Hedge Funds puzzles, solved step by step

The ten perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. Locker n is toggled once by each person whose number divides n, so its final state depends on how many divisors n has. Divisors come in pairs, d and n/d, which cancel out. Only a perfect square has an unpaired divisor, its square root, so only squares are toggled an odd number of times and end open.

What decides whether one locker ends open?

A light switch flipped an even number of times ends where it started; flipped an odd number of times, it ends the other way. Each locker is a switch, flipped once for every divisor of its number, so the question is which numbers from 1 to 100 have an odd number of divisors. Locker 12 is touched by persons 1, 2, 3, 4, 6 and 12, six times, and ends closed.

Why do only perfect squares have an odd number of divisors?

Pair every divisor d with n divided by d. For 12 the pairs are 1 and 12, 2 and 6, 3 and 4: six divisors, even. The pairing breaks only when a divisor is paired with itself, d = n/d, which happens exactly when n is a perfect square. For 36 the pairs are 1 and 36, 2 and 18, 3 and 12, 4 and 9, with 6 left over: nine divisors, odd, so locker 36 ends open. There are ten squares up to 100, so ten lockers.

Why does a fund ask a puzzle like this?

It tests whether you look for structure before you simulate. Walking through a hundred people toggling lockers is hopeless in an interview; turning it into a question about divisors takes one sentence and makes the answer obvious. The move, recasting a process as a property you can count, is the one that turns a messy trading rule into a quantity you can compute. Check it on a small case out loud: with 10 lockers, 1, 4 and 9 end open.

Yes: the direct quote is rich by 60 paise, so buy euros through dollars and sell them for rupees at 93.00, making Rs 6 lakh on EUR 1 million before costs. The implied rate is 1.10 x 84.00 = 92.40 rupees per euro. Spend Rs 9.24 crore on USD 1.1 million, turn that into EUR 1 million, and sell the euros for Rs 9.30 crore. You finish in rupees, where you started, with Rs 6 lakh more.

How do you spot the mispricing in one line?

If one stall sells mangoes at Rs 100 a dozen and the stall next to it buys them back at Rs 10 each, you would buy dozens and sell singles all day. Every pair of currencies can be priced two ways, directly or through a third currency, and when the two prices differ you buy on the cheap route and sell on the rich one. Through dollars a euro costs 1.10 dollars at Rs 84.00 each: Rs 92.40. The direct dealer pays Rs 93.00. That 60 paise gap is the whole trade.

What stops this from being free money in practice?

Three things. Real quotes have a bid and an offer, and the gap survives only if it is wider than the three spreads you cross plus the cost of dealing. A 60 paise gap on a 92 rupee price is about 0.65%, far wider than dealer spreads in major currencies, which is why a gap that size would be traded away almost at once. And the legs must be done together: if one price moves before you finish, you are left holding an open currency position instead of a locked-in profit.

Weight them 9/13 and 4/13, about 69% and 31%, and the blend's error falls to about 1.66%. For independent unbiased forecasts the best weights are proportional to one over each error variance: 1/4 for the 2% forecast and 1/9 for the 3% one. The blended error variance is 1 over (1/4 + 1/9), which is 36/13, so its standard deviation is 1.66%, better than either forecast alone.

Why does blending two forecasts beat the better one?

Ask two people to guess the weight of a pumpkin at a village fair. One is usually closer, but their mistakes are unrelated, so averaging tends to cancel part of each. Independent errors partly cancel when you average, so even a weaker forecast adds information, provided it gets a smaller weight. Throwing the 3% forecast away leaves you at 2%; blending it in well gets you to 1.66%.

Why inverse variance and not inverse error?

The blend's error variance is w squared times 4 plus (1 minus w) squared times 9. Setting its slope to zero gives w = 9/(4 + 9), so the weights follow one over the variance, which penalises the noisier forecast harder than one over the standard deviation would. Check the alternatives: equal weights give an error of 1.80%, weights of 3/5 and 2/5 give 1.70%, and the inverse-variance weights give the minimum, 1.66%.

What would you check before trusting the blend?

