Hedge Funds puzzles, solved step by step

Puzzles: 100
Traced to a firm: 38
Topics: 14
Hard: 30

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031You roll two fair dice and are paid the higher of the two faces, in rupees. What is the expected payout?Expected value and dice gamesWarm upWolverine TradingChicago · 2025
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Your quick estimate:
Show the worked solution
161/36, about Rs 4.47. The higher face equals k in 2k minus 1 of the 36 equally likely rolls: 1, 3, 5, 7, 9 and 11 rolls for k from 1 to 6. Multiply each value by its count, add to 161, and divide by 36. The lower face averages 91/36, about 2.53, and the two add to 7, the average total of two dice, which is a quick check.
How many of the 36 rolls give each maximum?
Think of two runners and a prize for the faster time: the winning time is better than a typical single runner's because you always keep the better of two. The higher face is at most k in k x k of the 36 rolls, so it equals exactly k in k squared minus (k minus 1) squared, which is 2k minus 1 rolls. That gives 1 roll with a maximum of 1, 3 with a maximum of 2, and on up to 11 with a maximum of 6. On the grid those cells form L shapes that grow as you move towards the corner.
Of the 36 equally likely rolls, the higher face is 1 in just one roll and 6 in eleven rolls, so the expected maximum is 161/36, about 4.47, well above the 3.5 of a single die.
The relationship
$E[\max] = \sum_{k=1}^{6} k \cdot \frac{2k-1}{36} = \frac{161}{36} \approx 4.47$
k the value of the higher face
2k - 1 the number of rolls, out of 36, whose higher face is exactly k
What it says in wordsWeight each possible maximum by how many of the 36 rolls produce it.
How do you check 4.47 in ten seconds?
Use the pair. The higher face plus the lower face always equals the total of the two dice, so their averages must add to 7. The lower face is at least k in (7 minus k) squared rolls, which gives an average of 91/36, about 2.53. 4.47 plus 2.53 is 7.00. A second method that lands exactly is what makes an interviewer stop checking your arithmetic and move on to the follow-up.
The follow-up is usually a game. If you could pay to roll one die or to roll two and keep the higher, the second is worth about Rs 0.97 more. That gap, the value of a free second look, is the same idea as an option: the right to choose after seeing the outcome is worth paying for.
Where candidates lose it
The common loss is answering 3.5 plus something vague, or 5, from instinct. Both skip the count of how often each maximum occurs, which is the whole question.
The second is listing all 36 rolls one by one under time pressure. Say the 2k minus 1 rule, give 161 over 36, and use the lower face check to show the number is right.
What the interviewer asks next
What is the expected higher face with three dice?
What would you pay to roll two dice and keep the higher, if you could reroll both once?
What is the expected value of the lower face, and why do the two add to 7?
Asked at Wolverine Trading, Equity Hedge, Chicago, 2025 (Wall Street Oasis): Typical dice questions that you can find in most probability textbooks
033Make a two-way market on the sum of three fair dice. Then one die is revealed to be a 6. Where do you move your market, and should it get wider or narrower?Market making and trading gamesCoreCitadelLondon · 2026
Try it first
After the 6 is shown, what happens to your market?
Show the worked solution
Move the mid from 10.5 to 13 and tighten the market by about a fifth. Each die averages 3.5, so three dice average 10.5. Once one die shows 6, the sum is 6 plus two unknown dice averaging 7, which is 13. The variance falls from 3 x 35/12 to 2 x 35/12, so the standard deviation drops from 2.96 to 2.42. If the first market was 9.5 at 11.5, the new one is about 12.2 at 13.8.
Where do you put the first market, and how wide?
Start from the fair value and then decide the width from how uncertain the outcome is. The mid is the expected sum, 3 x 3.5 = 10.5, and the width should scale with the standard deviation of the sum, because that is how far the answer typically lands from the mid. One die has variance 35/12, so three independent dice have 35/4 = 8.75, a standard deviation of 2.96. A market of 9.5 bid, 11.5 offered is a reasonable opening: tight enough to trade, with room for your edge.
What does revealing one die change?
A weather forecast for tomorrow is more precise than one for next week, because fewer things can still change. Once one die is known, it contributes a certain 6 and no uncertainty, so the mid rises by 2.5 and only two dice of variance remain. The mid becomes 6 + 7 = 13. The variance becomes 35/6, a standard deviation of 2.42, down from 2.96. Scale the width by the same ratio, about 0.82, and a 2.0 wide market becomes about 1.6 wide: 12.2 at 13.8.
