Hedge Funds puzzles, solved step by step

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.

About 31% of the ending wealth. Rs 1 lakh growing at 10% a year becomes Rs 6.73 lakh in 20 years; at 8% after the fee it becomes Rs 4.66 lakh. The fee takes Rs 2.07 lakh, about 31% of what the investor would otherwise have had, although each year it looks like only a fifth of the return.

Why does a fifth of each year's return become almost a third of the wealth?

Think of a mango tree whose fruit you replant. If a neighbour takes one mango in every five each year, you lose those mangoes and every tree they would have grown. A fee taken every year removes money that would itself have compounded for the remaining years, so its cost grows faster than the fee rate suggests. The share lost is 1 minus (1.08/1.10) to the 20th: each year the investor keeps 98.2% of what the gross path would have held, and 0.982 to the 20th is about 0.69.

How do you estimate it without a calculator?

Use the rule of 72. At 10% money doubles about every 7.2 years, so 20 years is about 2.8 doublings, near 7 times; at 8% it doubles every 9 years, about 2.2 doublings, near 4.6 times. The exact numbers are 6.73 and 4.66, so the estimate lands within a few per cent, and 4.6 over 6.9 already tells you roughly a third is gone. Say the estimate first, then the exact figure.

What does an allocator take from this?

That fees should be judged against the return they leave, over the holding period, not as a percentage in a single year. A manager charging 2% needs to beat a cheaper alternative by about 2 points a year, every year, just to leave the investor level; over 20 years the difference is 31% of the pot. Hedge fund fees often add a share of profits on top, which widens the gap further. The illustration assumes a steady 10%; real returns vary, but the compounding of the fee does not depend on that.

About 31 managers, 1,000 halved five times. Each year roughly half the unbeaten managers beat the market by luck, so 500 survive year one, 250 year two, then 125, 62.5 and 31.25. A perfect five-year record is something luck hands to about 3 managers in every 100, so in a large crowd it cannot on its own separate skill from chance.

Why does a crowd produce streaks even without skill?

Fill a stadium with a thousand people and ask each to toss a coin five times. Someone will throw five heads, and about 31 will. A result that is rare for one person is almost certain somewhere in a large group, so the question is never whether a flawless record exists but how many you would expect by chance. Each manager's chance is 1 in 32; across 1,000 managers that is 31.25 expected perfect records.

If some managers really are skilled, how much does a perfect record tell you?

Suppose 5% of the thousand are skilled and beat the market 60% of the time. They produce about 3.9 perfect records, while the 950 unskilled produce about 29.7, so a manager with five perfect years is skilled only about 12% of the time. A 60% manager has only a 8% chance of five perfect years, so most skilled managers do not have flawless records either. The record is weak evidence in both directions.

What should you look at instead?

Longer records, more decisions per year and a reason. Skill shows up more reliably in many independent decisions than in a handful of annual outcomes, and in a process that explains where the edge comes from. Allocators also check how many managers were in the starting pool, because the funds still reporting are the ones that survived; the ones that were closed after bad years have dropped out of the data. That is survivorship biasThe distortion that comes from studying only the survivors of a process, whose results look better than those of the whole starting group., and it makes every surviving record look stronger than it is.

About 8.5x, not 10x. Rs 300 crore of the Rs 1,000 crore market value is cash, so the operating business is valued at Rs 700 crore. The cash earns 6% after tax, Rs 18 crore, so the operating business earns Rs 82 crore. Rs 700 crore over Rs 82 crore is about 8.5 times earnings. The headline 10x blends a cheaper business with cash priced at about 16.7 times its income.

Why is the headline P/E misleading here?

Imagine buying a shop for Rs 10 lakh that has Rs 3 lakh sitting in its till. You are really paying Rs 7 lakh for the shop itself. A company's price and earnings both include its cash, and cash is valued very differently from a business: here it earns 6% after tax, so it is worth about 16.7 times its income. Blending the two gives the headline 10x, which is neither the price of the cash nor the price of the business.

How do you work it?

Strip the cash out of both the numerator and the denominator. The price of the operating business is 1,000 minus 300, Rs 700 crore; its earnings are 100 minus 6% of 300, Rs 82 crore; and 700 over 82 is 8.54x. Taking the cash out of only the price gives 7.0x, which is too low because the Rs 100 crore still contains the cash's interest. The check: 700 plus 300 is 1,000 and 82 plus 18 is 100.

When would you not strip out all the cash?

When the cash is not really free. Cash needed to run the business, cash trapped abroad that would be taxed on the way home, or cash a management team is likely to spend badly may deserve less than full value. A long-short analyst uses the ex-cash multiple to compare this business with peers that hold no cash, and then asks how much of the Rs 300 crore shareholders will actually see. That judgement moves the answer between about 8.5x and the headline 10x.

