Hedge Funds puzzles, solved step by step

Do not spin: you survive 75% of the time, against 66.7% if you spin. The empty click puts the cylinder on one of the four empty chambers, each equally likely. Because the two bullets sit together, three of those four empties are followed by another empty and only one is followed by a bullet. A fresh spin throws that information away and gives four chances in six.

What does the empty click tell you?

Picture six people in a queue where two friends always stand together. Pick someone at random who is not one of the friends and ask whether the person behind them is a friend. Only one of the four has a friend behind them: the one standing just in front of the pair. The empty click is information: it tells you which chambers you could be on, and using it is the whole puzzle. The cylinder has no memory, but you do.

Why does the answer flip if the bullets are not adjacent?

Separate the bullets, say into chambers 1 and 4. The four empties are now 2, 3, 5 and 6, and the chambers after them are 3, 4, 6 and 1: two empties and two bullets. Not spinning now survives only 2 times in 4, 50%, so spinning, at 66.7%, becomes the better choice. Adjacency is what bunches both bullets behind a single empty chamber. Ask where the bullets sit before you answer, and say that the answer depends on it.

Where does this reasoning show up on a desk?

The same move, updating on what you have just observed instead of resetting to the base rate, is how a trader reads a fill. Getting filled on your bid tells you something about who was selling, just as the empty click tells you which chamber you are on. Ignoring it is the equivalent of spinning the cylinder: it feels neutral, but it throws away an edge you were handed for free.

The probability is 7/8, and the density of X + Y is a triangle, a tent peaking at 1. Because X and Y are independent and uniform, every point of the unit square is equally likely, so probability is area. The line x + y = 1.5 slices off a corner triangle with legs of 0.5, area 1/8. The sum's density rises in a straight line from 0 to 1 and falls back to 0 at 2.

Why does probability become area here?

Throw a dart at a square board so that every point is equally likely to be hit. The chance it lands in a region is that region's share of the board. Two independent uniforms are exactly that dart: the pair (X, Y) lands evenly on the unit square, so any question about X and Y becomes a question about an area. The condition X + Y below 1.5 is everything under the line x + y = 1.5, which is the whole square except one corner.

How do you get the shape of the sum's density?

Slide the line x + y = s across the square and watch how long it is inside. Near s = 0 it barely clips the corner; at s = 1 it runs corner to corner, the longest it gets; past 1 it shortens again. The density of the sum at s is proportional to the length of that line inside the square, which gives a triangle rising from 0 to a peak at 1 and falling to 2. This is the convolutionThe density of a sum of independent variables, found by adding up every way the two parts can combine to the same total. of two flat densities, and the same reason two dice most often total 7.

Check the first answer with the tent. The area beyond 1.5 is a triangle with base 0.5 and height 0.5, which is 1/8 again. Two routes that agree is the check an interviewer wants to hear before you commit. Add a third uniform and the density becomes three joined curved pieces; add many and the sum looks normal, which is the central limit theorem arriving in slow motion.

997 x 1,003 = 999,991 and 67 squared = 4,489. The first is a difference of squares: the numbers sit 3 either side of 1,000, so the product is 1,000,000 minus 9. The second uses the same identity the other way round: 67 squared is 64 x 70 plus 3 squared, which is 4,480 plus 9. Both take one line once you spot the round number nearby.

Why does multiplying around a round number work?

Take a square garden 10 metres a side and reshape it to 13 by 7. The fence is the same length, but the plot shrinks from 100 square metres to 91. It always shrinks by the square of how far you moved each side, here 3 squared, 9. Two numbers spaced equally around a midpoint multiply to the midpoint squared minus the gap squared. For 997 x 1,003 the midpoint is 1,000 and the gap is 3, so the answer is 1,000,000 minus 9.

How do you square a number like 67 in your head?

Push it to a round neighbour and repair the difference. Move 3 down to 64 and 3 up to 70, multiply those, then add back the 3 squared you took away: 64 x 70 = 4,480, plus 9 is 4,489. You choose the gap so one factor is round. The expansion route agrees: 67 is 70 minus 3, so 67 squared is 4,900 - 420 + 9, the same 4,489. Two methods landing on one number is your check.