Before the reveal the sum is centred on 10.5 with a standard deviation of 2.96; after one die shows 6 it is centred on 13 with a standard deviation of 2.42, so the market moves up by 2.5 and tightens from 2.0 wide to about 1.6.
Say what would make you widen instead. If the person revealing the die can choose which die to show, or picks the moment, the reveal itself carries information and you should be more careful, not less. A 6 chosen as the highest of three tells you the other two are 6 or lower, and they no longer average 7. Interviewers like it when you ask who chose what to reveal before you requote.
Where candidates lose it
The common loss is widening after the 6 because it feels like a shock. A shock that is fully known removes uncertainty. The mid jumps, but the range of outcomes shrinks.
The second loss is moving the mid by the full 6, or to 16.5 as if all dice were sixes. Only the revealed die is known; the other two still average 3.5 each.
What the interviewer asks next
A second die is revealed as a 1. Where is your market now?
The revealer chose to show the highest of the three dice. Where do you quote?
Someone lifts your 13.8 offer straight away. What do you do next?
Asked at Citadel, Quantitative Research, London, 2026 (Wall Street Oasis): 3rd I got rejected it was different brainteasers and trading game
038X and Y are independent random variables with the same variance. What is the correlation between X and X + Y?Statistics and estimationWarm upSquarepoint CapitalMontreal · 2026
Try it first
Pick one:
Show the worked solution
1 over root 2, about 0.71. The covariance of X with X + Y is Var(X) plus Cov(X, Y), which is sigma squared plus zero. The standard deviation of X + Y is root 2 times sigma because the variances add. So the correlation is sigma squared over (sigma x root 2 sigma), which is 1/root 2. X explains half the variance of the sum, and the correlation is the square root of that half.
What is the fastest way to set it up?
A two-member team's score is the sum of both players' scores. If the players are equally good and play independently, knowing one player's score tells you something about the team total, but only half the story. Split the covariance: Cov(X, X + Y) = Cov(X, X) + Cov(X, Y) = sigma squared + 0. The variance of the sum is sigma squared + sigma squared = 2 sigma squared, because independent variances add. Correlation is covariance over the product of standard deviations: sigma squared over (sigma x root 2 sigma) = 1/root 2.
Drawn as arrows, independent X and Y sit at right angles and their sum lies at 45 degrees to X, so the correlation is cos 45, about 0.707; equivalently, X supplies half of the variance of X + Y, and the correlation is the square root of one half.
Why is the answer not 0.5?
Because 0.5 is the R squaredThe share of one variable variance explained by another; for a simple regression it is the correlation squared., not the correlation. X explains exactly half of the variance of X + Y, and correlation is the square root of the share of variance explained, so it is root 0.5, about 0.707. The geometric picture makes it stick: treat independent variables as arrows at right angles, and correlation as the cosine of the angle between arrows. X + Y sits at 45 degrees to X, and cos 45 is 0.707.
Give the general version to show you own it. If Y has variance k times X's, the correlation is 1/root(1 + k): the more noise you add, the lower it falls. That is the logic behind a noisy signal: a forecast that is half signal and half independent noise, by variance, correlates about 0.71 with the signal, not 0.5.
Where candidates lose it
The common loss is answering 0.5 because X is half of the sum. That is the share of variance, and correlation is its square root.
The other loss is saying zero because X and Y are independent. The sum contains X, so it cannot be independent of X. Split the covariance in one line and the answer falls out.
What the interviewer asks next
What is the correlation between X + Y and X - Y?
Y has four times the variance of X. What is corr(X, X + Y) now?
What is the correlation between the sum of the first 10 and the sum of the first 20 of a series of independent returns?
Asked at Squarepoint Capital, Desk Quant Analyst Interview, Montreal, 2026 (Wall Street Oasis): There were also 3-4 basic math/stats questions about mean, covariance, correlation, etc.
039Depreciation rises by Rs 10 and the tax rate is 25%. Walk the change through net income, the cash flow statement and the balance sheet.Valuation, accounting and macro riddlesWarm upMillennium ManagementNew York · 2024
Try it first
What happens to cash?
Show the worked solution
