Hedge Funds puzzles, solved step by step

One in three if you learn at least one is a girl, and one in two if you learn the elder is a girl. List the four equally likely families by birth order: boy-boy, boy-girl, girl-boy, girl-girl. At least one girl rules out only boy-boy and leaves three, one of which is girl-girl. Naming the elder rules out two orders and leaves girl-boy and girl-girl.

Why is the answer not simply one half?

Picture a friend tossing two coins behind a screen and telling you that at least one came up heads. She has told you something about the pair, not about a particular coin, so you cannot treat the other coin as a fresh toss. The pair had four equally likely outcomes and her remark removes only one of them, tails-tails. Two children work the same way, provided each birth is equally likely to be a boy or a girl and the two births are independent.

What changes when you learn that the elder is a girl?

Now the information is about a named child. Naming the child removes a whole row of the grid, both orders in which the elder is a boy, so two cells survive and the answer becomes one half. The younger child's sex is untouched by what you learned, which is why it now behaves like a fresh toss. The same arithmetic, 1/4 divided by 1/2, gives 1/2.

Why would a fund interviewer ask this?

Because every piece of market news arrives through a filter, and the filter changes what the news means. How you came to learn a fact is part of the fact. A screen that says at least one of two stocks beat estimates tells you less about each stock than a screen that names the one that did. Draw the grid, give both answers, and say the assumption out loud: independent births, each equally likely to be a boy or a girl.

40%. Measure everything in units of the average loss. Each trade wins 1.5 units with probability p and loses 1 unit otherwise, so the expected result per trade is 1.5p minus (1 - p). Setting that to zero gives p = 1/2.5 = 40%. At a 45% hit rate the book makes 0.125 units a trade, so a trader who is wrong more often than right can still run a good business.

Why is a hit rate below 50% not a problem on its own?

A street vendor selling umbrellas can stand idle most days and still do well, because the rainy days pay for the dry ones. What decides whether a trading book makes money is the hit rate times the size of the wins against the miss rate times the size of the losses, not the hit rate alone. A win-loss ratio of 1.5 moves the breakeven from 50% down to 40%.

What does the general rule look like?

For a win-loss ratio R the breakeven hit rate is 1 over (1 + R). At R = 1 it is 50%, at R = 1.5 it is 40%, at R = 2 it is 33% and at R = 3 it is 25%. This is why portfolio managers are reviewed on both numbers together. A falling hit rate is fine if the winners are running further, and a rising hit rate is a warning sign if it comes from cutting winners early and letting losers run.

What would you add before calling 45% a good business?

Costs and sample size. Commission, slippage and financing come off every trade, winners and losers alike, so they raise the breakeven. If costs run at 0.05 units a trade, the book needs 1.5p - (1 - p) = 0.05, a hit rate of 42%. And a ratio of 1.5 measured over twenty trades is noisy; say you would want a longer record before trusting either number.

They are exactly equal: 40 ml of wine sits in the water glass and 40 ml of water sits in the wine glass. After the first pour the water glass holds 200 ml of water and 50 ml of wine. A 50 ml pour of that mix is one fifth wine, so 10 ml of wine and 40 ml of water go back. Each glass ends at 200 ml, which forces the two amounts to match.

What is the argument that needs no arithmetic?

Two classrooms hold 30 students each. Five walk from room A to room B, and then any five people at all walk back. Both rooms hold 30 again. Every seat in room A left empty by a room A student who stayed away must now be filled by a room B student, so the room B students in room A always equal the room A students in room B. The glasses work the same way, because both finish at 200 ml.

How do the numbers confirm it?

After the first pour the water glass holds 250 ml, of which 50 ml, one fifth, is wine. Stirred evenly, a 50 ml pour back carries 10 ml of wine and 40 ml of water. The wine glass ends with 150 plus 10, 160 ml of wine, and 40 ml of water. The water glass keeps 200 minus 40, 160 ml of water, and 40 ml of wine. Forty and forty.

What does the interviewer learn from how you answer?