Two assumptions carry the answer. The forecasts must be unbiased and their errors independent; if both forecasters lean on the same survey, their errors are correlated and the gain from blending shrinks. With an error correlation of 0.5, the best blend gives the sharper forecast 6/7 of the weight and improves the error only from 2.00% to 1.96%. Ask where each forecast comes from before you average them.

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.

One in four. Three lengths make a triangle only if no piece is longer than the other two together, which for a stick of length 1 means no piece is longer than one half. Plot the two break points as a dot in the unit square. The dots that keep every piece under half fill two triangles of area 1/8 each, so the chance is 1/4.

What exactly has to be true for a triangle?

Try to make a triangle from three sticks where one is longer than the other two laid end to end: the short ones cannot reach each other across the long one. Three lengths form a triangle only if the longest is shorter than the sum of the other two, and with a total of 1 that means every piece must be under one half. That turns a geometry question into a single condition you can test on any pair of break points.

How do you turn two random breaks into an area?

Call the break points x and y, each uniform between 0 and 1. Every possible outcome is a dot in the unit square, and every dot is equally likely, so probability is area. Take the half where x is below y: the pieces are x, y minus x and 1 minus y, and all three are under half only when x is below 1/2, y is above 1/2 and y minus x is below 1/2. Those three lines cut out a triangle with corners (0, 1/2), (1/2, 1/2) and (1/2, 1), area 1/8. The half where y comes first gives its mirror image, another 1/8.

The red regions are also worth naming, because each one is a different piece being too long. Three failure regions of area 1/4 each are the first piece, the middle piece and the last piece exceeding half. They cannot overlap, since two pieces cannot both be longer than half, so the answer is 1 minus 3/4, which is 1/4, a second route to the same number.

Mention the variant, because interviewers often switch to it: break the stick once, then break the longer piece at a random point. The answer changes to about 0.386, which is 2 ln 2 minus 1, because the second break is no longer uniform over the whole stick. Knowing the setup changes the answer shows you are reading the question.

About 3.99, so roughly four. Give each face its own yes-or-no: did it appear at least once? A face is missed in all six rolls with chance (5/6)^6, about 33.5%, so it appears with chance about 66.5%. The expected count is the sum of the six chances: 6 x (1 minus (5/6)^6), which is 3.9906.

Why not work out the chance of seeing exactly 1, 2, up to 6 faces?

You could, but each of those needs a fiddly count. There is a shortcut that works whenever the question asks for an expected count. Think of a teacher checking attendance: the expected number present is simply the sum, over pupils, of each pupil's chance of turning up. Write the total as a sum of yes-or-no indicators, one per face, and the expected total is the sum of their probabilities, whether or not they are independent. That rule is {term('linearity of expectation', 'The expected value of a sum equals the sum of the expected values, even when the parts depend on each other.')}, and it turns a hard count into one line.

What is the chance that a given face shows up?

It is easier to find the chance it does not. Each roll misses the face with chance 5/6, and the six rolls are independent, so the face is missed every time with chance (5/6)^6, which is 15,625 over 46,656, about 33.5%. So each face appears at least once with chance about 66.5%, and six faces give 3.99. The indicators are not independent, since seeing face 1 a lot leaves less room for face 2, but linearity does not care.

Give the general form and a sense check. With n faces and n rolls the answer is n(1 minus (1 minus 1/n)^n), which tends to n(1 minus 1/e), about 63% of n. For a die that predicts 3.79, and the exact 3.99 is a little higher because six is small. The same logic tells a desk how many distinct names a random sample of trades touches.

Both end up side by side in the middle, and customers are worse off. From the quarter points, a seller who steps inward keeps everyone behind them and wins beach from the rival, so both drift to the centre. There they still split customers half and half, but the average walk doubles from 125 m to 250 m. Competition moves the sellers to the spot that maximises share, not the one that serves customers best.

Why can neither seller stay at the quarter points?

Picture two petrol pumps on a highway. Each wants the drivers on its own side plus as many from the middle as it can reach first. A seller keeps every customer on the far side of them wherever they stand, so moving towards the rival only ever adds customers. From 250 m, A steps to 400 m against B at 750 m: the dividing line moves to the midpoint, 575 m, and A's share rises from 50% to 57.5%. B then responds the same way, and the dance ends only when both stand at 500 m.