Net income falls Rs 7.5, cash rises Rs 2.5 and the balance sheet shrinks by Rs 7.5 on both sides. Pre-tax profit falls 10, tax falls 2.5, so net income falls 7.5. The cash flow statement starts at -7.5 and adds back the non-cash 10: cash up 2.5. On the balance sheet, cash is up 2.5 and fixed assets are down 10, so assets fall 7.5, matched by retained earnings down 7.5.
Why does cash go up when an expense goes up?
Imagine your employer lets you deduct the wear on your car from taxable income. No money leaves your pocket for the wear itself, but your tax bill falls. Depreciation is an expense that costs no cash but reduces tax, so the only cash effect is the tax saved: 25% of Rs 10, Rs 2.5. That is the {term('depreciation tax shield', 'The tax saved because depreciation is deductible even though it uses no cash; equal to depreciation times the tax rate.')}, and it is the one number the question is testing.
A Rs 10 rise in depreciation at a 25% tax rate cuts net income by Rs 7.5, raises cash by Rs 2.5 through the tax saved, and lowers fixed assets by Rs 10, so total assets and retained earnings both fall by Rs 7.5 and the balance sheet balances.
What order do you walk it in so nothing gets lost?
Income statement first, then cash flow, then balance sheet, one line each. Net income is the bridge: it closes the income statement, opens the cash flow statement, and lands in retained earnings on the balance sheet. Income statement: depreciation +10, pre-tax -10, tax -2.5, net income -7.5. Cash flow: -7.5 plus 10 added back, cash +2.5. Balance sheet: cash +2.5, fixed assets -10, so assets -7.5; retained earnings -7.5, so the two sides move together.
Add one sentence on why a hedge fund analyst cares. Two companies with identical operations can report different earnings because of depreciation choices, while their cash generation differs only by the tax effect. That is one reason investors look at cash flow alongside earnings before trusting a P/E.
Where candidates lose it
The common loss is saying cash is unchanged because depreciation is non-cash. That forgets the tax: depreciation is deductible, so the tax bill falls and cash rises by Rs 2.5.
The second loss is saying cash falls 7.5 by reading net income as cash. Walk the add-back out loud and check that assets and equity both fall by 7.5 before you stop.
What the interviewer asks next
Now the depreciation rise comes from a Rs 10 write-down of an asset that is not tax deductible. What changes?
What if the company is loss-making and pays no tax this year?
Walk a Rs 10 rise in inventory, bought with cash, through the three statements.
Asked at Millennium Management, Investment Research, New York, 2024 (Wall Street Oasis): Nothing as much, technical questions were super basic like $10 depreciation
040A bag holds four stones, each black or white. Before you look, every count of black stones from 0 to 4 is equally likely. You draw two stones without replacement and both are black. What is the chance the next stone is black, and at what price would you bet on it?Conditional probability and BayesHardCitadelNew York · 2025
Try it first
Chance the third stone is black:
Show the worked solution
3/4, so fair odds are 3 to 1 on black. Two black draws rule out bags with 0 or 1 black. The ways to draw two blacks in order are 2, 6 and 12 for bags with 2, 3 and 4 black, so those bags now carry 10%, 30% and 60%. The next stone is black with chance 0, 1/2 and 1 in them, which averages to 3/4. A contract paying 100 if black is worth 75.
How do the two black draws change your view of the bag?
If a friend pulls two red sweets from a jar you have never seen, you start to suspect it is mostly red. Each possible bag is reweighted by how likely it was to produce what you saw: equal priors times the chance of two blacks. Drawing two blacks in order has 0 ways from bags with 0 or 1 black, 2 x 1 = 2 ways with 2 black, 3 x 2 = 6 with 3 black and 4 x 3 = 12 with 4 black. Out of 20 in total, that is 10%, 30% and 60%: the all-black bag is now the favourite.
Before the draw each bag is 20% likely; after two black stones the bags with 2, 3 and 4 black carry 10%, 30% and 60%, and since the next stone is black with probability 0, one half and 1 in those bags, the chance it is black is 3/4.
How do you turn the posterior into a price?
Average the chance of black over the bags you still believe in. In the 2-black bag both remaining stones are white; in the 3-black bag one of two is black; in the 4-black bag both are, so the answer is 0.1 x 0 + 0.3 x 1/2 + 0.6 x 1 = 3/4. A contract paying 100 if the next stone is black is worth 75. You would buy it below 75 and sell it above; as a market maker you might quote 70 at 80. Offered even money on black, your expected profit per rupee staked is 0.75 minus 0.25, which is 50 paise.
The relationship
$P(B_3 \mid B_1 B_2) = \sum_{k} P(k \mid B_1 B_2)\,P(B_3 \mid k, B_1 B_2) = \tfrac{2}{20}\cdot 0 + \tfrac{6}{20}\cdot\tfrac{1}{2} + \tfrac{12}{20}\cdot 1 = \tfrac{3}{4}$
