 5 points?
   Answer and worked solution

   Answer: 16

   Key idea. Each chord adds one region, plus one for every chord it crosses.

   1. Start with 1 region (the disc).
   2. 5 points: C(5, 2) = 10 chords and C(5, 4) = 5 crossings (each set of 4 points gives one crossing).
   3. Regions = 1 + 10 + 5 = 16.

   Check. Draw it: 1, 2, 4, 8, 16 regions for 1 to 5 points.

2. (b) How many with 6 points?
   Answer and worked solution

   Answer: 31

   Key idea. Same count: regions = 1 + C(N, 2) + C(N, 4).

   1. 6 points: C(6, 2) = 15 chords, C(6, 4) = 15 crossings.
   2. Regions = 1 + 15 + 15 = 31.

   Check. For N = 5 the formula gives 16, matching the drawing.

   Trap. Guessing 32 from the doubling pattern. The pattern breaks at 6 points; that's the point of the question.

Interviewers may push further: Why does every crossing correspond to exactly 4 points?

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Quantitative Trader · Interview · Probability · reported by a candidate

You flip a fair coin 100 times.

1. What is the probability of an even number of heads?
   Answer and worked solution

   Answer: 0.5

   Key idea. Condition on everything except the last flip.

   1. Whatever the first 99 flips give, exactly one outcome of the 100th makes the total even.
   2. That outcome has probability ½.
   3. P(even) = ½.

   Check. It works for any number of fair flips n ≥ 1.

   Trap. Summing C(100, k) over even k. It works (2⁹⁹/2¹⁰⁰), but it's slow.

Interviewers may push further: What if the coin lands heads with probability p? (½ + ½(1 − 2p)¹⁰⁰.)

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Quantitative Trader · Interview · Probability · reported by a candidate

I flip a fair coin 100 times and you must guess the whole sequence. Before guessing you may ask me one yes/no question, which I answer truthfully.

1. With the best question your chance is 1 in 2^k. What is k?
   Answer and worked solution

   Answer: 99

   Key idea. One yes/no answer carries one bit of information: it can at most halve the possibilities.

   1. There are 2¹⁰⁰ equally likely sequences.
   2. Any question splits them into a "yes" group and a "no" group. You learn which, then guess one sequence in it.
   3. P(success) = Σ P(group) × 1/|group| = 2 × (1/2¹⁰⁰) = 1/2⁹⁹, for any split.
   4. So k = 99.

   Check. E.g. ask "was the first flip heads?", then guess the remaining 99 flips.

Interviewers may push further: What if you could ask two questions?

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Quantitative Trader · Interview · Game theory · reported by a candidate

You and a friend take turns flipping a fair coin. Whoever throws the first head wins $30 from the other.

The report doesn't say what happens after two tails; we assume you keep alternating until someone throws a head.

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

   Answer: 2/3 ≈ 0.66667

   Key idea. Use the fact that after a tail the roles swap.

   1. Let p = P(first player wins).
   2. They win immediately with ½; otherwise the opponent is now "first": p = ½ + ½(1 − p).
   3. So 1.5p = 1, p = 2/3.

   Check. Going first is clearly an advantage, so p > ½.

2. (b) What is going first worth to you, in expected dollars?
   Answer and worked solution

   Answer: 10

   Key idea. Expected dollars = P(win) × 30 − P(lose) × 30.

   1. 30 × 2/3 − 30 × 1/3 = $10.

   Check. Going second is worth −$10, so swapping from second to first is worth $20.

Interviewers may push further: How much would you pay to swap from second to first?

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

Four fair coins are tossed and you're told at least two show tails.

1. What is the probability that exactly three show tails?
   Answer and worked solution

   Answer: 4/11 ≈ 0.36364

   Key idea. Conditional probability: restrict to outcomes with at least two tails.

   1. Tails counts 0–4 occur 1, 4, 6, 4, 1 times out of 16.
   2. At least two tails: 6 + 4 + 1 = 11. Exactly three: 4.
   3. P = 4/11 ≈ 0.364.