Why would a fund ask arithmetic at all?

A trader checks prices, spreads and position sizes in their head all day, and a slip costs money before any spreadsheet catches it. Interviewers use speed on arithmetic like this as a proxy for how quickly you would catch a quote that does not add up. Say the identity as you use it, so that if you slip, the interviewer can see where and you can recover out loud.

Randomise: pay Rs 3.15 with probability 0.159 and Rs 3.14 otherwise. Pi is 3.14159..., which sits 0.159 of the way from 3.14 to 3.15. Paying the higher amount with exactly that probability makes the expected payment 3.14 + 0.01 x 0.1593..., which is pi. So the chance you pay Rs 3.15 is about 15.9%, and over many meals nobody is short-changed.

Why can no fixed amount be fair?

Always round to Rs 3.14 and the restaurant loses 0.159 paise every time; always pay 3.15 and you overpay 0.841 paise. Any fixed amount is unfair to one side, so the only way to be exactly fair is to be fair on average. Two friends who split a Rs 101 bill by taking turns to pay the odd rupee are doing the same thing: neither is exact on any one night, both are exact over time.

How do you actually draw a probability of 0.159?

Use any randomness you can split finely. Draw a uniform number between 0 and 1 and pay Rs 3.15 if it falls below 0.1593. With only a die, paying 3.15 on a six gives 1/6, an expected payment of Rs 3.141667: close, not exact. With only a fair coin you can be exact: toss it to generate the binary digits of a uniform number one at a time and stop as soon as the digits so far settle which side of 0.1593 it falls. Each toss settles it with probability one half, so on average it takes two tosses.

Where does randomised rounding show up in a fund?

Whenever a quantity has to be split in whole units. A fund allocating 1,003 shares across three accounts cannot give each 334.33; handing the odd share out by lottery, or in rotation, keeps each account fair on average. The principle is the same: when the exact amount is impossible, make the expected amount exact and keep the error unbiased. The limitation is that fair on average is not fair every time, which is why allocation policies also cap how far any account can drift.

One round trip makes at most 16; unlimited round trips make 26. For one trade, walk the prices once, carrying the lowest price so far and the best sale against it: buy at 96, sell at 112. For many trades, add every day-on-day rise and skip every fall: 8 + 11 + 7 = 26. Without short selling the falls are simply sat out, never profited from.

How do you find the best single trade without checking every pair?

Imagine walking down a street of shops that all sell the same phone, planning to buy once and sell once further along. You do not need to compare every pair of shops: carry the cheapest price seen so far in your head, and at each shop ask what selling here would make against it. One pass, keeping the running minimum and the best gap found so far, gives the best single trade. Here the running minimum drops to 96 on day 2 and the best gap appears on day 7: 112 - 96 = 16.

Why is the many-trade answer just the sum of the rises?

Any rise from a low to a later high is the sum of the daily steps inside it, and some of those steps may be falls. With no short selling and no costs, the most you can make is the total of every positive day-on-day move, 26 here, because trading only the up steps collects everything a longer trade would and skips its falls. In practice that is three round trips: buy 96, sell 104; buy 99, sell 110; buy 105, sell 112.

What does the interviewer add next?

Costs. Once each round trip costs something, the sum of rises overstates the profit, because small moves stop being worth trading. With a cost of 6 per round trip, the three separate trades net 2 + 5 + 1 = 8, the best two-trade split nets 9, and the single trade from 96 to 112 nets 10, so the single trade now wins. The general version is a short dynamic programme that tracks the best profit on each day while holding and while flat.

The price falls by about 2.94%. Spread duration of 6 says a 0.50 percentage point widening costs 6 x 0.50% = 3.00%. Convexity of 50 adds back one half x 50 x 0.005 squared, about 0.06%, because the price curve bends upwards. On a bond priced at 100 that is a move to about 97.06. At 50 basis points the convexity term is small; at 300 or 500 it is not.