An information ratio of about 1.0, and about 900 independent bets a year for 1.5. The fundamental law of active management says the information ratio is roughly the information coefficient times the square root of breadth: 0.05 x the square root of 400 = 0.05 x 20 = 1.0. To reach 1.5 the square root must be 30, so breadth must be 900, more than double, because breadth enters under a square root.

Why do many weak calls add up to a strong result?

Picture a cricket pundit who calls the winner right 52.5% of the time. On one match that is nearly useless; over hundreds of independent matches, the small edge becomes a steady record. With independent bets, the expected gain grows in proportion to the number of bets while the noise grows only with its square root, so the ratio of the two grows with the square root of the number of bets. An information coefficientThe correlation between a signal's forecasts and the returns that follow; for a simple up or down call it equals twice the hit rate minus one. of 0.05 is roughly that pundit's edge: a hit rate of 52.5%.

What does the square root mean for building a strategy?

Skill and breadth are not equal levers. Doubling the information coefficient doubles the information ratio; doubling breadth raises it only by about 41%, so matching a doubling of skill needs four times the bets. Going from 1.0 to 1.5 on breadth alone means 2.25 times as many independent bets, 900 against 400. That is why quant funds chase breadth across many stocks and short horizons, and why a small gain in forecast quality is worth so much.

What does the law leave out?

Two things that usually cut the answer. Independence is the hard part: 400 bets on stocks in one sector, or rebalanced so often that they repeat the same view, are far fewer than 400 independent bets. And constraints on position size, shorting and turnover stop a portfolio from fully expressing the signal; a transfer coefficientA number between 0 and 1 measuring how fully a constrained portfolio reflects the signal; it multiplies the fundamental law. of 0.6 would take the expected information ratio from 1.0 to 0.6. State the law, then say which of these you would check first.

4% of the book. If the stop is hit you lose 25% of the position, and that loss must not exceed 1% of the book. So the position times 25% equals 1%, and the position is 1% divided by 0.25, which is 4%. On a Rs 1,000 crore book that is a Rs 40 crore position and a Rs 10 crore loss at the stop.

Why is the answer not simply 1%?

Think of lending a friend money for a trip when you know the worst case is that a quarter of it never comes back. If you can stand to lose Rs 1,000, you can lend Rs 4,000, because only a quarter of the loan is at risk. The loss cap limits what you can lose, and the stop decides what fraction of the position you can lose, so the size is the cap divided by the stop distance. A trader who puts on 1% because the cap is 1% has confused the bet with the damage.

What happens to the size as the stop moves?

Tighten the stop and the position can grow; widen it and the position must shrink. With a 10% stop the same 1% cap allows a 10% position, and with a 50% stop only 2%. Every pair on the curve loses exactly 1% of the book when the stop is hit. This is why a trader whose thesis needs room to breathe, say through an earnings print, carries a smaller position than one working to a tight technical level.

What does the stop not protect you against?

The arithmetic assumes you get out at the stop price. A stock that gaps through the stop overnight, on results or a regulatory order, fills below it, and the loss exceeds the cap. If the stock opens 40% down, the 4% position loses 1.6% of the book, not 1%. Risk managers therefore size to the stop and then check the gap risk separately, often capping single-name positions whatever the stop says. Give that limitation straight after the number.

It makes no difference: switching and sticking each win half the time. List the six equally likely cases of car position and the host's random pick. Two of them reveal the car, and you have seen that they did not happen. The four that remain split two and two. The knowing host gives two thirds only because he never risks the car, which pushes those two cases into the switch column.

Where does the usual two-thirds answer come from?

In the standard game the host knows where the car is and always opens a goat door. Your first pick is right one time in three, and nothing the knowing host does can change that, so the other two thirds sit on the remaining closed door. His choice carries information because it is forced: when the car is behind door 2, he must open door 3, and when it is behind door 3, he must open door 2.

What changes when the host picks at random?

Picture a friend who does not know the answer to a quiz question and strikes out one option on a whim. If that option happens to be wrong, you have learned less than if someone who knew had struck it. The random host is that friend. Write out six cases: the car behind door 1, 2 or 3, each with the host's coin choosing door 2 or door 3. In two of the six the random host opens the car door, and the goat you saw rules those two out, which removes switch wins rather than stick wins.

Count what is left. The car behind door 1 survives both host choices: two cases where sticking wins. The car behind door 2 survives only when the host opened door 3, and the car behind door 3 only when he opened door 2: two cases where switching wins. Two against two.

Why would an interviewer want the intuitive answer broken rather than recited?

Because the lesson travels to every desk. The same observation carries different information depending on the process that produced it. A strong track record shown by a manager who launched ten funds and closed the nine that did badly is a host choosing which door to open for you. Before you update on evidence, ask whether the source could have shown you something else, and whether it chose what to show.

7/16, about 43.8%. Plot one trader's arrival on each axis of a 60 by 60 minute square. They meet when the arrivals are within 15 minutes, a band around the diagonal. They miss in the two corner triangles, each 45 by 45 over two, which together cover 45 squared over 60 squared, or 9/16. So they meet 1 minus 9/16, which is 7/16 of the time.