   Trap. Using 4/16 (forgetting the condition).

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

A fair coin is tossed until either HHT or HTT appears. A wins if HHT comes first, B if HTT does.

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

   Answer: 2/3 ≈ 0.66667

   Key idea. Nothing matters until the first H. From there, look one toss ahead.

   1. After "H", if the next toss is H you have HH, and the run of H's must end with a T, completing HHT. So HH means A wins.
   2. If the next toss is T you have HT: another T completes HTT (B wins); an H puts you back at "just seen H".
   3. Let p = P(A wins from "H"): p = ½ + ½ × ½ × p.
   4. So (3/4)p = ½, p = 2/3.

   Check. This is Penney's game: same-length patterns don't win equally often.

   Trap. Assuming ½ by symmetry: the patterns overlap with themselves differently.

Interviewers may push further: Expected number of tosses until the game ends?

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

A drawer holds endless socks in two colours, each equally likely. You pull socks out one at a time.

1. What is the expected number of socks drawn until you have a matching pair?
   Answer and worked solution

   Answer: 2.5

   Key idea. Pigeonhole: with two colours, you never need more than three socks.

   1. The first two match with probability ½: 2 draws.
   2. Otherwise you hold one of each colour and the third must match: 3 draws.
   3. E = 2 × ½ + 3 × ½ = 2.5.

Interviewers may push further: Three colours?

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Trader · Interview · Probability · reported by a candidate

An asset goes up 20% or down 20% each day, each with probability 1/2, independently.

1. (a) After one up day and one down day (either order), what fraction of its starting price is it worth?
   Answer and worked solution

   Answer: 0.96

   Key idea. Percentage moves multiply.

   1. Up then down: 1.2 × 0.8 = 0.96 of the start. The order doesn't matter.

   Check. A 20% gain followed by a 20% loss always loses 4%: volatility drag.

   Trap. Assuming they cancel because +20% − 20% = 0.

2. (b) For how many values of n ≥ 1 is it possible to be back at exactly the starting price after n days?
   Answer and worked solution

   Answer: 0

   Key idea. Write the price as a fraction and look at its prime factors.

   1. After u ups and d downs: (6/5)^u(4/5)^d = 6^u4^d/5^{u + d}.
   2. The numerator has no factor 5, so this equals 1 only if u + d = 0.
   3. It can never return exactly: 0 values of n.

   Check. Even the expected price returns exactly (E[move] = 1 each day), but no actual path does.

Interviewers may push further: What is the expected price after n days? (Exactly the start: E[move] = 1.) What is the median? (It falls.)

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Quant Trader · Interview · Logic · reported by a candidate

Fifteen couples are at a party. Nobody shakes their own partner's hand, and no pair shakes more than once. You ask 29 of the people how many hands they shook, and all 29 answers are different.

1. How many hands did the 30th person shake?
   Answer and worked solution

   Answer: 14

   Key idea. The 29 different answers must be exactly 0, 1, …, 28. Pair up the extremes.

   1. Each person shakes at most 28 hands (not themselves, not their partner), so the 29 answers are 0–28.
   2. The person with 28 shook everyone except their partner, so the only person who could have shaken 0 is their partner: they're a couple.
   3. Remove them. Everyone left loses one handshake, and the same argument pairs 27 with 1, 26 with 2, …, 15 with 13.
   4. 14 is left unpaired, so its partner, the 30th person, also shook 14.

   Check. If you are the host's partner, this is the classic Lewis/Halmos puzzle.

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Quant Trader · Interview · Logic · reported by a candidate

Five melons are weighed in every possible pair. The ten totals are 16, 17, 18, 19, 20, 21, 22, 24, 25 and 26.

1. (a) How heavy is the heaviest melon?
   Answer and worked solution

   Answer: 15

   Key idea. Each melon appears in 4 of the 10 pair sums.