k the number of black stones in the bag
B1 B2 the event that the first two draws are black
2, 6, 12 the ordered ways to draw two blacks from bags with 2, 3 and 4 black
What it says in wordsThe chance of another black is the average of each bag's chance, weighted by how much the evidence now favours that bag.
Check it with Laplace rule of successionWith a uniform prior, after s successes in n trials, the chance the next trial succeeds is (s + 1) / (n + 2).: after 2 blacks in 2 draws the next is black with chance (2 + 1)/(2 + 2) = 3/4, and the rule holds exactly for this finite bag. Two routes to 3/4 is what separates a solid answer from a lucky one. The limitation: everything rests on the flat prior. If you had reason to think mixed bags were more common, 3/4 would fall.
Where candidates lose it
The common loss is saying 1/2 because the remaining stones are unknown, which throws away the information in the two draws. The question is about updating, and the interviewer wants to see the reweighting.
The second loss is weighting the surviving bags equally, a third each, which gives 1/2. Each bag must be weighted by how likely it made two blacks: 2, 6 and 12. Then price it: a probability without a bet is half the answer at a trading firm.
What the interviewer asks next
The third stone is black too. What is the chance the fourth is black?
You quote 70 at 80 on a contract paying 100 if black and someone who has seen the bag lifts your offer. What now?
How does the answer change if the prior is that each stone is black with probability one half, independently?
Asked at Citadel, Quantitative Trading, New York, 2025 (Wall Street Oasis): Extended bayes derivative question about four stones in a bag (black and white stones).
041Two orders arrive one after the other. The first arrives after a wait that is exponential with a mean of one minute; the second arrives after a further, independent exponential wait with the same mean. What is the probability that both have arrived within one minute?Continuous probability and distributionsHardCitadelChicago · 2025
Try it first
Your estimate:
Show the worked solution
1 minus 2/e, about 26.4%. The total wait is the sum of two independent exponential waits. Convolving the two densities gives t e^-t, a gamma shape that starts at zero because two steps cannot both be instant. Its area from 0 to 1 is 1 minus e^-1 (1 + 1), which is 1 minus 2/e. A second route: it is the chance that a Poisson process with rate 1 produces at least two arrivals in one minute.
Why is this not the chance of one wait, squared?
Squaring would be right if both orders were racing from the same start line. Here they queue: the second clock only starts when the first order lands. It is like two buses where you must take the first to reach the stop for the second. The event is that the sum of the two waits is under one minute, which is stricter than each wait being under one minute. The chance one wait is under a minute is 1 minus 1/e, about 63%; squaring gives 40.0%, which answers a different question.
How do you get the density of the sum?
Add up every way to split the total t between the two waits. The density of a sum of independent waits is the convolutionThe density of a sum of two independent variables, found by integrating one density against the other shifted over every possible split of the total. of their densities, and for two exponentials it is t e^-t. Every split of t into s and t minus s has density e^-s times e^-(t minus s), which is e^-t whatever s is, and there is a length t of possible splits. Integrate t e^-t from 0 to 1 by parts and you get 1 minus 2/e.
The total of two independent one-minute exponential waits has density t e^-t, which starts at zero and peaks at one minute, so only 26.4% of its area, 1 minus 2/e, lies below one minute.
The relationship
$P(X_1 + X_2 \le 1) = \int_0^1 t e^{-t}\,dt = 1 - 2e^{-1} \approx 0.264$
X1, X2 the two independent exponential waits, each with mean one minute
t e^-t the density of their sum, from convolving the two exponential densities
What it says in wordsThe chance the total wait is under a minute is the area under the gamma density up to one minute.
Check it with counting. Exponential waits are the gaps of a Poisson process, so both orders arrive within a minute exactly when the process makes at least two arrivals in that minute. With one arrival expected per minute, the chance of zero is e^-1 and of exactly one is e^-1, so at least two is 1 minus 2/e, the same number. Say both routes and the interviewer will usually skip ahead.
Where candidates lose it
The common loss is squaring the single-wait probability, which answers the question of two independent orders racing in parallel. Read the setup again: one after the other means the waits add.
