Hedge Funds puzzles, solved step by step

21. The number of rolls until the first 6 averages 6, so there are on average 5 non-six rolls plus the 6. A roll known not to be a 6 is equally likely to be 1 to 5, so it averages 3, not 3.5, giving 5 x 3 + 6 = 21. Wald's identity confirms it: 6 expected rolls times 3.5 a roll is also 21, because a stop that looks only at past rolls does not bias the total.

Why is it tempting to get 23.5?

Suppose you keep buying scratch cards until one wins. Every card before the winner is, by definition, a loser, so those cards are worth less than an average card. The stopping rule changes the rolls before the stop: each one is known not to be a 6, so it averages 3, and valuing them at 3.5 overpays by 0.5 a roll, 2.5 in all. Five non-sixes at 3.5 plus a 6 is 23.5, which counts the high side twice: once in the 3.5 and again in the final 6.

How do the two routes agree?

Route one splits the sum: the expected number of non-six rolls times their average, plus the final 6. The count of rolls is a geometric wait with success chance 1/6, so it averages 6, of which 5 are non-sixes: 5 x 3 + 6 = 21. Route two is Wald's identityFor a stopping rule that uses only rolls already seen, the expected total equals the expected number of rolls times the average roll.: the expected total is the expected number of rolls times the average roll, 6 x 3.5 = 21. The low early rolls and the high final roll balance exactly.

Why would a trading firm ask this?

Stopping rules are everywhere on a desk: exit at the first stop-loss hit, rebalance at the first breach of a band. The question checks whether you can tell when a stopping rule biases what you observe, as it does for the early rolls, and when it does not, as for the total. Say the condition too: Wald's identity needs the decision to stop to use only rolls already seen, and the expected number of rolls to be finite. A rule that could peek at the next roll would break it.

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.

About 31% to finish above Rs 110, and about 62% to touch it during the year. A 10% move is half of one year's 20% standard deviation, and a normal variable ends more than half a standard deviation up 30.9% of the time. By the reflection principle, every path that touches 110 and ends below has a mirror twin that ends above, so touching is twice as likely as finishing above: 61.7%.

Why is finishing above 110 about a one in three chance?

Scale the move by the volatility. Over one year the price spreads out with a standard deviation of about Rs 20, so Rs 110 is half a standard deviation above the start, and a normal variable finishes more than half a standard deviation up 30.9% of the time. Treating the price as an arithmetic random walk is close enough for a 10% move; a lognormal model, in which prices cannot go negative, gives a slightly lower figure, about 28%. Say you are approximating, and say which way the error runs.

Why is touching twice as likely as finishing above?

Think of a walker on a foggy path who is equally likely to step forward or back. Once she reaches a marker post, her remaining steps are a fair coin again: from the post she is as likely to end past it as short of it. So for every path that touches 110 and ends below, reflecting the part after the touch gives an equally likely path that ends above; touching paths split evenly between the two. Every path that ends above must have touched on the way, so the chance of touching is twice the chance of ending above.

Where does the factor of two matter on a desk?

Anything that triggers on a touch rather than on the finish. A stop-loss set 10% away is hit about twice as often as the price ends beyond it, and an option that pays on a touch is worth roughly twice one that pays only if the price finishes past the same level. The limitation: the factor of two holds for a driftless, continuously watched walk. Drift, jumps and checking the price only at the daily close all move it, and a checked-daily barrier is touched a little less often than a continuous one.

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.

About 10.75% a year, for ever. In a constant-growth model the forward P/E equals the payout ratio divided by (required return minus growth). With a P/E of 40 and a payout of 0.5, r minus g must be 0.5/40 = 1.25%, so g = 12% - 1.25% = 10.75%. Retaining half its earnings, the company would need a return on equity of 21.5% for ever to fund that growth.

How does a P/E hide a growth assumption?

A flat that rents for Rs 30,000 a month and sells for Rs 1.2 crore is priced at 400 months of rent; a buyer paying that is quietly assuming the rent will grow. A price multiple is a compressed forecast: fix the return investors want and the share of earnings paid out, and the multiple pins down the growth the price needs. The constant-growth model, price equals next year's dividend over (r minus g), divided through by earnings, gives P/E = payout/(r - g).

