Hedge Funds puzzles, solved step by step

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

Showing 21–30 of 50 · filtered from 100Clear filters

035Three assets all have the same pairwise correlation, rho. What is the lowest value rho can take? What is the answer for n assets?Portfolio and risk mathsCoreMulti-manager platformsQuant and systematic funds
Try it first
Lowest possible common correlation for three assets:
Show the worked solution
Minus one half for three assets, and minus 1/(n minus 1) for n. Give each asset unit variance and add them up. The variance of the sum is 3 plus 6 rho, because there are three variances and six pairwise covariance terms. A variance cannot be negative, so rho is at least -1/2. With n assets the sum's variance is n plus n(n minus 1) rho, which gives rho at least -1/(n minus 1): -0.33 for four, -0.11 for ten.
Why can three assets not all be perfectly opposed?
Three friends cannot all disagree with each other on a yes-or-no question: if Ravi says yes and Meena says no, Arjun agrees with one of them. Perfect opposition is a relationship between two things; with three, two of them must lean the same way. The same limit holds for correlations. A set of numbers in a correlation matrix has to be internally consistent, and equal pairwise correlations become impossible well before -1 once there are three or more assets.
How do you find the exact floor?
Use the one fact that can never fail: a variance is zero or more. Add the three assets with unit variance; the sum has variance 3 from the three diagonal terms plus 6 rho from the six covariance terms, and 3 + 6 rho must be at least zero. That gives rho of at least -1/2. The picture is three arrows 120 degrees apart: each pair has a cosine of -1/2, and the three add to exactly zero, which is the boundary case. For n assets, n + n(n minus 1) rho at least zero gives -1/(n minus 1).
Three unit arrows 120 degrees apart add to zero, the arrangement where every pair has correlation minus one half, and the variance of the sum, 3 + 6 rho, turns negative below that value, which is impossible, so rho cannot fall below minus one half.
The relationship
$\operatorname{Var}\Big(\sum_{i=1}^{n} X_i\Big) = n + n(n-1)\rho \ge 0 \;\Rightarrow\; \rho \ge -\frac{1}{n-1}$
n the number of assets, each with variance 1
rho the common pairwise correlation
n(n-1) the number of ordered pairs, each contributing one covariance of rho
What it says in wordsThe variance of the equal-weighted basket must be non-negative, which caps how negative a shared correlation can be.
Say why a risk desk cares. A correlation matrix that breaks this rule is not a valid risk model: it implies some portfolio has negative variance, and an optimiser will pile into it. This happens in practice when correlations are estimated pair by pair from different data windows or overridden by hand in a stress test. The check is that the matrix is positive semi-definiteA matrix for which every weighted combination of the assets has a variance of zero or more., and the equal-correlation case is the cleanest example of the rule.
Where candidates lose it
The usual loss is answering -1, because that is the floor for any single pair. The interviewer is testing whether you see that the pairs constrain each other.
The second loss is getting -1/2 by intuition but having no proof. Say the variance of the sum in one line. It takes ten seconds, it generalises to n immediately, and it is the answer the follow-ups build on.
What the interviewer asks next
Two assets have correlation 0.9 with a third. What is the lowest possible correlation between the first two?
A stress test sets every pairwise correlation in a 20-asset book to -0.1. Is that a valid matrix?
How would you repair a correlation matrix that is not positive semi-definite?
036A regime model says a bull month is followed by another bull month 90% of the time, and a bear month by another bear month 80% of the time. In the long run, what share of months are bull months?Random walks and Markov chainsCoreQuant and systematic fundsProp and quant trading firms
Try it first
Long-run share of bull months:
Show the worked solution
Two thirds of months are bull months, whatever the starting state. In the long run the number of months switching from bull to bear must equal the number switching back. 10% of bull months switch out and 20% of bear months switch in, so 0.1 x bull = 0.2 x bear, which makes bull twice as common as bear: 2/3 against 1/3. A second route: bull spells last 10 months on average and bear spells 5.
What has to balance in the long run?
Think of a shop with people walking in and out all day. Once the crowd inside stops growing or shrinking, the number walking in each minute must equal the number walking out. In the long run, the flow from bull to bear must equal the flow from bear to bull, because otherwise one state would keep filling up. The flow out of bull is 10% of bull months; the flow out of bear is 20% of bear months. Setting 0.1 x bull equal to 0.2 x bear, with bull plus bear equal to 1, gives bull = 2/3.
Bull months turn bear 10% of the time and bear months turn bull 20% of the time, so the long-run share of bull months is two thirds, and a chain started in either state is within a few points of two thirds after about a year.