   1. Total of the sums = 208 = 4 × (total weight), so the melons weigh 52 in all.
   2. Sort a ≤ b ≤ c ≤ d ≤ e. Smallest sum a + b = 16, largest d + e = 26, so c = 52 − 42 = 10.
   3. Second smallest a + c = 17 gives a = 7, b = 9. Second largest c + e = 25 gives e = 15, d = 11.
   4. Heaviest: 15.

   Check. Check all ten sums of 7, 9, 10, 11, 15: 16, 17, 18, 19, 20, 21, 22, 24, 25, 26. They match.

2. (b) And the lightest?
   Answer and worked solution

   Answer: 7

   Key idea. From the same working.

   1. a = 17 − c = 17 − 10 = 7.

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Quant Trader · Interview · Statistics · reported by a candidate

Team A has a better shooting percentage than team B in each of two matches. Must A have the better percentage over both matches combined?

Interviewers may push further: Where would this matter in trading? (Comparing strategies or traders across regimes with different trade counts.)

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Assistant / Junior Trader · Interview · Probability · reported by a candidate

You roll a fair 12-sided die repeatedly, keeping a running total, and stop as soon as the total is odd.

1. Which final total is most likely?
   Answer and worked solution

   Answer: 11

   Key idea. The total stays even until the first odd roll, which ends the game. A final total s is reached from an even total within 11 below it.

   1. Let f(t) = P(the running total ever equals the even number t). f(0) = 1; every other f(t) is below 1.
   2. P(final total = s) = (1/12) × Σ f(t) over even t from max(0, s − 11) up to s − 1.
   3. For s ≤ 11 the sum includes f(0) = 1 plus all the f(t) below s, so it grows with s. For s ≥ 13 it loses f(0) and gains a smaller term.
   4. So the most likely final total is 11.

   Check. Simulate it: the game ends on 11 about 12.4% of the time, and 1, 3, … 9 less often.

   Trap. Answering 1 (the most likely single first roll). Every first roll is equally likely; it's the paths through even totals that add up.

2. (b) Follow-up (ours): what is the probability the game ends on 11?
   Answer and worked solution

   Answer: 0.12435

   Key idea. Add up the paths into 11: P(11) = (1/12)(f(0) + f(2) + … + f(10)).

   1. f(0) = 1; f(2) = 1/12; f(4) = 13/144; in general f(2k) = (13/12)^{k − 1}/12.
   2. Sum of f(0) … f(10) = 1 + (1/12)(1 + 13/12 + … + (13/12)⁴) = (13/12)⁵.
   3. P(11) = (13/12)⁵/12 = 13⁵/12⁶ ≈ 0.124.

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

An n-sided die pays its face in dollars. You may reroll as often as you like for $1/n per reroll; each reroll replaces your result.

Reported for an n-sided die. We fix n = 10 for the first part.

1. (a) For n = 10 (rerolls cost $0.10), what is the game worth?
   Answer and worked solution

   Answer: 9.1

   Key idea. Threshold strategy: stop on the top k faces, and optimise k.

   1. You roll n/k times on average and pay 1/n for each of the n/k − 1 rerolls.
   2. Your stopping roll averages the top k faces: (2n − k + 1)/2.
   3. V(k) = (2n − k + 1)/2 − (1/n)(n/k − 1). For k = 1 and k = 2 this is n − 1 + 1/n; k = 3 is lower.
   4. n = 10: V = 9 + 0.1 = 9.1.

   Check. Rerolls are so cheap you only stop on the very top faces.

2. (b) For n = 100?
   Answer and worked solution

   Answer: 99.01

   Key idea. Same formula, n = 100.

   1. V = 100 − 1 + 1/100 = 99.01.

   Check. The value approaches n − 1 as n grows.

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

At each of three stops, three quarters of the passengers on a bus get off and then 7 get on. The number of passengers is always a whole number.