The second loss is freezing on the convolution integral. If the integral will not come, switch to the Poisson count: at least two arrivals in one minute. Candidates who know one route and not the other are the ones interviewers push hardest.
What the interviewer asks next
What is the probability that three orders in sequence all arrive within two minutes?
Given that both orders arrived within one minute, what is the expected arrival time of the first?
The two waits have means of one and two minutes. What is the density of their sum?
Asked at Citadel, Quant Research Interview, Chicago, 2025 (Wall Street Oasis): if i knew this was about convolutions, i would have answered better.
042A bowl holds 100 noodles. You repeatedly pick two free ends at random and tie them together, until no free ends remain. What is the expected number of loops?Counting and combinatoricsHardD.E. ShawNew York · 2026
Try it first
Roughly how many loops?
Show the worked solution
About 3.28 loops. With k strands in the bowl there are 2k free ends. Pick any end; the other end you pick is one of the remaining 2k minus 1, and exactly one of those belongs to the same strand, so this tie closes a loop with chance 1/(2k minus 1). Either way the number of strands falls by one. Adding 1/199 + 1/197 + ... + 1/3 + 1 gives about 3.284.
What does one tie do, whatever happens?
Start with the bookkeeping, because it makes the rest easy. Every tie reduces the number of loose strands by exactly one: either it closes a strand into a loop, or it joins two strands into one longer strand. So there are always exactly 100 ties, and with k strands left there are 2k free ends. The question becomes how many of those 100 ties happen to close a loop.
What is the chance a given tie closes a loop?
Think of a room of dancers holding hands in lines: grab one free hand, then pick a second free hand at random, and a circle forms only if the second hand is at the other end of the same line. With 2k free ends, the second end is one of 2k minus 1, and exactly one of them is the other end of the strand you picked, so the chance is 1/(2k minus 1). Give each tie an indicator that is 1 if it closes a loop; by linearity of expectation the expected number of loops is the sum of the chances, from 1/199 for the first tie up to 1 for the last.
The first tie closes a loop with chance 1 in 199 and the chances stay tiny until the last few ties, 1/5, 1/3 and 1, so the expected number of loops from 100 noodles is only 3.28, about a third of it from the last three ties.
The relationship
$E[\text{loops}] = \sum_{k=1}^{100} \frac{1}{2k-1} = 1 + \tfrac13 + \tfrac15 + \dots + \tfrac{1}{199} \approx 3.28$
k the number of strands in the bowl before a tie
1/(2k-1) the chance that tie joins the two ends of one strand
What it says in wordsAdd each tie's chance of closing a loop to get the expected number of loops.
For a sense check without a calculator: the sum of odd reciprocals up to 1/(2n minus 1) is about half of ln n plus ln 2 plus half of Euler's constant, which for n = 100 gives 3.28. The number of loops grows only like the logarithm of the number of noodles: a million noodles would give only about 7.9 loops.
Where candidates lose it
The first loss is trying to track the lengths of the strands, which quickly becomes impossible. The length of a strand never matters; only the count of strands does.
The second loss is getting 1/(2k minus 1) right but summing it wrong, for example as 100 x 1/199. Write the sum out from the last tie backwards, 1 + 1/3 + 1/5, and the size of the answer becomes obvious.
What the interviewer asks next
What is the variance of the number of loops?
What is the probability that you end with exactly one big loop?
How does the answer grow with the number of noodles, roughly?
Asked at D.E. Shaw, Research, New York, 2026 (Wall Street Oasis): What is the expected number of loops from tying 100 noodles' ends together randomly
043You may roll a fair die up to three times. After each roll you either stop and are paid the face in rupees, or throw that roll away and roll again; if you reach the third roll you must take it. What is the best stopping rule, and what is the game worth?Expected value and dice gamesHardSquarepoint CapitalLondon · 2026
Try it first
On the first roll you get a 4. What do you do?
Show the worked solution
Stop on the first roll only with a 5 or 6, on the second with a 4 or more, and the game is worth 14/3, about Rs 4.67. Work backwards. The last roll is worth 3.5. With two rolls left, keep 4, 5 or 6 and reroll otherwise: worth (4 + 5 + 6)/6 + 3/6 x 3.5 = 4.25. With three rolls left, keep only what beats 4.25, a 5 or 6: worth (5 + 6)/6 + 4/6 x 4.25 = 14/3.
Why start from the last roll?