How do you check two thirds another way?
Use the length of each spell. A state you leave with probability p each month lasts 1/p months on average, so bull spells last 10 months and bear spells 5. Spells alternate, so over a long stretch the market spends 10 months bull for every 5 bear: 10 out of 15 is two thirds. Two methods that agree is what the interviewer is listening for.
The relationship
$\pi_{\text{bull}} = \frac{0.2}{0.1 + 0.2} = \frac{2}{3} \qquad \text{gap after } t \text{ months} \propto (0.9 + 0.8 - 1)^t = 0.7^t$
pi_bull the long-run share of bull months
0.1, 0.2 the chances of leaving bull and leaving bear each month
0.7 how much of any starting gap survives each month
What it says in wordsThe long-run share of a state is the chance of entering it divided by the total chance of switching, and the start is forgotten at a rate of 0.7 a month.
Answer the part of the question people skip: why the starting state does not matter. The gap between today's odds and two thirds shrinks by a factor of 0.7 every month, so after 12 months only 1.4% of it is left. A model this sticky still forgets its starting point within about a year, which is why regime forecasts beyond a few months mostly return the long-run average.
Where candidates lose it
The first loss is answering 90%, the one-step persistence, as if it were the long-run share. The second is saying it depends on today's state, which is true for next month and false for the long run.
Set up the flow balance in one line, give two thirds, then check it with spell lengths of 10 and 5 months. If you have time, say how fast the start is forgotten: 0.7 a month.
What the interviewer asks next
Today is a bear month. What is the chance that the month after next is a bull month?
How long does the average bear spell last, and what is the chance one lasts more than a year?
How would you estimate the two transition probabilities from 20 years of monthly data, and how wide would the error be?
046A fund has annual volatility of 12% and its benchmark index has 15%. The correlation between them is 0.9. What is the fund's tracking error?Portfolio and risk mathsCoreMulti-manager platformsQuant and systematic funds
Try it first
Your estimate of the tracking error:
Show the worked solution
About 6.7%. Tracking error is the volatility of the fund's return minus the benchmark's. The variance of a difference is the two variances minus twice the covariance: 0.12 squared plus 0.15 squared minus 2 x 0.9 x 0.12 x 0.15, which is 0.0144 + 0.0225 - 0.0324 = 0.0045. The square root is 6.71%. A correlation of 0.9 sounds tight but still leaves a sizeable gap.
What exactly is tracking error measuring?
Two friends walking to the same station take slightly different routes; tracking error is how far apart they typically are, not how fast either walks. Tracking errorThe standard deviation of the difference between a fund return and its benchmark return, usually quoted per year. is the volatility of the return difference, fund minus benchmark, so it depends on both volatilities and on how closely the two move together. The variance of a difference is Var(F) + Var(B) - 2 Cov(F, B), and the covariance is the correlation times the two volatilities.
Drawn as arrows of length 12 and 15 set 25.8 degrees apart, the angle whose cosine is 0.9, the fund and benchmark tips sit 6.7 apart, which is the tracking error; it would be 3.0% only at a correlation of 1 and rises to 13.7% at 0.5.
Why is the answer so much bigger than 3%?
Because the correlation is below 1, the gap is not just the difference in size. The variance of the difference is 144 + 225 - 324 = 45 in squared percent, and the square root of 45 is 6.7%: the 10% of correlation that is missing contributes more than the 3 point difference in volatility. The arrow picture shows it: two arrows 12 and 15 long, set about 26 degrees apart, have tips further apart than 3. That is just the law of cosines.
The relationship
$TE = \sqrt{\sigma_F^2 + \sigma_B^2 - 2\rho\,\sigma_F\sigma_B} = \sqrt{0.0144 + 0.0225 - 0.0324} \approx 6.7\%$
sigma_F, sigma_B the fund and benchmark volatilities, 12% and 15%
rho the correlation between their returns, 0.9
What it says in wordsTracking error is the volatility of the gap between fund and benchmark returns.
Say what it means for an allocator. A fund with 0.9 correlation to its index can still trail or beat it by 6 to 7 points in an ordinary year, so a single year of underperformance tells you very little. The limitation: the calculation assumes the correlation and volatilities are stable, and they tend to shift in stressed markets, which is when tracking error matters most.
Where candidates lose it
The fast wrong answer is 3%, the difference in volatilities, which is only true at a correlation of exactly one. The interviewer chose 0.9 precisely because it sounds close to one.
The second loss is forgetting the factor of 2 on the covariance term, which gives root(207), over 14%. Write the variance of a difference in full before plugging in numbers.
What the interviewer asks next
What correlation would give a tracking error of 5%?
If the fund had a beta of 0.72 to the index, what is its information ratio if it beats the index by 2% a year?
