Hedge Funds puzzles, solved step by step
- Puzzles
- 100
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- 14
- Hard
- 30
071You start with Rs 10 and bet Rs 1 at a time on a game you win with probability 0.55, winning or losing Rs 1 each round. What is the chance you reach Rs 20 before you go broke?Two SigmaNew York · 2023
Try it first
Roughly what is the chance of reaching Rs 20 first?
Show the worked solution
About 88%. Let r be the ratio of losing to winning odds, 0.45/0.55 = 9/11. The chance of reaching Rs 20 from Rs 10 is (1 - r to the 10th)/(1 - r to the 20th), which simplifies to 1/(1 + r to the 10th). Since r to the 10th is about 0.134, the answer is 1/1.134, about 88.1%. A fair game would give exactly 50%.
Why does a small edge per bet become a large edge on the game?
Think of a tug of war between two teams, one a shade stronger. A single pull is close to a coin toss, but the rope has to travel a long way before either side wins, and every pull leans the same way. Reaching Rs 20 or Rs 0 takes many Rs 1 bets, and the 0.55 edge applies to every one of them, so the chance of winning the whole game rises far above 0.55. Staking all Rs 10 on one bet would give only 55%; betting Rs 1 at a time gives about 88%.
How do you get the formula?
Let P(i) be the chance of reaching 20 from a stake of i. One bet later you are at i + 1 with chance p or i - 1 with chance q, so P(i) = p P(i + 1) + q P(i - 1), with P(0) = 0 and P(20) = 1, and the solution is P(i) = (1 - r to the i)/(1 - r to the 20), where r = q/p. For a fair game the formula collapses to a straight line, P(i) = i/20, because a fair game keeps your expected wealth at 10, so 20 times P must equal 10. Checking the fair case is the fastest way to trust the biased one.
The relationshipP(10) the chance of reaching Rs 20 before Rs 0 from Rs 10 r the ratio of the losing chance to the winning chance r^20 the same ratio over the full distance of Rs 20 What it says in wordsThe chance of success from the middle is one over one plus the odds ratio raised to the distance to either end.Betting Rs 1 a time from Rs 10, a fair game reaches Rs 20 first 50% of the time, a 0.55 edge lifts that to 88.1% and a 0.45 disadvantage cuts it to 11.9%, because the small edge on each bet compounds over the many bets the game takes. How do you compute r to the tenth in your head, and what is the lesson?
Square repeatedly: 9/11 squared is 81/121, about 0.669; squared again about 0.448; again about 0.201; times 0.669 gives about 0.134. The desk lesson is about sizing: with an edge, make many small bets so the edge compounds; without one, the same arithmetic works against you, and at 0.45 the Rs 1 strategy reaches Rs 20 only 12% of the time against 45% for one bold bet. This is the classic gambler's ruinThe problem of a gambler betting fixed amounts until reaching a target or losing everything, solved as a random walk with two absorbing ends. result, and it is why a trader with a real but thin edge wants volume, not size.
Where candidates lose it
The common wrong answer is 55%, the chance of winning one bet, carried over to the whole game. It ignores that the game takes many bets and the edge applies to each of them.
The second loss is setting up the recursion and getting lost in the algebra. Write the answer in the form 1/(1 + r to the 10th), check it on the fair case, and compute the power by repeated squaring.
What the interviewer asks next
- With a win probability of 0.45, should you bet Rs 1 at a time or everything at once, and why?
- How many rounds do you expect the game to last at p = 0.5?
- The casino has unlimited money and you never stop at Rs 20. What is your chance of eventual ruin at p = 0.55?
Asked at Two Sigma, Research, New York, 2023 (Wall Street Oasis):
Biased gamblers ruin problems; Markov Chain problems; sampling uniformly from triangle
095Daily returns have an autocorrelation of 0.1 at a one-day lag, decaying geometrically: 0.01 at two days, 0.001 at three, and so on. By how much is the true volatility over a 21-day month higher than daily volatility times the square root of 21?AQR Capital ManagementGreenwich · 2022
Try it first
How much higher is the true monthly volatility?
Show the worked solution
About 10% higher. The variance of a sum of 21 daily returns is 21 daily variances plus twice every covariance between pairs of days. With autocorrelation 0.1 at lag one, 0.01 at lag two and so on, the covariance terms add about 4.42 daily variances, lifting the total from 21 to about 25.4, a 21% rise. Volatility, the square root, rises by about 10%.
Why does a small autocorrelation matter over a month?
Think of a queue where each person who joins makes it slightly more likely that the next person joins too. Each nudge is small, but over a month of days they add up to longer queues than pure chance would give. The variance of a sum counts every pair of days, and there are 20 neighbouring pairs in a 21-day month, each counted twice, so a lag-one correlation of 0.1 alone adds 4 daily variances to the 21.
