Hedge Funds puzzles, solved step by step

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

Showing 61–70 of 100

061How many flips of a fair coin do you expect to need before you see two heads in a row? Why is the answer different if you wait for a head followed by a tail?Random walks and Markov chainsCoreSquarepoint CapitalLondon · 2025
Try it first
What are the expected waits for HH and for HT?
Show the worked solution
Six flips on average for two heads in a row, and four for a head then a tail. Track how far along the pattern you are. For HH, a tail at any point sends you back to the start, including right after a head. For HT, a head after a head keeps you one step away, so progress is never lost. Solving the two small chains gives 6 and 4.
Why do two equally likely patterns take different times?
Think of two ladders where a slip costs you differently. On one, slipping from the first rung drops you to the ground; on the other, you can only ever slip back to the first rung. Both patterns are equally likely in any given pair of flips, but after one head the wrong next flip costs you everything for HH and nothing for HT. A head then a tail breaks HH and restarts it; a head then a head is still a perfect start for HT.
In the HH chain a tail from the one-head state falls back to the start, so the expected wait is 6 flips; in the HT chain a head from the one-head state stays where it is, so progress is never lost and the wait is 4 flips.
How do you solve the chain?
Let E0 be the expected flips still needed from the start and E1 after one head. For HH, E0 = 1 + E1/2 + E0/2 and E1 = 1 + E0/2, because a tail from one head sends you back, and these solve to E1 = 4 and E0 = 6. For HT, the one-head state just waits for a tail, which takes 2 flips on average, and reaching the first head takes 2 more, giving 4. Say the states out loud before the algebra; the interviewer wants to hear them named.
The relationship
$E_0 = 1 + \tfrac12 E_1 + \tfrac12 E_0, \qquad E_1 = 1 + \tfrac12 \cdot 0 + \tfrac12 E_0 \;\Rightarrow\; E_0 = 6$
E0 expected flips still needed from the start
E1 expected flips still needed after one head
1 the flip you are about to make
What it says in wordsEach state's expected wait is one flip plus the average wait from wherever that flip sends you.
What is the general pattern?
Patterns that can fail back to nothing take longer. The expected wait for n heads in a row is 2 to the power n + 1, minus 2: 2, 6 and 14 flips for one, two and three heads. This is a Markov chainA process whose next step depends only on the current state, not on how it got there, so it can be solved state by state. at heart, and the same state-by-state method handles anything that depends on a path: a streak, a barrier, a drawdown rule on a trading book.
Where candidates lose it
The usual wrong answer is 4 for both, reached by noting that each pattern has a one-in-four chance in a pair of flips. That treats the flips as separate pairs, which they are not: a pattern can start at any flip, and what happens after a failure depends on the pattern.
The second loss is setting up one equation instead of two. Name the states, start and one head, and write one equation for each.
What the interviewer asks next
How many flips do you expect to need for three heads in a row?
Two players race: one wins at the first HH, the other at the first HT. Who is more likely to win?
With a coin that lands heads 60% of the time, how long do you expect to wait for HH?
Asked at Squarepoint Capital, Quantitative Research, London, 2025 (Wall Street Oasis): statistical problems e.g. # of throws expected to get 2 heads in a row
062A fund makes 1.5% every month for a year. What is its return for the year? Another fund made 40% in total over three years. What is its annual rate of return?Returns, compounding and feesWarm upFund of funds and allocatorsMulti-manager platforms
Try it first
What does 1.5% a month for twelve months come to?
Show the worked solution
1.5% a month compounds to about 19.6% a year, and 40% over three years is about 11.9% a year. Compounding multiplies growth factors rather than adding rates: 1.015 to the twelfth is 1.196. Going the other way, take the cube root of 1.40, which is 1.119. Simple arithmetic gives 18% and 13.3%, understating the first answer and overstating the second.
Why is twelve times 1.5% not the annual return?