Why might a fund's realised tracking error jump in a sell-off?
047An index rises 10% one day and falls 10% the next. What happens to the index over the two days, and to a fund that delivers exactly twice the index's daily return?Returns, compounding and feesCoreFund of funds and allocatorsMulti-manager platforms
Try it first
Over the two days, the 2x fund is
Show the worked solution
The index ends down 1% and the 2x fund down 4%, not 2%. The index goes 100, 110, 99. The fund goes 100, 120, 96, because it doubles each day's move: up 20%, then down 20% of a bigger number. Two days of index returns of plus and minus r cost the index r squared, 1%, but cost the 2x fund 4r squared, 4%. The drag comes from resetting leverage daily, and it grows with volatility.
Why is the index down at all after up 10% and down 10%?
A salary cut of 10% after a 10% raise leaves you below where you started, because the cut is taken on the bigger salary. An up move of r followed by a down move of r multiplies to (1 + r)(1 - r) = 1 - r squared, so the index loses r squared, here 1%. This is the same volatility drag that makes compound returns lower than average returns. It is small for the index because r squared is small.
Why does the 2x fund lose four times as much?
Because it resets its leverage each day. The fund multiplies to (1 + 2r)(1 - 2r) = 1 - 4r squared, so its drag is four times the index's, not two times. With r = 10% that is 4%. A buyer expecting twice the index return over the period expected 2 x (-1%) = -2% and got -4%. The extra 2 points is the price of re-levering after the gain, which buys more exposure at the top, and de-levering after the loss.
Over two days of plus and minus 10%, the index ends at 99, down 1%, while the fund that doubles each daily move ends at 96, down 4%, twice the naive 2% loss, because leverage is reset every day.
The relationship
$(1+Lr)(1-Lr) = 1 - L^2 r^2 \qquad L = 2,\; r = 0.10:\; 1 - 0.04 = 0.96$
L the daily leverage multiple, 2
r the size of each day's index move, 10%
What it says in wordsA leveraged fund's drag in a choppy market grows with the square of the leverage.
Say where the drag does not apply. In a steady trend, daily resetting helps: two days of +10% take the index to 121 and the 2x fund to 144, more than the naive 142. Daily-reset leveraged funds win in trends and lose in chop, so holding one for months is a bet on the path, not just the direction. That is why allocators treat them as trading tools rather than long-term holdings.
Where candidates lose it
The common loss is saying the 2x fund is down 2%, doubling the index's two-day result. The fund doubles each daily return, and the compounding does the rest.
The second loss is saying the index is flat. Up 10% and down 10% is always a loss; say 1 minus r squared and the two answers come together.
What the interviewer asks next
What is the two-day return of a fund delivering minus twice the daily return?
Over a year with 20% index volatility and no trend, roughly how much does a 2x daily fund lag twice the index return?
Two days of +10%: how does the 2x fund compare with twice the index return?
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
049You short a stock at Rs 100 at 5% of NAV. It rises to Rs 150 while the rest of the book is flat. What is your loss, and what share of NAV is the short position now?Valuation, accounting and macro riddlesCoreLong-short equity fundsGlobal macro funds
Try it first
The short's share of NAV at Rs 150 is about
Show the worked solution
You lose 2.5% of NAV, and the short is now about 7.7% of the book. On NAV of 100, the short is worth 5. A 50% rise makes it a liability of 7.5, a loss of 2.5, so NAV falls to 97.5. The position is now 7.5 over 97.5, which is 7.7%. A losing short grows as a share of the book, so risk rises exactly when the trade is going wrong, the opposite of a losing long.
How do the loss and the new weight work out?
Keep NAV at 100 so every number is a percentage. The short is a liability that rises with the price: 5 at Rs 100 becomes 7.5 at Rs 150, a loss of 2.5, which takes NAV from 100 to 97.5. The weight is the liability over the new NAV: 7.5 / 97.5 = 7.69%. Both parts of the fraction move against you: the top grows and the bottom shrinks.
Why is a losing short more dangerous than a losing long?
A losing long is like a debt that shrinks as the thing you own loses value: a 5% long falling to half is worth 2.5 of NAV 97.5, 2.6%, and it can never lose more than the 5 you put in. A losing short does the reverse: the more it loses, the bigger it gets, and there is no ceiling on the price, so there is no cap on the loss. Left alone, a short that doubles is 10 of NAV 95, 10.5%, and one that triples is 16.7%. This is why long-short funds size shorts smaller and cut them faster than longs.
A 5% long whose price halves shrinks to 2.6% of the book and can lose at most 5% of NAV, while a 5% short whose price rises to 150 grows to 7.7% and keeps growing, with no cap on the loss.
The relationship
$w_{\text{short}} = \frac{0.05 \times 1.5}{1 - 0.05 \times 0.5} = \frac{7.5}{97.5} \approx 7.7\%$
0.05 the starting weight, 5% of NAV
1.5 the price relative, 150 over 100