Is 10.75% for ever plausible?

Test it against the business. Growth funded by retained earnings is return on equity times the share retained, so 10.75% growth with half the earnings kept needs a return on equity of 21.5%, held for ever. Few businesses hold returns like that for decades, and no company can outgrow the economy it sells into indefinitely, so compare the figure with the nominal growth you expect for that economy and say it as your assumption. The price is not wrong by arithmetic; it is demanding by assumption.

How sensitive is the answer?

Very. Because r - g is only 1.25%, every point on the required return moves the implied growth by a full point: at 11% the price implies 9.75%, at 13% it implies 11.75%. Using trailing rather than forward earnings shifts it too: 40 = 0.5(1 + g)/(0.12 - g) gives 10.62%. A high multiple rests on a thin gap between two large numbers, so small changes in either swing the value.

A proposes 98 for himself, 0 for B, 1 for C, 0 for D and 1 for E. Work backwards. With two pirates, D's own vote is half, so he keeps all 100. With three, C buys E with 1 coin. With four, B buys D with 1 coin. With five, A needs two votes beyond his own and buys the two pirates who get nothing in the four-pirate split, C and E, for one coin each.

Why start from the end?

Planning a train journey, you work back from when you must arrive, not forward from when you wake up. Each pirate votes by comparing the offer with what he would get if the proposal failed, so you can only price a vote once you know the next round's outcome, which means solving the smallest game first and working upwards. That method, backward induction, is the whole puzzle.

How does each round play out?

Two pirates, D and E: D proposes 100 for himself, and his own vote is half, so it passes. Three pirates: C needs one more vote and buys E, who gets nothing in the two-pirate game, for 1 coin: 99, 0, 1. Four pirates: B needs one more vote and buys D, who gets nothing in the three-pirate game: 99, 0, 1, 0. Five pirates: A needs two more votes and buys C and E, both empty-handed in the four-pirate game: 98, 0, 1, 0, 1.

What is the interviewer really testing?

Whether you reason about the alternative each party faces rather than about fairness. A vote costs exactly one coin more than what the voter gets if the deal fails, so the cheapest supporters are the ones with the worst fallback. The same logic runs through any negotiation: a creditor backs a restructuring plan when it beats their recovery in liquidation, and support is cheapest from those whose alternative is worst. State the assumptions: perfect rationality, and a pirate who is indifferent votes against, which is why one coin, not zero, is needed.

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.

About 25 trading days, roughly five weeks. The market trades 0.2% of the shares a day and you take at most a fifth of that, so you can sell 0.04% of the company a day. A 1% stake divided by 0.04% a day is 25 days. In the desk's language the position is five days of volume, and exiting it without moving the price takes a month.

What is the one division that answers it?

Emptying a water tank through a tap that you are only allowed to open a fifth of the way: how long it takes is the tank size divided by the flow you actually use. Days to exit equals position size divided by your daily selling capacity, and capacity is market volume times your participation cap. Here that is 1% over (0.2% x 20%), or 1% over 0.04%: 25 days. Keeping everything in percent of shares means you never need the share count or the price.

Why does a portfolio manager care about this number?

Because the price can move a long way in 25 days. A position you cannot exit quickly carries more risk than its daily volatility suggests: if the stock falls 2% a day for a week, you have sold only a fifth of it. That is why many desks cap a position at a set number of days of volume, often a few days, and why liquidity sits beside volatility in position sizing. The measure has a name, days to liquidatePosition size divided by the volume you can realistically trade in a day; a common liquidity limit on hedge fund books., and interviewers like hearing it.

Say the limitations. Volume is not steady; it dries up exactly when you most want to sell, and in a sell-off everyone is trying to do the same thing. A 20% participation rate also moves the price against you, so the real exit costs more than the screen price. A stronger answer adds that you would stress the calculation with half the normal volume, which gives 50 days.