A savings account that credits interest every month pays interest on last month's interest. Each month's 1.5% is earned on a base that already includes every earlier month's gain, so the growth factors multiply: 1.015 times itself twelve times. That is 1.196, a 19.6% year. The extra 1.6 points over 18% are interest on interest, tiny in any one month and not tiny over a year.
The relationship
$1.015^{12} - 1 = 19.6\%, \qquad 1.40^{1/3} - 1 = 11.9\%$
1.015 the monthly growth factor, 1 plus 1.5%
1.40 the three-year growth factor, 1 plus 40%
1/3 the cube root, which undoes three years of compounding
What it says in wordsRaise the growth factor to the number of periods to go forward, and take the matching root to go back.
Twelve months at 1.5% compound to 19.6% rather than 18.0%, and a 40% three-year gain is 11.9% a year rather than 13.3%, because 13.3% compounded for three years would give 45.6%.
How do you go back from a total to an annual rate?
Take the root, not the division. A 40% total over three years means the yearly growth factor cubed is 1.40, so the factor is the cube root of 1.40, about 1.119, an annual rate of 11.9%. Dividing 40 by 3 gives 13.3%, and 1.133 cubed is 1.456, a 45.6% total: division overstates the yearly rate because later years grow on earlier gains. Allocators compare managers on the compound annual growth rateThe single yearly rate that, compounded over the period, turns the starting value into the ending value., so dividing can misrank two funds.
How do you do it in your head?
Add the square-term correction. Compounding adds roughly n(n - 1)/2 times r squared to n times r: for 12 months at 1.5% that is 66 x 0.000225, about 1.5 points, taking 18% to about 19.5%. That is close enough to show you know the direction and the size. For the cube root, guess and check: 1.12 cubed is about 1.405, a shade over 1.40, so the answer sits just under 12%. Checking by cubing is faster and safer than estimating a root directly.
Where candidates lose it
The trap is simple arithmetic: 12 x 1.5% = 18% and 40 / 3 = 13.3%. Both come out fast and both are wrong, in opposite directions, which is why the interviewer asks the pair together.
The second loss is getting 19.6% and then dividing on the second half out of habit. State the rule once, multiply going forward and take the root going back, and apply it to both halves.
What the interviewer asks next
A fund loses 1.5% every month for a year. What is its annual return?
Which pays more: 1% a month, or 12.5% paid once a year?
A manager reports a three-year return of 40% and an average annual return of 13.3%. What is wrong with the second number?
063The sample variance computed with n minus 1 in the denominator is an unbiased estimator of the population variance. Is its square root an unbiased estimator of the standard deviation?Statistics and estimationCoreSquarepoint CapitalLondon · 2026
Try it first
Is the square root of the unbiased sample variance unbiased for the standard deviation?
Show the worked solution
No. The square root of the unbiased variance underestimates the standard deviation on average. The square root is concave, so by Jensen's inequality the average of the square roots is below the square root of the average. For normal data with two observations the estimate averages about 0.80 sigma; the bias shrinks as the sample grows, to about 6% at five observations and under 1% at thirty.
Why does taking a square root break unbiasedness?
Two square rooms have floor areas of 4 and 16 square metres, so their sides are 2 and 4 metres. Average the areas, 10, and take the root: 3.16 metres. Average the sides instead: 3 metres. Averaging and then taking a square root gives a bigger answer than taking square roots and then averaging, because the square root bends downwards. The sample variance is right on average, so the average of its square roots must fall short of the true standard deviation.
Two equally likely variance estimates of 0.04 and 1.96 average to the true variance of 1.0, but their square roots, 0.2 and 1.4, average only 0.8, below the true standard deviation of 1.0, because the square-root curve bends downwards.
How big is the bias?
It depends on the sample size and on the distribution. For normal data the expected sample standard deviation is c4 times sigma, with c4 about 0.80 at n = 2, 0.94 at n = 5, 0.97 at n = 10 and 0.99 at n = 30. At n = 2 you can check it directly: the sample standard deviation is the gap between the two draws divided by the square root of 2, and the average gap between two normal draws is 2 sigma over the square root of pi, which leaves the square root of 2/pi, about 0.798.
The relationship
$E[s] = c_4(n)\,\sigma < \sqrt{E[s^2]} = \sigma, \qquad c_4(2) = \sqrt{2/\pi} \approx 0.798$