0.05 x 0.5 the loss as a share of starting NAV, 2.5%
What it says in wordsThe short's new weight is its grown liability over the NAV that the loss has reduced.
Add what a desk does about it. Many funds set a stop or a maximum weight for each short and cover part of it as it rises, so that a losing short is brought back towards its original risk. The discipline exists because the position sizes itself up without anyone deciding to. Short squeezes make this worse, since the price can gap up on little news when many holders cover at once.
Where candidates lose it
The common loss is saying the weight is 7.5%, forgetting that NAV fell. It is 7.5 over 97.5, not over 100. A smaller trap is saying 5%, as if the short's size were fixed at the entry value.
The bigger miss is not drawing the lesson. The interviewer wants to hear that shorts grow when they lose and longs shrink, so short risk compounds against you and needs its own limits.
What the interviewer asks next
At what price does the short reach 10% of NAV?
How much would you buy back to bring the short to 5% of NAV at Rs 150?
The stock pays a dividend while you are short. What happens to your P&L?
051You start with Rs 1,000 and can bet any amount on a coin that wins 60% of the time at even money, as many times as you like. What fraction of your money do you bet each time, and what happens if you bet twice that?Betting and sizingCoreMillennium ManagementAtlanta · 2025
Try it first
Which fraction of your money gives the fastest long-run growth?
Show the worked solution
Bet 20% of your current money each time; bet 40% and the long-run growth is gone. For an even-money bet the Kelly fraction is the win probability minus the loss probability, 0.6 minus 0.4. At 20% the typical path grows about 2.0% a bet. At double that, the average log growth is about -0.2% a bet, so the same edge now earns nothing over time.
Why not bet everything when the odds are in your favour?
Think of a shopkeeper who puts all of today's takings into tomorrow's stock. Six days in ten the stock sells and she doubles her money; on the other four she is back to zero, and one zero ends the shop. Wealth compounds, so a single wipe-out is permanent, and the stake that maximises the average payoff of one round is not the stake that maximises what you end with after many. Betting 100% has the highest expected value per round and a certainty of eventual ruin.
How do you find the best fraction?
Maximise the growth rate, which is the average of the log of what each bet does to your money. Bet a fraction f and a win multiplies your wealth by 1 + f, a loss by 1 - f. The growth rate 0.6 ln(1 + f) + 0.4 ln(1 - f) peaks where f equals the edge, 0.6 - 0.4 = 20%, at about 2.01% a bet. Over 100 bets that turns Rs 1,000 into about Rs 7,490 on the typical path. This rule is the Kelly criterionA sizing rule that picks the bet size maximising the long-run growth rate of wealth, the expected log return per bet..
The relationship
$g(f) = p\ln(1+f) + q\ln(1-f), \qquad f^{*} = p - q = 0.6 - 0.4 = 0.2$
p the chance of winning, 0.6
q the chance of losing, 0.4
f the fraction of current money bet
g(f) the expected log growth per bet
What it says in wordsGrowth per bet is the probability-weighted average of the log gain and the log loss, and it is highest when you bet the edge.
Expected log growth per bet rises to about 2.0% at a 20% stake, falls back to slightly below zero, -0.2%, at a 40% stake and turns sharply negative beyond it, so over-betting a real edge can remove all long-run growth.
What happens at twice the Kelly bet?
Near the peak the growth curve is close to a parabola centred on 20%. Double the Kelly fraction and you sit as far down the far side as not betting at all sits on the near side: growth of about zero, -0.24% a bet. Over 100 bets the typical path ends near Rs 783, below where you started, although every single bet had a positive expected value. The extra swings cost more in compounding than the extra stake earns.
Say the limitation too. Kelly assumes you know the 60% exactly. On a desk the edge is an estimate, and overestimating it pushes you to the right of the peak, where the curve falls fastest. That is why many traders bet half Kelly: 10% here keeps about 75% of the growth with half the swings.
Where candidates lose it
The fast wrong answer is to bet big because the odds favour you, often 60% because that is the chance of winning. A candidate who only computes expected value per round will always pick the largest stake, and the interviewer is waiting to see whether you notice that money compounds.
The second loss is reaching 20% and being unable to say what happens beyond it. Have the shape ready: growth peaks at the edge, is roughly zero at twice the edge, and is negative after that.
What the interviewer asks next
The coin now pays 2 to 1 when you win and still wins 60% of the time. What is the Kelly fraction?
You are only 80% sure the coin is 60/40 rather than fair. How does that change your bet?
What would you pay to play 100 rounds of this game starting with Rs 1,000?
Asked at Millennium Management, Software Engineering Intern Interview, Atlanta, 2025 (Wall Street Oasis): Start with 1000 and bet each round. Kelly criterion would be very useful for this step.