Why turn two arrival times into a square?

Think of two friends who say they will meet at a cafe sometime in the lunch hour. Every possible pair of arrival times is one point: the first friend's time across, the second's up. Because both times are uniform and independent, every point in the square is equally likely. A probability question about two uniform times becomes an area question, and the areas of triangles need no calculus.

How do you get the area without integrating?

Find the region where they miss, because it is two clean triangles. They miss when A arrives more than 15 minutes after B, or B more than 15 minutes after A. Each miss region is a right triangle with legs of 45 minutes, so each covers 45 x 45 / 2 = 1,012.5 square minutes out of 3,600. Together that is 2,025 out of 3,600, or 9/16, and the band where they meet is the remaining 7/16.

How does the answer move with the waiting time?

The formula shows the shape. Waiting 30 minutes gives 1 minus (30/60) squared, which is 75%, not double 7/16. The chance rises quickly at first and then flattens, because the band is squeezed at the ends of the hour, where someone arriving at 9:55 has only 5 minutes left to be joined. The same geometry answers questions about two orders arriving within a latency window, or two news releases landing in the same trading hour.

About 34.1%. Count the opposite. A hand with no ace is five cards from the 48 non-aces: 1,712,304 hands out of 2,598,960, or 65.9%. So at least one ace comes up about 34.1% of the time. Adding up exactly one, two, three and four aces reaches the same 886,656 hands, but takes four calculations instead of one.

Why count the hands without an ace?

If someone asks whether at least one of your five friends will be late to dinner, you do not add the chances of exactly one late, exactly two late and so on. You ask the chance that everyone is on time and subtract it from one. At least one is a collection of four separate cases, while none is a single case, so the complement turns four calculations into one.

Deal the cards one at a time. The first card misses the aces with chance 48/52, the second with 47/51, and so on down to 44/48. Multiply the five fractions and you get 0.6588. The same number is C(48,5) over C(52,5), which is 1,712,304 over 2,598,960.

Why is 5 x 4/52 wrong, and what does it actually measure?

Five times 4/52 is 38.5%, and it sounds reasonable. It adds up the chance that each card is an ace, which counts a hand with two aces twice and a hand with four aces four times. What it really gives is the expected number of aces in the hand, 0.385, which is always at least the chance of seeing one. The two drift further apart as the hand grows: deal fourteen cards and the same method gives more than 100%.

How do you check the answer the long way?

Exactly one ace: 4 ways to pick the ace times C(48,4) = 194,580 for the rest, 778,320 hands. Two aces: 6 x 17,296 = 103,776. Three: 4 x 1,128 = 4,512. Four: 48. The four cases sum to 886,656 hands, 34.1% of 2,598,960, matching the complement. Offer this as the check if there is time, never as the first route.

About Rs 12,500 crore. Each stock allows 5% of Rs 50 crore, Rs 2.5 crore a day. Across 200 stocks that is Rs 500 crore a day, and across 250 days Rs 1,25,000 crore a year. A book turned over 10 times a year needs 10 rupees of trading for every rupee of capital, so the book can be Rs 1,25,000 crore divided by 10. Treat it as a ceiling, not a working size.

What does capacity actually mean?

Picture a juice stall that can squeeze 500 glasses a day. How many regular customers can it serve? That depends on how often each one comes. If every regular drinks a glass a day, 500; if each comes once in ten days, 5,000. A strategy's capacity is the capital it can run before its own trading exceeds what the market will absorb, so it is tradable volume divided by how often the book must be traded.

How do you build the number out loud in the room?

Go one multiplication at a time and say each one. 5% of Rs 50 crore is Rs 2.5 crore per stock per day. Two hundred stocks make Rs 500 crore a day. Two hundred and fifty days make Rs 1,25,000 crore a year. Dividing by a turnover of 10 gives Rs 12,500 crore, and the step people drop is the last one: capacity is not a day's or a year's volume.

Why is turnover the lever that matters most?

Every input enters in proportion, but the market and the risk team set the others, while the strategy itself sets its turnover, and turnover varies far more across strategies. A strategy that turns over 50 times a year has a capacity of Rs 2,500 crore on this universe; one that turns over twice a year has Rs 62,500 crore. That is why fast strategies close to new money early and slow ones can run large books.

Why is the real capacity lower than this ceiling?

The estimate assumes the strategy trades evenly every day and in every stock. It does not. Trades cluster when signals fire, which is often when others trade too, and the thinner names in the list hit the 5% limit long before the larger ones. The definition of turnover matters as well: if 10 times means buying 10 times the book and also selling it 10 times, the traded value doubles and capacity halves to Rs 6,250 crore. Say which definition you used, and add that trading costs rise before the hard limit, so returns fade well before the ceiling.