Deciding whether to take a job offer is easier if you know what your fallback is worth. Each keep-or-reroll decision compares the roll in hand with the value of the rolls still to come, so you need the value of the future first, and the only stage with no future is the last one. On the last roll you must take whatever comes, which is worth 3.5 on average. That single number lets you solve the stage before it, and so on backwards. This is {term('backward induction', 'Solving a sequence of decisions from the last one to the first, using the value of each later stage to make the earlier decision.')}, the core of dynamic programming.
With one roll left the game is worth 3.5, so with two left you keep 4 or more and the game is worth 4.25; with three left you keep only 5 or 6, and the whole game is worth 14/3, about 4.67.
How do the thresholds come out?
With two rolls left, a roll of 4, 5 or 6 beats the 3.5 you expect from rerolling, and 1, 2 or 3 does not. So the two-roll game is worth the average of the kept faces times their chance, plus the chance of rerolling times 3.5: 15/6 + 1.75 = 4.25. On the first roll the fallback is now 4.25, so a 4 is no longer good enough: only 5 or 6 is kept. That gives 11/6 plus 4/6 x 4.25, which is 1.833 plus 2.833, or 14/3.
The relationship
$V_n = \frac{1}{6}\sum_{f=1}^{6} \max\left(f,\, V_{n-1}\right), \quad V_1 = 3.5,\; V_2 = 4.25,\; V_3 = \tfrac{14}{3}$
V_n the value of the game with n rolls left
f the face you just rolled
max(f, V_{n-1}) keep the roll or throw it away, whichever is worth more
What it says in wordsEach stage is worth the average, over the six faces, of the better of keeping the face or playing on.
Add the pattern. More rolls always raise the value, but by less each time: 3.5, 4.25, 4.67, then about 4.94 with four rolls. An extra option is always worth something and never worth more than what it can still improve. That is the same logic as valuing a trade you can exit early: the right to wait is priced by what the future is worth, not by the average outcome.
Where candidates lose it
The common loss is using 3.5 as the threshold at every stage, which keeps a 4 on the first roll. The fallback on the first roll is the two-roll game, 4.25, not a single roll.
The second loss is solving forwards and getting lost. Say you will start from the last roll, compute 3.5, 4.25 and 14/3 in that order, and the thresholds fall out.
What the interviewer asks next
What is the game worth with four rolls, and what is the first-roll threshold?
You must pay Rs 1 for every reroll. How do the thresholds change?
You are paid the square of the face instead. What is the optimal rule?
Asked at Squarepoint Capital, Quantitative Research, London, 2026 (Wall Street Oasis): Dynamic programming questions with focus on probability at the end.
045You make a market on the number of heads in 10 fair coin flips: 4.5 bid, 5.5 offered. A counterparty who has already seen the first three flips lifts your offer. What does the trade tell you, and where do you requote?Market making and trading gamesHardCitadelNew York · 2025
Try it first
Given that they bought at 5.5, the fair value is about
Show the worked solution
The lift says they saw at least two heads, so the fair value is now at least 5.75, not 5; requote around 5.75 bid, 6.5 offered. The other seven flips are worth 3.5 heads, so the buyer's value is heads seen plus 3.5. Paying 5.5 only makes sense with two heads (5.5) or three (6.5). Those are 3 to 1 likely, giving 5.75; a buyer who needs a strict edge saw three heads, worth 6.5.
What is the trader's view before they trade?
A friend offers to buy your raffle ticket after the first few numbers are drawn. The offer itself is the warning. The informed trader values the contract at heads already seen plus 3.5, the expected heads in the seven unseen flips, so their value is 3.5, 4.5, 5.5 or 6.5 with chances 1, 3, 3 and 1 in 8. Against your market of 4.5 bid and 5.5 offered, they buy only if their value is at least 5.5, and sell to you at 4.5 only if it is 4.5 or less. The flat 5 you quoted around is right only for someone who has seen nothing.
The informed trader's value is 3.5, 4.5, 5.5 or 6.5 depending on how many heads they saw, so a lift at 5.5 means two or three heads, and weighting those 3 to 1 puts the fair value given the trade at 5.75, above your 5.5 offer.
How do you turn the trade into a new fair value?
Condition on the fact that they traded. Only the two-head and three-head worlds produce a buy at 5.5, and they are 3/8 and 1/8 likely, so given a lift the value is (3 x 5.5 + 1 x 6.5) / 4 = 5.75. If you assume they would not bother trading at zero edge, only the three-head world is left and the value is 6.50. Either way you sold too cheaply: this is adverse selectionThe tendency of a market maker to trade most with the people who know more, so the trades that happen are the ones that lose money for the market maker., and it is the cost every market maker prices into the spread.