053Two independent numbers are drawn uniformly at random between 0 and 1. What is the expected value of the smaller one, and of the larger one?Continuous probability and distributionsCoreQuant and systematic fundsProp and quant trading firms
Try it first
What is the expected value of the smaller of the two draws?
Show the worked solution
The smaller draw averages 1/3 and the larger averages 2/3. Two random points cut the stretch from 0 to 1 into three pieces, and nothing distinguishes one piece from another, so each has expected length 1/3. The minimum ends the first piece and the maximum ends the second. The two averages add to 1, as they must, because the minimum plus the maximum is just the sum of the two draws.
What is the one-line argument?
Imagine two friends who each arrive at a random minute within the same hour. The wait until the first arrival, the gap between the two arrivals and the time from the second arrival to the end of the hour are three stretches of time. No stretch is special, so each averages a third of the hour, and the first friend arrives, on average, 20 minutes in. The same symmetry puts the minimum of two uniform draws at 1/3 and the maximum at 2/3.
The diagonal splits the square of possible draws into two mirror-image halves; the density of the smaller draw falls in a straight line from 2 at zero to 0 at one with its average at 1/3, and the density of the larger draw mirrors it with its average at 2/3.
How do you check it with a calculation?
Work from the chance that the minimum sits above some level x. Both draws must be above x, which happens with probability (1 - x) squared. For a quantity that cannot be negative, the average equals the area under its chance of exceeding each level, and the area under (1 - x) squared from 0 to 1 is 1/3. The maximum then follows without a second integral: minimum plus maximum equals X plus Y, which averages 1, so the maximum averages 2/3.
The relationship
$E[\min] = \int_0^1 P(\min > x)\,dx = \int_0^1 (1-x)^2\,dx = \tfrac{1}{3}, \qquad E[\max] = 1 - \tfrac{1}{3} = \tfrac{2}{3}$
P(min > x) the chance both draws exceed x, (1 - x) squared
E[min] the expected value of the smaller draw
E[max] the expected value of the larger draw
What it says in wordsIntegrate the chance that the minimum clears each level to get its average, then use min plus max equals the sum of the draws.
Where does the pattern lead next?
With n draws, the points cut the interval into n + 1 pieces, so the smallest averages 1/(n + 1) and the largest n/(n + 1). The ordered values of uniform draws are evenly spaced on average, and that single fact answers a whole family of follow-ups. With ten bids drawn at random across a range, the highest averages 10/11 of the way up. The quick route here is also the one to say first: the order statisticsThe draws sorted from smallest to largest: the minimum is the first order statistic, the maximum the last. of uniform draws split the interval evenly.
Where candidates lose it
The common wrong answer is 1/4, reached by reasoning that the smaller draw lies in the bottom half and so sits at the middle of it. The smaller draw is not confined to the bottom half; it only has to be below the other draw, and it lands above 0.5 a quarter of the time.
The second loss is grinding through the integral in silence and never giving the symmetry argument. Say the three-pieces picture first, then offer the integral as the check.
What the interviewer asks next
What is the expected distance between the two draws?
With three draws, what is the expected value of the middle one?
What is the probability that the smaller draw is below 0.5?
054You flip a fair coin five times. What is the probability of seeing a run of at least three heads in a row somewhere in the five flips?Counting and combinatoricsCoreQuant and systematic fundsProp and quant trading firms
Try it first
What is the chance of at least three heads in a row in five flips?
Show the worked solution
The chance is 8 in 32, exactly 1/4. Sort the qualifying sequences by where their first run of three heads starts. Starting at flip 1: HHH then anything, 4 sequences. Starting at flip 2: flip 1 must be tails, then HHH, then anything, 2 sequences. Starting at flip 3: flip 2 must be tails and flip 1 is free, 2 sequences. Four plus two plus two is eight.
Why does the obvious count go wrong?
Suppose you count guests at a party by asking each room how many people it holds, while some guests stand in doorways. They get counted twice. A run of three can start in three places, but one sequence can contain a run starting at more than one of them, so three places times four fillings, 12, counts HHHHH three times and two other sequences twice. The fix is to give each sequence exactly one home: the position where its first run starts.
The eight qualifying sequences sort into four whose first run starts at flip 1, two at flip 2 and two at flip 3, so the chance is 8 of 32, one quarter; the naive count of 12 counts overlapping runs more than once.
How do you make every sequence count once?