s the square root of the unbiased sample variance
sigma the true standard deviation
c4(n) the correction factor for normal data, below 1 for every n
What it says in wordsThe average sample standard deviation is a fixed fraction of the true one, and that fraction is below one.
Does it matter in practice?
Sometimes. With a year of daily returns the bias is a rounding error; with a handful of monthly returns for a new fund it is not. A manager with five monthly returns has a volatility estimate that averages about 6% too low under normality, which flatters a Sharpe ratioAverage excess return divided by the standard deviation of returns, a measure of return per unit of risk. before anyone has looked at fat tails. Dividing by c4 removes the bias for normal data, but the fix depends on the distribution, so name the assumption. And unbiased is not the same as most accurate.
Where candidates lose it
The trap is assuming unbiasedness carries through any function of an estimate. It carries through straight-line transformations only; the square root is curved, so the property is lost.
The second loss is saying it is biased without the direction or the size. Say biased low, give the Jensen reason in one sentence, and quote about 0.80 at two observations, shrinking towards 1 as the sample grows.
What the interviewer asks next
Is the square of an unbiased estimator of the standard deviation unbiased for the variance?
Why does the sample variance divide by n minus 1 rather than n?
Which estimator of sigma has the lowest mean squared error for normal data?
Asked at Squarepoint Capital, Quantitative Research, London, 2026 (Wall Street Oasis): Is the square root of the unbiased estimator for sample variance unbiased for standard deviation?
064An exporter earns Rs 100 of revenue, 60% of it billed in dollars, and has Rs 85 of costs, all paid in rupees. The rupee weakens from 80 to 88 to the dollar. What happens to its operating margin?Valuation, accounting and macro riddlesCoreLong-short equity fundsGlobal macro funds
Try it first
What is the operating margin after the move?
Show the worked solution
The margin rises from 15% to about 19.8%, and operating profit rises 40%. Each dollar now buys 10% more rupees, which turns Rs 60 of dollar revenue into Rs 66, so revenue reaches Rs 106. Costs are all in rupees and stay at Rs 85, so profit climbs from Rs 15 to Rs 21, and 21 over 106 is 19.8%. A 6% rise in revenue becomes a 40% rise in profit.
Why does a weaker rupee help this company so much?
Think of a tutor in Pune who teaches some students abroad for fees in dollars and pays rent and a salary in rupees. When the rupee weakens, each dollar fee converts into more rupees while the rent does not move. A currency move lands on the revenue billed in the foreign currency and on nothing else, so when every cost is local the whole gain drops into profit. Here the gain is Rs 6, on a profit that started at only Rs 15.
Before the move, Rs 40 of rupee revenue and Rs 60 of dollar revenue less Rs 85 of rupee costs leave Rs 15 of profit, a 15% margin; at 88 to the dollar the dollar revenue is worth Rs 66, costs stay at Rs 85 and profit rises to Rs 21, a 19.8% margin.
How do you size the profit move quickly?
Chain three numbers. The dollar buys 10% more rupees, 60% of revenue is exposed, so revenue rises 6%; a thin 15% margin turns that Rs 6 into a 40% rise in profit. The ratio of the profit change to the revenue change, 40 over 6, almost 7 times, is the company's operating leverageHow much operating profit moves for a given move in revenue, high when most costs are fixed or unaffected by the change. to the currency. An analyst on a long-short book uses that multiplier to judge how much of a stock's move a currency view is really driving.
The relationship
$\text{margin} = \frac{40 + 60 \times \tfrac{88}{80} - 85}{40 + 60 \times \tfrac{88}{80}} = \frac{21}{106} \approx 19.8\%$
40 revenue billed in rupees, unchanged
60 x 88/80 dollar revenue restated at the new rate, Rs 66
85 costs, all in rupees, unchanged
What it says in wordsRestate only the dollar-billed revenue at the new rate, hold rupee costs fixed, and divide profit by the new revenue.
What would you check before trusting the new margin?
Three things that often shrink the gain. Hedges, pricing and imported inputs: a company that sold its dollars forward keeps the old rate until the hedges roll off, buyers abroad may push for lower dollar prices, and any imported materials get dearer too. Each one eats into the Rs 6. The 19.8% is the unhedged, first-round answer; naming those three tells the interviewer you would not stop at it.
