Hedge Funds puzzles, solved step by step

Stop on the first roll only with a 5 or 6, on the second with a 4 or more, and the game is worth 14/3, about Rs 4.67. Work backwards. The last roll is worth 3.5. With two rolls left, keep 4, 5 or 6 and reroll otherwise: worth (4 + 5 + 6)/6 + 3/6 x 3.5 = 4.25. With three rolls left, keep only what beats 4.25, a 5 or 6: worth (5 + 6)/6 + 4/6 x 4.25 = 14/3.

Why start from the last roll?

Deciding whether to take a job offer is easier if you know what your fallback is worth. Each keep-or-reroll decision compares the roll in hand with the value of the rolls still to come, so you need the value of the future first, and the only stage with no future is the last one. On the last roll you must take whatever comes, which is worth 3.5 on average. That single number lets you solve the stage before it, and so on backwards. This is {term('backward induction', 'Solving a sequence of decisions from the last one to the first, using the value of each later stage to make the earlier decision.')}, the core of dynamic programming.

How do the thresholds come out?

With two rolls left, a roll of 4, 5 or 6 beats the 3.5 you expect from rerolling, and 1, 2 or 3 does not. So the two-roll game is worth the average of the kept faces times their chance, plus the chance of rerolling times 3.5: 15/6 + 1.75 = 4.25. On the first roll the fallback is now 4.25, so a 4 is no longer good enough: only 5 or 6 is kept. That gives 11/6 plus 4/6 x 4.25, which is 1.833 plus 2.833, or 14/3.

Add the pattern. More rolls always raise the value, but by less each time: 3.5, 4.25, 4.67, then about 4.94 with four rolls. An extra option is always worth something and never worth more than what it can still improve. That is the same logic as valuing a trade you can exit early: the right to wait is priced by what the future is worth, not by the average outcome.

14 drops. With k drops available, the first egg should go from floor k: if it breaks, the second egg checks the k minus 1 floors below one at a time. If it survives, you have k minus 1 drops left, so the next gap is one smaller. k drops therefore cover k + (k minus 1) + ... + 1 = k(k + 1)/2 floors. 13 drops cover 91 floors, 14 cover 105, so 14 is the minimum: drop from 14, 27, 39, 50 and so on.

Why does binary search fail here?

Binary search assumes you can keep testing after a failure. With two eggs, the first break leaves you one egg, and one egg can only be used safely by walking up one floor at a time. Once the first egg breaks, every floor below it that has not been ruled out costs one drop of the second egg, so large jumps with the first egg are expensive. Dropping the first egg at floor 50 and seeing it break could cost 49 more drops. It is like searching for a leak with one spare pipe: once the first one bursts, you test the rest slowly.

How do you balance the worst cases?

Make every worst case take the same number of drops. If you allow k drops in total, the first drop should be from floor k, the next k minus 1 floors higher, the next k minus 2 higher, because each first-egg drop used leaves one fewer drop for the second egg's walk. With k = 14 the first egg goes from 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99 and 100. If it breaks at 27, the second egg tests 15 to 26: 2 plus 12 is 14 drops. The same count holds at every step.

Say what an interviewer wants beyond the number. The move is to fix the budget of drops and ask how many floors it can cover, rather than fixing the building and searching for a strategy. That reversal is what makes the problem easy, and it generalises: with three eggs and k drops, the floors covered are the two-egg coverage for each smaller budget, plus one per drop, added up, which is why three eggs need only 9 drops for 100 floors.

The lift says they saw at least two heads, so the fair value is now at least 5.75, not 5; requote around 5.75 bid, 6.5 offered. The other seven flips are worth 3.5 heads, so the buyer's value is heads seen plus 3.5. Paying 5.5 only makes sense with two heads (5.5) or three (6.5). Those are 3 to 1 likely, giving 5.75; a buyer who needs a strict edge saw three heads, worth 6.5.

What is the trader's view before they trade?

A friend offers to buy your raffle ticket after the first few numbers are drawn. The offer itself is the warning. The informed trader values the contract at heads already seen plus 3.5, the expected heads in the seven unseen flips, so their value is 3.5, 4.5, 5.5 or 6.5 with chances 1, 3, 3 and 1 in 8. Against your market of 4.5 bid and 5.5 offered, they buy only if their value is at least 5.5, and sell to you at 4.5 only if it is 4.5 or less. The flat 5 you quoted around is right only for someone who has seen nothing.

How do you turn the trade into a new fair value?

Condition on the fact that they traded. Only the two-head and three-head worlds produce a buy at 5.5, and they are 3/8 and 1/8 likely, so given a lift the value is (3 x 5.5 + 1 x 6.5) / 4 = 5.75. If you assume they would not bother trading at zero edge, only the three-head world is left and the value is 6.50. Either way you sold too cheaply: this is adverse selectionThe tendency of a market maker to trade most with the people who know more, so the trades that happen are the ones that lose money for the market maker., and it is the cost every market maker prices into the spread.

Now requote. Your bid should not be below what you now believe the floor is, and your offer should sit where even the best informed buyer has no edge. Something like 5.75 bid, 6.5 offered does both: a buyer who saw three heads is indifferent at 6.5, and you are no longer selling below value. Cut your size too, because you know someone is trading with more information than you, and say you would ask whether they could see the flips before quoting again.

Roughly 1.2, not 4.0. Consecutive 12-month returns share 11 months, so 240 monthly rows over 20 years hold only about 20 independent observations. OLS standard errors assume independence and come out too small by roughly the square root of the overlap, root 12, about 3.5. Dividing 4.0 by 3.46 gives about 1.15: the result is no longer significant. A Newey-West or Hansen-Hodrick standard error does this properly.

What does the overlap do to the regression?

Asking twelve friends for restaurant advice sounds like twelve opinions, but if eleven of them only repeat what the first one said, you have heard about one. Each 12-month return shares 11 months with its neighbour, so the rows are mostly the same data counted again, and the regression thinks it has twelve times more independent evidence than it does. The slope estimate is not biased by the overlap. What breaks is the standard error, because the residuals are strongly correlated from one row to the next, and that breaks one of the {term('OLS assumptions', 'The conditions under which ordinary least squares standard errors are correct, including residuals that are uncorrelated across observations.')}.

Why divide by root 12 and not by 12?

The standard error scales with one over the square root of the number of independent observations. If the effective sample is twelve times smaller, the standard error is root 12, about 3.46, times larger, and the t-statistic is 3.46 times smaller: 4.0 becomes about 1.15. This is a rough correction. The exact factor depends on how persistent the signal is: for a slow-moving signal, such as a valuation ratio, it is close to root 12; for a fast-moving one it can be smaller.

Say how you would fix it properly: use Newey-West standard errors with at least 11 lags, or Hansen-Hodrick errors built for exactly this overlap, or run the regression on non-overlapping annual data and accept the smaller sample. Any of those should give a t-statistic well below 4.0, and a researcher who reports only the OLS number has not yet shown the signal works.

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.

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..

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.

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.

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.

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.

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..

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.

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.

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.

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.

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.