Force the flip just before the run to be tails. If the first run starts at flip 2, flip 1 must be tails; if it starts at flip 3, flip 2 must be tails, and flip 1 can be anything. That gives 4, then 2, then 2. Then check a second way by counting the complement: sequences with no run of three follow a Tribonacci-style rule, each term the sum of the three before, giving 1, 2, 4, 7, 13, 24 for 0 to 5 flips. And 32 minus 24 is 8.
The relationship
$P = \frac{2^2 + 2 + 2}{2^5} = \frac{8}{32} = \frac{1}{4} \qquad \text{check: } 1 - \frac{24}{32} = \frac14$
2^2 first run at flip 1: the last two flips are free
2 first run at flip 2 or 3: one free flip each
24 sequences with no run of three heads
What it says in wordsCount each qualifying sequence once, by the start of its first run, and confirm by counting the sequences that fail.
Why would a quant fund ask this?
Streaks are what people notice in trading records and backtests. A run of three heads turns up in a quarter of all five-flip sequences of a fair coin, so a short winning streak is weak evidence of skill. The counting habit matters as much as the number: giving each outcome one home is the same discipline that stops you double counting overlapping signals when you tally how often a strategy fired.
Where candidates lose it
The trap is 12 out of 32: three places a run can start times four fillings of the other flips. It sounds rigorous, and it is wrong because a sequence with a long run is counted once for every place a run of three fits inside it.
The second loss is giving 1/4 with no check. Offer the complement count, 24 sequences with no run of three, as the second route to the same answer.
What the interviewer asks next
What is the chance of at least three heads in a row in ten flips?
In five flips, what is the chance of a run of exactly three heads, no longer?
How many flips do you expect to make before you first see three heads in a row?
056Game A: roll two fair dice and be paid the product of the faces in rupees. Game B: roll one fair die and be paid the square of its face. Which game is worth more, and by exactly how much?Expected value and dice gamesCoreCitadelMiami · 2022
Try it first
Which game would you rather play, and by roughly how much?
Show the worked solution
The one-die game is worth more: Rs 15.17 against Rs 12.25, a gap of 35/12, about Rs 2.92. With two independent dice the average product is the product of the averages, 3.5 x 3.5 = 12.25. With one die the average square is (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6. The gap between the two is exactly the variance of one die.
Why is averaging a square not the same as squaring an average?
Take two students who score 2 and 8 in a test. Their average is 5, and 5 squared is 25. Square each score first and then average, and you get (4 + 64)/2 = 34. Squaring rewards the high value more than it penalises the low one, so the average of the squares is always at least the square of the average, and the gap is the spread. Here the gap, 9, is exactly the variance of the two scores.
How do the two games come out?
In game A the dice are independent, so the average of the product is the product of the averages. Two independent dice give 3.5 x 3.5 = Rs 12.25, because a high roll on one die is as likely to meet a low roll on the other as a high one. In game B one number is multiplied by itself, so a 6 always meets a 6 and a 1 always meets a 1. The average of the six squares is 91/6, about Rs 15.17.
The relationship
$E[X^2] = \mathrm{Var}(X) + E[X]^2 = \tfrac{35}{12} + 12.25 = \tfrac{91}{6} \approx 15.17, \qquad E[XY] = E[X]\,E[Y] = 12.25$
X, Y the faces of two independent dice
Var(X) the variance of one die, 35/12
E[X] the average face, 3.5
What it says in wordsThe average square is the square of the average plus the variance; the average product of independent dice has no variance term.
The six squares from 1 to 36 average 15.17, above the 12.25 that squaring the average roll gives; the two-dice product game is worth 12.25, so the one-die square game is worth 2.92 more, exactly the variance of one die.
What is the desk lesson?
A payoff that is the square of a move gains from dispersion; one built from two independent pieces does not. Correlation is what turns a product into a square: if the second die always copied the first, game A would be game B. Flip it and make the second die show 7 minus the first, and the average product falls to 9.33. Say that range out loud and the interviewer knows you see the payoff as a bet on varianceThe average squared distance of an outcome from its mean; for one fair die it is 35/12, about 2.92. as well as on the average.
Where candidates lose it
The quick wrong answer is that the games are worth the same, because both feel like three and a half times three and a half. That is true only for independent dice. Squaring one die ties the factors together.
The second loss is getting 15.17 and 12.25 and not naming the gap. Say that 2.92 is the variance of a die; that one sentence is what the interviewer is waiting for.
What the interviewer asks next
What is game A worth if you are paid the sum of the two dice instead of the product?
The second die always shows 7 minus the first. What is the average product now?
What is the expected value of the larger of two dice?
Asked at Citadel, Sales and Trading, Miami, 2022 (Wall Street Oasis): No, pretty typical interview questions, dice questions, etc.