Where candidates lose it
The first slip is saying the margin is unchanged because revenue and costs both grow. Only the dollar-billed revenue moves; rupee costs stay where they were, and that asymmetry is the whole point.
The second is applying the 10% to all Rs 100 of revenue, which gives Rs 110 and a 22.7% margin. Only 60% of revenue is billed in dollars, so revenue rises 6%, not 10%.
What the interviewer asks next
Half of the costs are imported and paid in dollars. What is the new margin?
The company hedged half its dollar revenue at 80. What margin does it report this year?
What does the same move do to an importer with the mirror-image structure?
065A stock rises 50% or falls 40% each year with equal probability. Its expected return is positive, but what happens to a buy-and-hold investor over time? And what fraction of wealth should sit in the stock if the rest is held in cash and the mix is rebalanced every year?Betting and sizingHardMulti-manager platformsProp and quant trading firms
Try it first
Over many years, what happens to the typical buy-and-hold investor?
Show the worked solution
The typical buy-and-hold investor loses about 5.1% a year, yet a 25% stake rebalanced yearly grows about 0.6% a year. The average year returns +5%, but a good year and a bad year multiply wealth by 1.5 x 0.6 = 0.9. With 25% in the stock the two years multiply wealth by 1.125 x 0.9 = 1.0125. That 25% is the Kelly fraction, the stake that maximises the average log return.
How can a positive average return shrink your wealth?
Imagine a shop whose sales rise 50% in a good year and fall 40% in a bad one. After one of each it is at 90% of where it began, whatever the order. Wealth compounds by multiplying, so over many years what matters is the typical growth factor, the square root of 1.5 x 0.6, about 0.949, not the average return of +5%. The average is real, but it is carried by rare paths with long lucky streaks. After 20 years the typical investor holds about 0.35 of the starting money while the average across all paths is 2.65 times it.
The relationship
$g(f) = \tfrac12\ln(1 + 0.5f) + \tfrac12\ln(1 - 0.4f), \qquad g'(f) = 0 \;\Rightarrow\; f^{*} = 0.25$
f the fraction of wealth held in the stock, the rest in cash
g(f) the expected log growth per year of the rebalanced mix
f* the stake that maximises it
What it says in wordsPick the stake that makes the average log return per year as large as possible; here that is a quarter of your wealth.
Over 20 alternating years the all-stock investor falls to about 0.35 of the starting wealth while the average across all paths climbs to 2.65, and a 25% stake rebalanced every year grows to about 1.13, because growth depends on the log return, not the average return.
Why does holding less of the stock help?
Rebalancing to a fixed mix sells after gains and buys after losses, and a smaller stake shrinks the swings. With a fraction f in the stock, a good year multiplies wealth by 1 + 0.5f and a bad year by 1 - 0.4f; typical growth peaks where 0.5/(1 + 0.5f) equals 0.4/(1 - 0.4f), which gives f = 25%. At 25% a pair of years gives 1.125 x 0.9 = 1.0125, about 0.6% a year. The quick check is return over variance: 0.05 divided by 0.45 squared is 0.247.
What is the limit of this answer?
The 25% rests on knowing both outcomes and their odds exactly, on cash earning nothing and on free rebalancing. Change any of those and the fraction moves, and because a real edge is only an estimate, desks size well below the full Kelly number. The lesson to lead with in the room is the gap itself: a positive average return is not a positive growth rate, and position size decides which one you earn. That gap is called volatility dragThe shortfall of the compound growth rate below the average return, roughly half the variance of returns..
Where candidates lose it
Most candidates answer that the investor earns 5% a year, because that is the average. The interviewer built the numbers so the average and the typical outcome point in opposite directions, and wants to see you notice.
The second loss is concluding the stock is simply bad and putting nothing in it. Zero earns nothing; the point is that a small, rebalanced stake turns the same gamble into positive growth.
What the interviewer asks next
Cash now earns 3% a year. How does the best stake change?
What changes if you rebalance every two years instead of every year?
Two such stocks move independently. What happens if you hold half in each and rebalance yearly?