What do duration and convexity each measure?

Picture a playground slide that curves and flattens towards the bottom. Judge the drop from the steepness at the top and you overstate it, because the slide levels off as you go. Spread durationThe percentage change in a bond price for a one percentage point change in its credit spread, holding the risk-free rate fixed. is the steepness at today's spread; convexity is the flattening, so the straight-line estimate always overstates the loss when spreads widen. Duration gives the first-order move, 6 x 0.50% = 3.00% down; convexity corrects it by a term that depends on the square of the move.

When does the convexity term start to matter?

It grows with the square of the move. At 50 basis points convexity is worth 0.06% against a 3.00% duration loss; at 300 basis points duration says -18% and convexity adds back 2.25%, which is no longer small. That is why a credit desk can run duration-only risk for everyday moves but needs convexity for stress scenarios. One more distinction marks a strong answer: for a fixed-coupon bond spread duration and rate duration are close, but a floating-rate note has almost no rate duration and still carries several years of spread duration.

Say the limitation plainly. Both numbers are local, measured at today's spread, and a distressed bond stops behaving like this long before default, when its price starts tracking the expected recovery instead. For a bond trading near par, as here, the two-term estimate is good to a few hundredths of a per cent for moves of this size.

The book's volatility is about Rs 23.1 crore a year, and the Rs 60 crore position carries about 74% of it on 60% of the capital. Stand-alone risks are Rs 18 crore and Rs 8 crore. Book variance is 18 squared plus 8 squared plus 2 x 0.5 x 18 x 8, which is 532, so volatility is Rs 23.07 crore. Each position's contribution is its covariance with the book over the book's volatility: Rs 17.17 crore and Rs 5.90 crore, which add back to the total.

Why is risk not shared out like capital?

Two friends share a taxi. One rides twice as far, straight through the traffic jam; the other gets off after a short hop. Splitting the fare by the number of bags each carries would be absurd. Risk belongs to a position in proportion to how much it moves and how much it moves with everything else, not to how much money sits in it. Here the first stock is larger, more volatile and positively correlated with the second, so it carries far more than its 60% of the capital.

Why do the contributions add up exactly to the total?

Split the variance. The cross term, 2 x 0.5 x 18 x 8 = 144, is shared equally, 72 to each position. So position A owns 324 + 72 = 396 of the 532 of variance and position B owns 64 + 72 = 136, and dividing each by the book's volatility of 23.07 gives rupee contributions that sum exactly to Rs 23.07 crore. The diversification benefit is the gap between the stand-alone total of Rs 26 crore and the book's Rs 23.07 crore.

What does a risk manager do with the split?

Cut where the risk is, not where the money is. Each rupee in position A carries 28.6 paise of marginal risk against 14.7 paise in position B, so trimming Rs 10 crore from A lowers book volatility by roughly Rs 2.9 crore. Recomputing exactly gives Rs 2.84 crore, close to the estimate. The limitation: the split is a snapshot at one correlation, and when correlations move, both the total and the split move with them.

The MLE is 9.0, the largest observation; it is biased low, and the unbiased correction is 5/4 x 9.0 = 11.25. The likelihood is 1 over theta to the fourth for any theta of at least 9.0 and zero below it, so it peaks at the sample maximum. But the maximum of n draws averages n/(n + 1) of theta, here 4/5, so scaling by (n + 1)/n removes the bias.

Why is the MLE the largest observation?

A friend draws four raffle tickets numbered from 1 up to some unknown top number, and the highest you see is 90. The top number is at least 90; guessing higher only spreads your belief over tickets nobody drew. The likelihood, 1 over theta to the n, is zero for any theta below the largest observation and falls as theta rises above it, so it is maximised exactly at the sample maximum, 9.0. This is a case where you do not differentiate: the maximum sits on a boundary, not where a slope is zero.

Why is it biased, and by how much?