n	the number of assets, each with variance 1
rho	the common pairwise correlation
n(n-1)	the number of ordered pairs, each contributing one covariance of rho

pi_bull	the long-run share of bull months
0.1, 0.2	the chances of leaving bull and leaving bear each month
0.7	how much of any starting gap survives each month

sigma_F, sigma_B	the fund and benchmark volatilities, 12% and 15%
rho	the correlation between their returns, 0.9

L	the daily leverage multiple, 2
r	the size of each day's index move, 10%

Try it first

Why can three assets not all be perfectly opposed?

How do you find the exact floor?

Where candidates lose it

What the interviewer asks next

Try it first

What has to balance in the long run?

How do you check two thirds another way?

Where candidates lose it

What the interviewer asks next

Try it first

What exactly is tracking error measuring?

Why is the answer so much bigger than 3%?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the index down at all after up 10% and down 10%?

Why does the 2x fund lose four times as much?

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

Try it first

How do the loss and the new weight work out?

Why is a losing short more dangerous than a losing long?

Where candidates lose it

What the interviewer asks next

Try it first

Why not bet everything when the odds are in your favour?

How do you find the best fraction?

What happens at twice the Kelly bet?

Where candidates lose it

What the interviewer asks next

Try it first

What is the one-line argument?

How do you check it with a calculation?

Where does the pattern lead next?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the obvious count go wrong?

How do you make every sequence count once?

Why would a quant fund ask this?

Where candidates lose it

What the interviewer asks next

Try it first

Why is averaging a square not the same as squaring an average?

How do the two games come out?

What is the desk lesson?

Where candidates lose it

What the interviewer asks next