066You roll a fair die repeatedly until you have seen every even number, 2, 4 and 6. Given that the last new even number to appear is a 2, what is the probability that your first roll was a 1? Why is the intuitive answer of 1/5 wrong?Conditional probability and BayesHardSquarepoint CapitalLondon · 2026

Try it first

Given the game ends on a 2, what is the chance the first roll was a 1?

Show the worked solution

The probability is 1/6, not 1/5. Ending on a 2 rules out a first roll of 2, but it does not leave the other five faces equally likely. A first roll of 4 or 6 has already cleared one rival, so 2 then finishes last half the time; after an odd first roll it finishes last only a third of the time. Weighting the faces by those chances gives 1/6 for a 1.

Why does ruling out one face not spread its weight evenly?

Suppose you hear that a friend arrived late to a meeting. Before, the bus, the train and the car were equally likely ways she travelled. Learning how things ended shifts weight towards the starts that make that ending more likely, in proportion to how strongly each one leads to it. If the bus is late twice as often as the train, the bus now carries twice the train's weight. The 1/5 answer treats the ending as if it only ruled a face out and said nothing else.

How likely is a 2 to finish last after each first roll?

Odd rolls never change the order in which the even numbers first appear, so after an odd first roll the three evens are still symmetric and 2 is last with chance 1/3; after a first roll of 4 or 6 only two evens remain and 2 is last with chance 1/2; after a first roll of 2 it can never be the last new even. Multiply each by the 1/6 chance of that first roll: the joint chances are 1/18 for each odd face, 0 for a 2 and 1/12 for each of 4 and 6. They add to 1/3.

An odd first roll leaves 2 last among the evens a third of the time, a 4 or 6 leaves it last half the time and a 2 never does, so given the game ends on a 2 each odd face has chance 1/6 and each of 4 and 6 has chance 1/4, not 1/5 each.

The relationship

P(\text{first}=1 \mid 2 \text{ last}) = \frac{P(\text{first}=1)\,P(2 \text{ last} \mid \text{first}=1)}{P(2 \text{ last})} = \frac{\tfrac16 \cdot \tfrac13}{\tfrac13} = \frac16

P(first = 1)	the chance of rolling a 1 first, 1/6
P(2 last \| first = 1)	the chance 2 is the last even to appear after an odd first roll, 1/3
P(2 last)	the overall chance the game ends on a 2, 1/3 by symmetry

What it says in wordsBayes' rule: the prior chance of a 1, times how strongly a 1 leads to ending on a 2, divided by the overall chance of ending on a 2.

How do you check the answer?

Make the six posterior chances add up. Three odd faces at 1/6 each and two even faces, 4 and 6, at 1/4 each give 1/2 plus 1/2, which is 1, with nothing left for a 2. The odd faces keep exactly their starting weight because an odd roll tells you nothing about the evens; all the weight removed from the 2 goes to 4 and 6. Saying that sentence shows the interviewer you understand Bayes ruleThe rule for updating a probability after new information: the prior times the likelihood of the information, divided by the overall chance of the information. rather than recite it.

Where candidates lose it

The trap is to condition only by elimination: the game ends on a 2, so the first roll was not a 2, so the five other faces share the weight at 1/5 each. That treats the ending as a filter when it is also evidence about the start.

The second loss is getting 1/6 and being unable to say where the missing weight went. Name it: 4 and 6 each rise to 1/4, because they make ending on a 2 more likely.

What the interviewer asks next

Given the game ends on a 2, what is the chance the first roll was a 4?
What is the expected number of rolls in this game?
Given the game ends on a 2, what is the chance the second roll was a 1?

Asked at Squarepoint Capital, Quant Research Intern Interview, London, 2026 (Wall Street Oasis): why is the probability of seeing a 1 on our first roll, given that we end on a 2, not 1/5