The largest draw can never be above theta and is almost always below it. Four points dropped at random on 0 to theta cut it into five gaps of equal expected length, so the largest point sits on average four fifths of the way up, and the MLE underestimates theta by a fifth on average. Multiplying by 5/4 fixes it: 9.0 becomes 11.25. The bias shrinks as n grows, since n/(n + 1) tends to 1, but with four points it is large.

How does it compare with the obvious alternative?

The method of moments doubles the sample mean, since a uniform on 0 to theta averages theta/2: the mean here is 6.175, giving 12.35. Both 11.25 and 12.35 are unbiased, but the corrected maximum has a much smaller variance, theta squared over n(n + 2) against theta squared over 3n, because the largest draw carries the most information about the top of the range. With four points that is theta squared over 24 against theta squared over 12: half the variance.

Take B, C and D: 10 units for Rs 21 crore. Ranking by profit per unit of risk picks B (2.33), then A (2.25), then D (2.00), which uses 9 units for Rs 20 crore and leaves one unit idle. Swapping A for C costs a little ratio but fills the budget and adds Rs 1 crore. With whole ideas, you check the few sets that fill the budget rather than trust the ratio alone.

Why is profit per unit of risk the right place to start?

Packing a suitcase for a flight with a weight limit, you favour the things that give the most use per kilo. When the budget is the scarce thing, the useful measure is profit per unit of budget, not profit alone. E earns the most in rupees but only 1.83 per unit, the worst of the five. B earns 2.33 per unit, A 2.25, and C and D 2.00 each. If ideas could be split, you would simply fill from the top of that list and the answer would be exact.

Why does the ratio ranking fail here?

Ideas come whole. Take B and A and you have used 7 units; D fits for 9 units, and nothing else fits in the last one. A unit left idle earns nothing, so a set with a slightly lower average ratio that uses the full budget can beat the greedy pick. Replace A (4 units) with C (5 units) and the budget is full: B, C and D earn 7 + 10 + 4 = Rs 21 crore against Rs 20 crore. This is the knapsack problemChoosing whole items, each with a size and a value, to maximise total value without exceeding a capacity., and the fix is to check the handful of sets that nearly fill the budget.

With five ideas there are only 31 possible sets, so enumerate the full ones out loud: B, C, D for 21; A, B, D for 20; A, E for 20; A, C for 19. On a real book with hundreds of positions a desk uses an optimiser, but the logic is the same, and interviewers want to hear that you know the ratio rule is a starting point and not a proof.

Say the limitation too. Expected profit is an estimate, and the gap here is Rs 1 crore on Rs 21 crore. If C's estimate is shakier than A's, a portfolio manager could reasonably prefer the greedy set. The arithmetic tells you the best set on the stated numbers; confidence in each number decides whether that edge is real.

About 15%, not 90%. Picture 10,000 trades. 100 are suspicious and the screen flags 90 of them. 9,900 are clean and it wrongly flags 5%, which is 495. So 585 trades are flagged and only 90 are suspicious: 90 over 585 is 15.4%. The false flags swamp the true ones because clean trades are so common.

Why is 90% the wrong number?

A smoke alarm that goes off whenever there is a fire is good. But if it also goes off every time someone makes toast, most of its alarms are toast. The 90% tells you how often a suspicious trade gets flagged; the question asks how often a flag is suspicious, and those run in opposite directions. Mixing them up is called the {term('base rate', 'How common something is before any test is run. Here, one trade in a hundred is suspicious.')} fallacy, and it is exactly what this question is built to catch.

How do you get the number without the formula?

Use counts, not percentages. Start with 10,000 trades because it makes every number whole. Split by the truth first, then by what the screen says, and then read only the flagged column. 1% of 10,000 is 100 suspicious trades; 90% of those, 90, are flagged. 9,900 are clean; 5% of those, 495, are flagged anyway. The flagged column holds 585 trades, and 90 of them are the real thing.

Add the desk point after the number. A second, independent check changes things fast: run the 585 flagged trades through a second screen with the same error rates and the 15.4% prior becomes about 77%. A weak test is still useful as a first filter; it is only misleading when its hit rate is read as its accuracy.