What does the middle cost the customers?

With both sellers in the centre, a sunbather is on average a quarter of the beach away, 250 m. With sellers at 250 m and 750 m, nobody is more than 250 m away and the average is 125 m. The shares are identical in both setups; the only thing that changed is how far customers walk, so the stable outcome is strictly worse for them and no better for the sellers. This is the {term('Hotelling model', 'A model of competition on a line, set out by Harold Hotelling in 1929, in which rivals crowd towards the centre to win the middle ground.')}, and it is why rival shops cluster and why two parties often converge on the middle voter.

Say the equilibrium idea in one line: a pair of positions is stable when neither player can do better by moving alone, and the middle is the only such pair here. Then say where the model breaks: if customers stop buying when the walk is too long, or if prices can differ, the sellers have a reason to spread out again. Interviewers ask this to see whether you can reason about another player's best response, which is most of trading.

Move the mid from 10.5 to 13 and tighten the market by about a fifth. Each die averages 3.5, so three dice average 10.5. Once one die shows 6, the sum is 6 plus two unknown dice averaging 7, which is 13. The variance falls from 3 x 35/12 to 2 x 35/12, so the standard deviation drops from 2.96 to 2.42. If the first market was 9.5 at 11.5, the new one is about 12.2 at 13.8.

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

Start from the fair value and then decide the width from how uncertain the outcome is. The mid is the expected sum, 3 x 3.5 = 10.5, and the width should scale with the standard deviation of the sum, because that is how far the answer typically lands from the mid. One die has variance 35/12, so three independent dice have 35/4 = 8.75, a standard deviation of 2.96. A market of 9.5 bid, 11.5 offered is a reasonable opening: tight enough to trade, with room for your edge.

What does revealing one die change?

A weather forecast for tomorrow is more precise than one for next week, because fewer things can still change. Once one die is known, it contributes a certain 6 and no uncertainty, so the mid rises by 2.5 and only two dice of variance remain. The mid becomes 6 + 7 = 13. The variance becomes 35/6, a standard deviation of 2.42, down from 2.96. Scale the width by the same ratio, about 0.82, and a 2.0 wide market becomes about 1.6 wide: 12.2 at 13.8.

Say what would make you widen instead. If the person revealing the die can choose which die to show, or picks the moment, the reveal itself carries information and you should be more careful, not less. A 6 chosen as the highest of three tells you the other two are 6 or lower, and they no longer average 7. Interviewers like it when you ask who chose what to reveal before you requote.

About Rs 69, so the stock must move about 6.9% either way. Scale the 30% annual volatility to one month by the square root of time: 30% x root(1/12) is about 8.7%. An at-the-money straddle is worth about 0.8 times that move times the price: 0.8 x 8.7% x 1,000 is about Rs 69. At expiry the buyer profits only below about Rs 931 or above Rs 1069.

Where does the 0.8 x sigma x root T rule come from?

A straddle pays the size of the move, whichever way it goes, so its value is the expected absolute move. For a normal distribution, the average absolute move is the standard deviation times root(2/pi), which is about 0.8. Over one month the standard deviation of the price is S x sigma x root T = 1,000 x 0.30 x root(1/12), about Rs 87. Times 0.8 gives about Rs 69. Black-Scholes with zero rates gives Rs 69.1, so the shortcut is good to within a rupee for short-dated at-the-money options.

Why is the breakeven move bigger than the typical move traders expect?

Buying a straddle is like buying insurance against a big move in either direction: you pay the average claim up front. The breakeven at expiry is the premium, about 6.9%, while the average absolute move is the same 6.9% by construction, so a buyer at fair implied volatility breaks even only on average. The buyer makes money when realised volatility turns out higher than the 30% priced in, and the seller makes money when it turns out lower. That is why desks talk about straddles as a bet on volatility, not on direction.

Say the limits. The rule assumes the option is at the money and short dated, and rates are small; for long-dated or out-of-the-money options it drifts. A single call or put is half the straddle, about 0.4 x sigma x root T x S, which is a useful second number to have ready.