067Daily returns are normal with 1% volatility on 80% of days and normal with 4% volatility on the other 20%, both with zero mean. What are the overall daily volatility and the kurtosis of this mixture?Continuous probability and distributionsHardTwo SigmaNew York · 2025
Try it first
What is the overall daily volatility of the mixture?
Show the worked solution
Overall volatility is 2% a day and the kurtosis is 9.75, against 3 for a normal distribution. Variances mix in proportion: 0.8 x 1 + 0.2 x 16 = 4, so volatility is 2%. Fourth moments mix the same way, and each normal contributes 3 times its volatility to the fourth: 0.8 x 3 + 0.2 x 768 = 156. Dividing by the variance squared, 16, gives 9.75.
Why is a mixture of normals not normal?
Think of a commute that takes 30 minutes on most days and two hours on strike days. The average trip hides the shape: most days cluster tightly and a few days sit far out. Mixing a calm regime with a wild one gives more small moves than a normal with the same overall spread, fewer medium ones, and far more big ones. The peak is taller, the shoulders thinner and the tails fatter, which is what {term('kurtosis', 'The fourth moment of a distribution divided by the variance squared; 3 for a normal distribution, higher when tails are fatter.')} measures.
How do you get the two numbers?
Work with moments, because they mix in proportion to the weights. The variance is 0.8 x 1 + 0.2 x 16 = 4, so volatility is 2%; the fourth moment is 0.8 x 3 x 1 + 0.2 x 3 x 256 = 2.4 + 153.6 = 156, and kurtosis is 156 / 4 squared = 9.75. Look at where the 156 comes from: 153.6 of it is the wild days, which occur only one day in five. The fourth power makes rare large moves dominate.
The relationship
$\sigma^2 = \sum_i w_i\sigma_i^2 = 4, \qquad \kappa = \frac{\sum_i w_i \cdot 3\sigma_i^4}{\left(\sum_i w_i\sigma_i^2\right)^2} = \frac{156}{16} = 9.75$
w_i the share of days in each regime, 0.8 and 0.2
sigma_i the volatility in each regime, 1% and 4%
3 sigma_i^4 the fourth moment of a zero-mean normal with volatility sigma_i
What it says in wordsAverage the second and fourth moments across regimes, then divide the fourth moment by the variance squared.
Against a normal with the same 2% volatility, the mixture has a taller peak, thinner shoulders and fatter tails: a daily move beyond 6% either way happens on 2.7% of days under the mixture but only 0.27% under the normal, about 10 times as often.
What does this mean for a risk model?
A model that fits a normal to the 2% volatility is right about the average day and wrong about the days that matter. It says a move beyond 6% happens on about 0.27% of days, roughly once in 370 trading days; the mixture says 2.7%, roughly once in 37. That is the usual story of market returns: calm stretches and volatile stretches, each close to normal, adding up to fat tails. A model that lets volatility change over time captures much of it.
Where candidates lose it
The first slip is averaging the volatilities, 0.8 x 1% + 0.2 x 4% = 1.6%. Variances average in a mixture, not standard deviations, so the answer is 2%.
The second is guessing that a mixture of normals has kurtosis 3 because each piece does. Mixing different variances always pushes kurtosis above 3, and here the fourth-power weight on the wild days takes it to 9.75.
What the interviewer asks next
What mixture weight on the 4% regime maximises the kurtosis?
What is the probability density of the mixture at zero, compared with the normal?
If the two regimes had different means but the same volatility, what would happen to skew and kurtosis?
Asked at Two Sigma, Quantitative Research, New York, 2025 (Wall Street Oasis): They asked a couple questions involving Mixture Gaussians (e.g., probability density and moments).
068You and an opponent each secretly show heads or tails. If both show heads you win Rs 3, if both show tails you win Rs 1, and if they differ you pay Rs 2. The payoffs look balanced. What mix should each player use, and what is the game worth to you?Expected value and dice gamesHardCitadelsydney · 2025
Try it first
If both of you play well, what is the game worth to you per round?
Show the worked solution
Both players should show heads 3/8 of the time, and the game is worth minus Rs 0.125 a round to you. Choose your mix so the opponent gains nothing by switching: 3p - 2(1 - p) = -2p + (1 - p) gives p = 3/8. The opponent's mix solves the same balance from your side, also 3/8. At those mixes you lose an eighth of a rupee a round, although the payoffs look even.
Why is a fair coin the wrong strategy?
Think of a penalty taker and a goalkeeper. If the taker always shoots left, the keeper dives left; the taker's only defence is to mix so the keeper cannot profit from guessing. A mix is right only if it leaves the other side indifferent, and a fair coin here does not: against it the opponent's tails pays you 0.5 x (-2) + 0.5 x 1 = -0.50 a round. So the opponent always shows tails, and you lose Rs 0.50 a round rather than breaking even.
How do you find the equilibrium mix?