The square-root rule counts 21 daily variances for a month, but one-day lags add 4.0, two-day lags 0.38 and longer lags 0.04, for 25.42, so monthly volatility is 5.04 daily volatilities rather than 4.58, about 10% higher. How do you do the sum?
Write the monthly variance as the daily variance times 21, plus twice the sum over every lag. Lag one appears in 20 pairs with correlation 0.1, lag two in 19 pairs with 0.01, lag three in 18 with 0.001: twice (2.0 + 0.19 + 0.018 and so on) is about 4.42. The total is 25.42 daily variances instead of 21, a ratio of 1.210, and the volatility ratio is its square root, 1.100. Over a long horizon the ratio tends to (1 + 0.1)/(1 - 0.1) = 1.222, so a month is already close to the limit.
The relationshipsigma^2 the variance of one day's return n days in the month, 21 rho_k the autocorrelation at a lag of k days, 0.1 to the power k n - k how many pairs of days in the month are k days apart What it says in wordsA month's variance is the sum of the daily variances plus twice every covariance between pairs of days, weighted by how many such pairs the month holds.What does this mean for risk and performance numbers?
Scaling daily volatility by the square root of time silently assumes zero autocorrelation. Positive autocorrelation, which shows up in trend-following returns and in portfolios of thinly traded assets priced from stale quotes, makes the square-root rule understate longer-horizon risk. Negative autocorrelation, typical of mean-reverting strategies, does the reverse: at minus 0.1 the monthly variance is only 0.83 times the rule's. The same effect is why correlations between two assets measured on daily returns can differ from those measured on monthly returns: prices that react to the same news on different days look less related day by day than month by month.
Where candidates lose it
Candidates either dismiss 0.1 as too small to matter or report 21% as the answer. The first ignores that the correlation enters twenty times over; the second forgets that volatility is the square root of variance.
The other loss is refusing to estimate without doing every lag. The lag-one term alone gives 21 + 4.0, close to the full 25.42; say that the higher lags add about 0.42 and that you checked.
What the interviewer asks next
- What is the ratio if the lag-one autocorrelation is minus 0.1 instead?
- A fund holds thinly traded assets priced from stale quotes. Which way does its reported volatility err?
- Why can the correlation between two assets look higher on monthly returns than on daily returns?
Asked at AQR Capital Management, Quantitative Research, Greenwich, 2022 (Wall Street Oasis):
describe what covariance means what would be the difference between the correlation of daily vs monthly returns of a given year
098With orthonormal regressors, ordinary least squares gives coefficients of 0.6 and 0.15. What do ridge and lasso with a penalty of 0.2 give for each, and why does only lasso set a coefficient to zero?CitadelLondon · 2026
Try it first
Using half the residual sum of squares plus the penalty, what do the two methods give?
Show the worked solution
Ridge gives 0.5 and 0.125; lasso gives 0.4 and exactly 0. With orthonormal regressors each coefficient is shrunk on its own. Ridge divides each by 1 + 0.2 = 1.2, so it scales both down and never reaches zero. Lasso subtracts 0.2 from each size and stops at zero, so the small coefficient, 0.15, is removed. This uses the scaling of half the residual sum of squares plus 0.2 times the penalty.
Why do orthonormal regressors make this a one-line problem?
When the regressors are uncorrelated and scaled to unit length, the fit for each coefficient does not depend on the others, so the penalised problem splits into separate one-variable problems. Think of adjusting the volume on two speakers that are not wired together: turning one down does not change the other. For each coefficient you minimise half of (beta minus b) squared plus the penalty, where b is its least squares value, and the answer depends only on that one number. Ridge's penalty is half of beta squared times 0.2; lasso's is the size of beta times 0.2.
Ridge scales every least squares coefficient by 1/1.2, so 0.6 becomes 0.5 and 0.15 becomes 0.125, while lasso subtracts 0.2 and stops at zero, so 0.6 becomes 0.4 and 0.15, inside the dead zone below 0.2, becomes exactly 0. What does each penalty do to a coefficient?
Ridge's squared penalty pulls hard on big coefficients and gently on small ones. Setting the slope to zero gives beta x (1 + 0.2) = b, so ridge divides by 1.2: 0.6 becomes 0.5 and 0.15 becomes 0.125, shrunk but never zero. Lasso's absolute-value penalty pulls with the same force, 0.2, whatever the size. Its answer is the sign of b times the larger of (the size of b minus 0.2) and zero: 0.6 becomes 0.4, and 0.15, weaker than the pull of 0.2, lands exactly on 0.
The relationshipb the least squares coefficient, 0.6 or 0.15 lambda the penalty weight, 0.2 What it says in wordsRidge divides every coefficient by the same factor; lasso takes the same amount off every coefficient and never goes past zero.Why does only lasso select variables?