Whether you look for something that stays fixed before you reach for arithmetic. The conservation argument survives every variation: uneven stirring, several pours back and forth, any spoon size, as long as both glasses finish at their starting volumes. Desks use the same move on inventory: if two accounts hold the same totals after a string of transfers, whatever left one must have been replaced from the other, whatever the route. Give the one-line argument first, then the 40 and 40 as the check.

Let X be -1, 0 or 1 with equal chances and let Y = X squared. Y is fixed completely once you know X, yet the correlation is exactly zero. The covariance is E[XY] minus E[X]E[Y]; E[X] is 0 and E[XY] is the average of -1, 0 and 1, which is also 0. Correlation measures only straight-line co-movement, and a symmetric U shape has none.

What does correlation actually measure?

Think of a household's electricity bill against the outside temperature. It is high in the coldest months, when the heater runs, high in the hottest, when the air conditioner runs, and low in between. Correlation asks only whether one variable tends to rise along a straight line as the other rises, so a U-shaped link, however tight, can score zero. Over a year balanced around a mild middle, temperature explains the bill almost completely and a straight-line measure misses all of it.

How do you prove the covariance is zero?

Write out the three cases. X averages 0, Y averages 2/3, and XY takes the values -1, 0 and 1, which also average 0, so the covariance E[XY] - E[X]E[Y] is exactly 0. Symmetry does the work: every point to the right of the axis has a mirror image on the left with the same Y, so the upward slope on one side cancels the downward slope on the other. A continuous version works the same way: X normal with mean zero and Y equal to X squared.

Where does this bite on a desk?

Anywhere a payoff depends on the size of a move rather than its direction. A long straddleA call and a put bought at the same strike and expiry, which gains from a large move in either direction. gains from a big move either way, so over moves balanced around zero its return shows little correlation with the stock's return while being driven entirely by it. A risk report built only on correlations would call that position unrelated to the stock. Zero correlation means no straight-line link; only independence means no link at all.

1.5% a month compounds to about 19.6% a year, and 40% over three years is about 11.9% a year. Compounding multiplies growth factors rather than adding rates: 1.015 to the twelfth is 1.196. Going the other way, take the cube root of 1.40, which is 1.119. Simple arithmetic gives 18% and 13.3%, understating the first answer and overstating the second.

Why is twelve times 1.5% not the annual return?

A savings account that credits interest every month pays interest on last month's interest. Each month's 1.5% is earned on a base that already includes every earlier month's gain, so the growth factors multiply: 1.015 times itself twelve times. That is 1.196, a 19.6% year. The extra 1.6 points over 18% are interest on interest, tiny in any one month and not tiny over a year.

How do you go back from a total to an annual rate?

Take the root, not the division. A 40% total over three years means the yearly growth factor cubed is 1.40, so the factor is the cube root of 1.40, about 1.119, an annual rate of 11.9%. Dividing 40 by 3 gives 13.3%, and 1.133 cubed is 1.456, a 45.6% total: division overstates the yearly rate because later years grow on earlier gains. Allocators compare managers on the compound annual growth rateThe single yearly rate that, compounded over the period, turns the starting value into the ending value., so dividing can misrank two funds.

How do you do it in your head?

Add the square-term correction. Compounding adds roughly n(n - 1)/2 times r squared to n times r: for 12 months at 1.5% that is 66 x 0.000225, about 1.5 points, taking 18% to about 19.5%. That is close enough to show you know the direction and the size. For the cube root, guess and check: 1.12 cubed is about 1.405, a shade over 1.40, so the answer sits just under 12%. Checking by cubing is faster and safer than estimating a root directly.

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.

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.

It cannot be done at any finite speed. Averaging 60 mph over the 120-mile round trip means finishing in 2 hours. The outward 60 miles at 30 mph already took 2 hours, so the return leg would have to take no time at all. Driving back at 90 mph, the tempting answer, gives an average of only 45 mph.

Why is 90 mph the wrong instinct?

Averaging 30 and 90 to get 60 treats the two speeds as if they counted equally. They do not, because the car spends far longer at the slow speed. Think of a student who scores 30% on a three-hour paper and 90% on a ten-minute quiz: nobody would call that a 60% performance. Average speed is total distance over total time, so the slow leg carries more weight because it takes up more of the clock.