Now requote. Your bid should not be below what you now believe the floor is, and your offer should sit where even the best informed buyer has no edge. Something like 5.75 bid, 6.5 offered does both: a buyer who saw three heads is indifferent at 6.5, and you are no longer selling below value. Cut your size too, because you know someone is trading with more information than you, and say you would ask whether they could see the flips before quoting again.
Where candidates lose it
The common loss is staying at 5 because the coin is fair. The coin is fair; the counterparty is not uninformed. The trade itself carries information and you must update on it.
The other loss is overreacting and moving to 8 or 9, as if the trader knew all ten flips. They saw three. Condition on what could have made them trade, weight those worlds, and move by exactly that much.
What the interviewer asks next
The same trader then hits your new bid. What do you conclude?
How wide should your first market have been if you knew one counterparty could see three flips?
What if the trader had seen the first three flips but traded a small size and then a large size?
Asked at Citadel, Quantitative Trading, New York, 2025 (Wall Street Oasis): Superday was more market-making but requires very sold foundation in math and statistics.
048A strategy's true annual Sharpe ratio is 1.0. How many years of monthly returns do you need before its average return shows a t-statistic of 2? What if the true Sharpe is 0.5?Statistics and estimationCoreViking Global InvestorsNew York · 2014
Try it first
Years needed for a Sharpe of 0.5:
Show the worked solution
About 4 years for a Sharpe of 1.0 and about 16 years for a Sharpe of 0.5. The t-statistic of a mean return is the mean over its standard error, which works out to the annual Sharpe ratio times the square root of the number of years, whatever the data frequency. Setting Sharpe x root(years) = 2 gives years = (2 / Sharpe) squared: 4 for 1.0, 16 for 0.5 and just 1 for 2.0.
Why does the t-statistic grow with the square root of time?
A coin that lands heads 55% of the time looks fair after 20 tosses; you need hundreds before the bias shows through the noise. The average return grows in proportion to time, but the noise around it grows only with the square root of time, so the signal-to-noise ratio, the t-statistic, grows with root time. With monthly data, the t-statistic is the monthly Sharpe times root(12 x years), and the monthly Sharpe is the annual Sharpe divided by root 12, so the twelves cancel: t = annual Sharpe x root(years).
Because the t-statistic equals the Sharpe ratio times the square root of years, a Sharpe of 2.0 clears t = 2 after 1 year, a Sharpe of 1.0 after 4 years and a Sharpe of 0.5 only after 16 years.
Why does monthly data not shorten the wait?
More frequent data gives more observations but each is noisier relative to its mean. Sampling the same years more often does not add information about the mean return; only more years do. This is why a {term('t-statistic', 'An estimate divided by its standard error; a value around 2 is the usual threshold for saying an effect is unlikely to be pure noise.')} on the average return depends on the span of the data, not the number of rows. Frequency helps you estimate volatility, not the mean.
The relationship
$t \approx SR_{\text{annual}}\sqrt{Y} \;\Rightarrow\; Y = \left(\frac{2}{SR}\right)^2: \quad SR = 1 \to 4, \quad SR = 0.5 \to 16$
SR the true annual Sharpe ratio
Y years of data
2 the target t-statistic
What it says in wordsThe years needed to prove a strategy grow with the inverse square of its Sharpe ratio.
Say the practical point. Most real strategies have Sharpe ratios well below 1, so their track records are too short to separate skill from luck with any confidence. An allocator looking at a three-year record with a Sharpe of 0.8 sees a t-statistic of about 1.4. The limitation of the rule: it assumes returns are independent and stable over the whole sample, and fat tails or regime changes make the real uncertainty larger.
Where candidates lose it
The common loss is thinking monthly data gives twelve times the evidence, which leads to answers like four months. The twelve cancels, because the monthly Sharpe is smaller by root 12.
The second loss is saying a Sharpe of 0.5 needs twice as long as 1.0. The dependence is on the square: half the Sharpe, four times the data.
What the interviewer asks next
How many years for a Sharpe of 0.3?
You test 20 strategies and pick the best one with t = 2.2. How much do you trust it?
Would daily data change the answer for estimating the Sharpe ratio itself rather than the mean?
Asked at Viking Global Investors, Quantitative Research, New York, 2014 (Wall Street Oasis): how to reject a hypothesis test, what's your structure of your code, what's the sample size