At zero, the squared penalty is flat: its slope is zero, so any small coefficient still earns its place by improving the fit a little. The absolute-value penalty has a corner at zero with a slope of 0.2 on each side, so a coefficient stays at zero unless the fit improves by more than 0.2 per unit, and 0.15 does not. That is why lasso gives sparseHaving many coefficients exactly equal to zero, so the model uses only a few of the available variables. models and ridge does not. The limitation to state: with correlated regressors lasso tends to keep one of a group arbitrarily and drop the rest, which is why desks often blend the two penalties in an elastic net.
Where candidates lose it
The common slip is to swap the two, saying ridge sets small coefficients to zero because it penalises harder. Ridge's penalty is heavy on big coefficients and almost nothing on small ones, which is exactly why it never zeroes them.
The second loss is getting the lasso numbers off by a factor of two. Written as the full residual sum of squares plus 0.2 times the absolute values, without the half, the threshold is 0.1, giving 0.5 and 0.05. State your scaling before you give numbers.
What the interviewer asks next
- At what penalty does lasso set the 0.6 coefficient to zero as well?
- The two regressors now have a correlation of 0.9. How do ridge and lasso behave differently?
- How would you choose the penalty in practice without fitting it to noise?
Asked at Citadel, Quantitative Research, London, 2026 (Wall Street Oasis):
very detailed and difficult questions about regularisation ridge and lasso
100You have n cars, each with enough fuel to drive 1,000 miles, and cars can transfer fuel to each other on the road. How far can one car get? What happens as n grows?Millennium ManagementLondon · 2024
Try it first
With 4 cars, how far can one car get?
Show the worked solution
1,000 x (1 + 1/2 + 1/3 + ... + 1/n) miles, which grows without limit but only like 1,000 x ln n. Drive all n cars together for 1,000/n miles: together they have burned one tankful, so one car tops up the others and stops. Then n - 1 cars drive 1,000/(n - 1), and so on, until the last car drives a full 1,000. With 4 cars that is about 2,083 miles; with 100, about 5,187.
Why can the fuel not simply be pooled into one car?
Each tank holds exactly 1,000 miles of fuel, so one car can never carry more than that at once. Think of porters carrying water across a desert: the helpers walk part of the way, hand over what they can spare, and drop back. The helper cars exist to keep the lead car's tank full for as long as possible, and they can only do that by travelling with it and burning fuel themselves.
Four cars drive 250 miles together before one refills the other three and stops, three drive 333 more, two drive 500 more and the last drives a full 1,000, reaching 2,083 miles; ten cars reach 2,929 and a hundred reach 5,187. How long is each leg?
With k cars travelling together on full tanks, drive until the group has burned exactly one tankful, which takes 1,000/k miles. Each tank is then 1/k empty, so the k - 1 cars that continue have (k - 1)/k of a tank of space between them, and the car that stops has exactly (k - 1)/k of a tank left to fill it. The legs are 1,000/n, then 1,000/(n - 1), and so on to 1,000 for the last car alone. With four cars: 250 + 333.3 + 500 + 1,000 = 2,083.3 miles. Dropping each helper the moment its fuel can refill the rest keeps as few cars as possible burning fuel at every mile.
The relationshipH_n the harmonic number, 1 + 1/2 + ... + 1/n 1000/k the leg driven while k cars are still moving 0.577 Euler's constant, the gap between H_n and ln n for large n What it says in wordsThe distance is 1,000 miles times the sum of one over each number of cars still driving, which grows like the natural logarithm of the number of cars.What happens as n grows?
The harmonic series never stops growing, so with enough cars there is no ceiling on the distance. But it grows only like the logarithm of n: 10 cars reach about 2,929 miles, 100 cars about 5,187, and each further tenfold increase in cars adds only about 2,303 miles. Reaching 5,000 miles takes 83 cars. That is the pattern worth naming in the room: unlimited in principle, very expensive in practice, the same diminishing return you meet whenever each extra unit of effort adds less than the one before.
Where candidates lose it
The quick wrong answer is n x 1,000 miles, pooling all the fuel, which ignores that no tank holds more than 1,000 and that helpers burn fuel just keeping up. The opposite slip is 1,000 miles, forgetting that fuel can be passed forward at all.
The second loss is reaching the harmonic series and then saying the distance levels off, or that it grows in proportion to n. Name the growth rate: like ln n, unbounded but slow.
What the interviewer asks next
- With 3 cars, exactly how far can one car get?
- Cars may now turn back and refuel at the start. Can the lead car get further?
- Roughly how many cars do you need for one car to travel 5,000 miles?
Asked at Millennium Management, Investments, London, 2024 (Wall Street Oasis):
Suppose you have n cars, each fueled so that they can drive for 1000 miles.