How do you prove it cannot be done?

Work in time, not speed. At 60 mph, 120 miles takes exactly 2 hours, and the first leg has already spent those 2 hours. Any return speed, however fast, adds some time, which pushes the average below 60 mph. At 90 mph the return takes 40 minutes and the average is 45 mph; at 300 mph it takes 12 minutes and the average is 54.5 mph. The average creeps towards 60 but never reaches it.

Where does the same mistake show up on a desk?

It appears whenever numbers are averaged without the right weights. A position that falls 50% and then rises 50% does not break even, because the second move works on a smaller base; the average that matters is the one weighted the way the thing actually compounds. Speeds over equal distances call for the harmonic mean, which sits below the simple average whenever the numbers differ. It is the same reason that putting a fixed rupee amount into a fund each month buys units at an average cost below the average price over those months.

Maximum profit is Rs 62 and the breakeven is Rs 488. Above the Rs 550 strike the stock is called away, so you keep the Rs 50 rise from 500 to 550 plus the Rs 12 premium. Below 550 the call expires worthless and you keep the premium, which cushions the first Rs 12 of any fall. You lose money only below 500 minus 12, which is Rs 488.

What have you actually sold?

Think of renting out a flat you own with an agreement that the tenant may buy it at a fixed price within the year. You collect rent now, but if flat prices soar, the tenant buys at the agreed price and the extra gain is theirs. A covered call swaps the upside above the strike for cash today: the premium is the rent, and the strike is the agreed sale price.

How do you find the two numbers quickly?

Take the two regions separately. At or above Rs 550 the position is worth 550 plus the 12 kept, against 500 paid, so Rs 62 whatever the stock does. Below 550 the call is worthless and the position is just the stock plus 12, so it loses money only once the stock falls more than 12 below the purchase price, at Rs 488. At a stock price of 520 you make 32: 20 on the stock and 12 of premium. At 450 you lose 38 instead of 50.

What is the trade-off in plain terms?

The premium improves every outcome below Rs 562 and worsens every outcome above it. Above Rs 562 the plain stock position beats the covered call, because the gain you gave away exceeds the premium you took in. On the downside the cushion is thin: the stock can fall all the way from 500 and the premium covers only 12 of it. That is the limitation to state: a covered call is income with a cap, not protection.

About Rs 31.6 crore: Rs 10 crore times the square root of 10. If daily P&L is independent from day to day, with the same volatility and a mean near zero, variances add, so ten-day volatility is root 10 times daily volatility, and a normal quantile scales the same way. The rule also needs the positions held unchanged for ten days and a distribution that keeps its shape over the horizon.

Why not ten times the one-day number?

Picture ten friends each tossing a coin for Rs 100. The worst case is the group losing Rs 1,000, but the typical spread of the group's total is nowhere near ten times one person's, because some win while others lose. Independent daily moves partly cancel, so their variances add while their volatilities do not, and volatility grows with the square root of the number of days. Ten times would need every bad day to line up in the same direction, which is exactly what independence rules out. Value at riskThe loss a book should not exceed over a set horizon at a set confidence level, for example one day at 99%. inherits that square root when the distribution is normal.

What has to be true for the rule to hold?

List the assumptions, because that is the real question. Returns must be independent across days, volatility constant, the mean close to zero, the positions unchanged, and the distribution one that keeps its shape when summed, as the normal does. Break any one and the rule drifts. Positive autocorrelation, where bad days follow bad days, makes the true ten-day number larger. A book that is cut after losses makes it smaller. Fat tails make the one-day 99% quantile a poor guide to the ten-day one.

Which way does the error usually run?

In a calm market the rule is a fair approximation. In stress it tends to understate, because volatility rises and losses cluster just when the ten-day horizon matters. The square root of time is a scaling convenience, not a law, so a risk team checks it against ten-day P&L measured directly. The same assumption sits behind the desk habit of multiplying daily volatility by 16 to get an annual figure, 16 being roughly the square root of the trading days in a year; stretch it to 250 days here and you get Rs 158 crore, a number few would trust.