k	the value of the higher face
2k - 1	the number of rolls, out of 36, whose higher face is exactly k

k	the number of black stones in the bag
B1 B2	the event that the first two draws are black
2, 6, 12	the ordered ways to draw two blacks from bags with 2, 3 and 4 black

X1, X2	the two independent exponential waits, each with mean one minute
t e^-t	the density of their sum, from convolving the two exponential densities

k	the number of strands in the bowl before a tie
1/(2k-1)	the chance that tie joins the two ends of one strand

Try it first

How many of the 36 rolls give each maximum?

How do you check 4.47 in ten seconds?

Where candidates lose it

What the interviewer asks next

Try it first

Where do you put the first market, and how wide?

What does revealing one die change?

Where candidates lose it

What the interviewer asks next

Try it first

What is the fastest way to set it up?

Why is the answer not 0.5?

Where candidates lose it

What the interviewer asks next

Try it first

Why does cash go up when an expense goes up?

What order do you walk it in so nothing gets lost?

Where candidates lose it

What the interviewer asks next

Try it first

How do the two black draws change your view of the bag?

How do you turn the posterior into a price?

Where candidates lose it

What the interviewer asks next

Try it first

Why is this not the chance of one wait, squared?

How do you get the density of the sum?

Where candidates lose it

What the interviewer asks next

Try it first

What does one tie do, whatever happens?

What is the chance a given tie closes a loop?

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What the interviewer asks next

Try it first

Why start from the last roll?

How do the thresholds come out?

Where candidates lose it

What the interviewer asks next

Try it first

What is the trader's view before they trade?

How do you turn the trade into a new fair value?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the t-statistic grow with the square root of time?

Why does monthly data not shorten the wait?

Where candidates lose it

What the interviewer asks next