Set up your payoff for each of your choices as the opponent's chance of heads, q, varies. Showing heads pays 3q - 2(1 - q) = 5q - 2; showing tails pays -2q + (1 - q) = 1 - 3q; they are equal at q = 3/8, where both pay -1/8. The opponent plays 3/8 heads so that nothing you do beats -1/8. By the same balance from the opponent's side, you play 3/8 heads so that nothing the opponent does pushes you below -1/8. That pair is the Nash equilibriumA pair of strategies in which neither player can do better by changing only their own choice..
The relationship
$5q - 2 = 1 - 3q \;\Rightarrow\; q = \tfrac{3}{8}, \qquad V = 5 \cdot \tfrac38 - 2 = -\tfrac18$
q the opponent's chance of showing heads
5q - 2 your expected payoff if you show heads
1 - 3q your expected payoff if you show tails
V the value of the game to you
What it says in wordsThe opponent's mix makes your two choices pay the same, and that common payoff is what the game is worth.
Your payoff from showing heads rises with the opponent's chance of heads and your payoff from tails falls; the lines cross at 3/8, where either choice pays minus Rs 0.125, so the opponent plays 3/8 heads and the game is worth minus Rs 0.125 a round to you.
Why is the game negative when the payoffs look even?
Because the balance is in the totals, not in the play. Your two winning cells need coordination the opponent will not give you, while their winning cells, the mismatches, pay the same 2 either way. The opponent can lean towards tails, starving your big Rs 3 cell, and the only price is feeding your small Rs 1 cell. Say what a desk would do with it: ask to be paid about 13 paise a round to play, or ask to swap sides.
Where candidates lose it
The trap is adding up the payoffs, 3 and 1 against 2 and 2, calling the game fair and playing a fair coin. That ignores that the opponent chooses too, and against a fair coin their best reply costs you Rs 0.50 a round.
The second slip is solving for your own mix by making yourself indifferent. Each player's mix is chosen to make the other player indifferent; set up the equation from the opponent's payoffs.
What the interviewer asks next
Change the tails-tails payoff to Rs 2. What are the mixes and the value now?
What would you pay per round to play this game from the opponent's side?
The opponent is known to show heads half the time. What do you do, and what do you earn?
Asked at Citadel, Quantitative Trading, sydney, 2025 (Wall Street Oasis): many probability questions for OA. mix of prob and game theory for technical
069You have 12 coins that look identical. One is either heavier or lighter than the others, and you do not know which. Using a two-pan balance only three times, find the odd coin and say whether it is heavy or light.Logic and brainteasersHardProp and quant trading firmsLong-short equity funds
Try it first
Why is three weighings enough, in principle?
Show the worked solution
Weigh four against four first, then mix suspects with coins you already know are genuine so every later weighing splits the cases three ways. There are 24 possibilities, 12 coins each heavy or light, and three weighings have 27 outcomes. If 1 to 4 balances 5 to 8, weigh 9, 10, 11 against three good coins; if not, weigh 1, 2, 5 against 3, 6, 9. The third weighing settles what is left.
How do you know three weighings can be enough?
A game of twenty questions works because each yes or no halves what is left. A balance is better than a yes or no: it answers left heavy, right heavy or balanced. Three weighings give 3 x 3 x 3 = 27 outcomes, and there are 24 cases to tell apart, 12 coins each possibly heavy or light, so a procedure can exist only if every weighing splits the remaining cases into three near-equal groups. That counting sets the design: the first weighing must leave at most 9 cases on every branch.
Weighing four against four splits the 24 cases into three groups of 8; the second weighing mixes suspects with known good coins to split each 8 into 3, 2 and 3; the third weighing then separates what is left, so all 24 cases fit inside the 27 outcomes.
What do you do after the first weighing tips?
Say the left pan was heavy: the odd coin is 1, 2, 3 or 4 and heavy, or 5, 6, 7 or 8 and light. Weigh 1, 2 and 5 against 3, 6 and 9, moving some suspects across and bringing in a known good coin, so each outcome points to a different small group. Left heavy again means 1 heavy, 2 heavy or 6 light: weigh 1 against 2, and a balance means 6. Right heavy means 3 heavy or 5 light: weigh 3 against a good coin. A balance means 4 heavy, 7 light or 8 light: weigh 7 against 8.
What if the first weighing balances?
Then coins 1 to 8 are genuine and the odd coin is among 9 to 12, still heavy or light. Weigh 9, 10 and 11 against three good coins: a tip tells you both that the odd coin is among the three and whether it is heavy or light, and a balance points to coin 12. After a tip, weigh 9 against 10: if the odd coin is heavy the heavier of the two is it, if light the lighter, and a balance means 11. After a balance, weigh 12 against a good coin to learn heavy or light.