1. What is the smallest possible number of passengers at the start?
   Answer and worked solution

   Answer: 52

   Key idea. Each stop maps x to x/4 + 7, so x must be divisible by 4 every time.

   1. Write x = 4a. After stop 1: a + 7, which needs a ≡ 1 (mod 4).
   2. Write a = 4b + 1. Then a + 7 = 4b + 8, which becomes b + 2 + 7 = b + 9 after stop 2: needs b ≡ 3 (mod 4).
   3. Smallest b = 3, so a = 13 and x = 52.

   Check. Check: 52 → 13 + 7 = 20 → 5 + 7 = 12 → 3 + 7 = 10. All whole numbers.

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Junior Quantitative Trader · Interview · Decision making · reported by a candidate

A 6-sided and a 10-sided die are rolled. You guess the total beforehand; if you're right you're paid that total in dollars, otherwise nothing.

1. (a) What should you guess?
   Answer and worked solution

   Answer: 11

   Key idea. Maximise expected payout s × P(s), not P(s) alone.

   1. d6 + d10: totals 7 through 11 can each be made 6 ways out of 60 (a flat top).
   2. Among those, the payout is largest at the top: 11 × 6/60 = 1.1.
   3. 12 can be made 5 ways: 12 × 5/60 = 1.0. Lower.
   4. Guess 11.

   Check. Higher totals pay more, so you want the highest total that's still on the plateau.

   Trap. Guessing the most likely total. Several totals tie for most likely, and they pay differently.

2. (b) What is that guess worth?
   Answer and worked solution

   Answer: 1.1

   Key idea. Expected payout = guess × P(guess).

   1. 11 × 6/60 = $1.10.

Interviewers may push further: Which total is most likely? (Any of 7–11; the most likely total is not the best guess.)

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

You start with one amoeba. Each minute every amoeba independently dies, stays as one, splits into two or splits into three, each with probability 1/4.

The report only says "amoeba extinction". We use the standard version: each minute an amoeba dies, stays one, splits into two or into three, each with probability 1/4.

1. What is the probability the population eventually dies out?
   Answer and worked solution

   Answer: 0.41421

   Key idea. Condition on the first minute: the population dies out only if every amoeba's line dies out.

   1. Let p = P(extinction from one amoeba). p = ¼(1 + p + p² + p³).
   2. Rearrange: p³ + p² − 3p + 1 = 0.
   3. p = 1 is a root: (p − 1)(p² + 2p − 1) = 0.
   4. The roots of p² + 2p − 1 are −1 ± √2. The answer is the smallest root in [0, 1]: √2 − 1 ≈ 0.414.

   Check. Mean offspring = (0 + 1 + 2 + 3)/4 = 1.5 > 1, so survival is possible and p < 1.

   Trap. Stopping at p = 1 because it solves the equation.

Interviewers may push further: Why is p = 1 also a root, and why is it not the answer here? (Mean offspring 1.5 > 1.)

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Assistant / Junior Trader · Interview · Mental maths · reported by a candidate

Mental maths, asked out loud.

1. What is 37²?
   Answer and worked solution

   Answer: 1369

   Key idea. Square a nearby round number, then correct.

   1. 37 = 40 − 3.
   2. (40 − 3)² = 1600 − 240 + 9 = 1369.

   Check. Ends in 9 (7² = 49), and 37² is a bit under 40² = 1600.

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Trade Desk Operations Engineer · Interview · Probability · reported by a candidate

Two people each arrive at a uniformly random time within the same hour, and each waits 20 minutes for the other.

1. What is the probability they meet?
   Answer and worked solution

   Answer: 5/9 ≈ 0.55556

   Key idea. Plot the two arrival times as a point in the unit square.

   1. They meet when |x − y| ≤ 1/3.
   2. They miss in two corner triangles with legs 2/3: total area (2/3)² = 4/9.
   3. P(meet) = 1 − 4/9 = 5/9.

   Check. With 15-minute waits it would be 1 − (3/4)² = 7/16.

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