Where candidates lose it
The usual loss is weighing six against six first. It wastes the balance outcome, because the odd coin is always in one of the pans, and it leaves 12 cases on a branch that only two weighings, 9 outcomes, must resolve.
The second is forgetting that heavy or light is part of the answer. Candidates find the coin and stop; the counting argument, 24 cases in 27 outcomes, is the proof that you have not left anything to luck.
What the interviewer asks next
What is the largest number of coins you can handle with three weighings if you must also say heavy or light?
How does the problem change if you have one extra coin known to be genuine?
Can you design all three weighings in advance, without looking at the earlier results?
070A portfolio is split equally between two assets, each with 15% annual volatility. Their correlation is 0.2 in normal markets and 0.8 in a crisis. What is the portfolio's volatility in each regime?Portfolio and risk mathsCoreMulti-manager platformsQuant and systematic funds
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What happens to the portfolio's volatility when correlation rises from 0.2 to 0.8?
Show the worked solution
About 11.6% in normal markets and 14.2% in a crisis. With equal weights and equal volatilities, portfolio variance is 15% squared times (1 + rho)/2. At a correlation of 0.2 that is 225 x 0.6 = 135, a volatility of 11.6%; at 0.8 it is 225 x 0.9 = 202.5, a volatility of 14.2%. About 77% of the diversification benefit disappears just when it is needed.
Where does correlation enter the arithmetic?
Picture an umbrella shop and an ice-cream stall owned by one family. On ordinary days one does well when the other is quiet, and the family income is smoother than either shop's. A city-wide power cut shuts both at once. Portfolio variance is each asset's own variance, weighted, plus a co-movement term that carries the correlation, so when correlation jumps the co-movement term grows and the smoothing shrinks. Here the own terms are 56.25 each and the co-movement term is 2 x 0.5 x 0.5 x rho x 225.
The relationship
$\sigma_p^2 = w_1^2\sigma_1^2 + w_2^2\sigma_2^2 + 2w_1w_2\rho\,\sigma_1\sigma_2 = 225 \times \frac{1+\rho}{2}$
w1, w2 the weights, 0.5 each
sigma1, sigma2 the volatilities, 15% each
rho the correlation between the two assets
What it says in wordsPortfolio variance is the weighted own variances plus a cross term that grows with correlation.
At a correlation of 0.2 the co-movement term adds 22.5 to a variance of 112.5, giving 11.6% volatility; at 0.8 it adds 90.0, giving 14.2%, so the benefit of holding both assets falls from 3.4 points to 0.8.
What are the two numbers?
Normal markets: 112.5 plus 2 x 0.25 x 0.2 x 225 = 22.5 gives 135, and the square root is 11.6%. Crisis: the co-movement term rises to 90, variance to 202.5, and volatility to 14.2%, a 22.5% jump in risk with no change in either asset's own volatility. The benefit of holding two assets instead of one falls from 3.4 points to 0.8. In a real crisis each asset's volatility usually rises too, so this is the milder case.
What would a risk manager do with this?
Stop sizing positions on correlations measured in calm markets. A risk limit set on a normal-market correlation understates crisis risk by about 22% here, before any rise in the assets' own volatility. Platforms that run many books run stressed-correlation scenarios for this reason, and ask each manager what the book looks like if everything moves together. The honest limitation: nobody knows the crisis correlation in advance; 0.8 is an assumption, and the answer should say so.
Where candidates lose it
The common loss is treating correlation as if it scaled volatility directly, or quoting 15% in both regimes because each asset's volatility has not changed. Correlation enters only through the cross term in the variance.
The second is computing both numbers and missing the point of the question: the benefit you were counting on shrinks exactly in the regime where you need it. Say that sentence.
What the interviewer asks next
At what correlation does the portfolio's volatility reach 13%?
With three equally weighted assets, what is the volatility at a correlation of 0.8?
Why might correlations between assets rise in a sell-off?

1.015	the monthly growth factor, 1 plus 1.5%
1.40	the three-year growth factor, 1 plus 40%
1/3	the cube root, which undoes three years of compounding

s	the square root of the unbiased sample variance
sigma	the true standard deviation
c4(n)	the correction factor for normal data, below 1 for every n

40	revenue billed in rupees, unchanged
60 x 88/80	dollar revenue restated at the new rate, Rs 66
85	costs, all in rupees, unchanged

f	the fraction of wealth held in the stock, the rest in cash
g(f)	the expected log growth per year of the rebalanced mix
f*	the stake that maximises it

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What would a risk manager do with this?

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