Hedge Funds puzzles, solved step by step

Do not spin: you survive 75% of the time, against 66.7% if you spin. The empty click puts the cylinder on one of the four empty chambers, each equally likely. Because the two bullets sit together, three of those four empties are followed by another empty and only one is followed by a bullet. A fresh spin throws that information away and gives four chances in six.

What does the empty click tell you?

Picture six people in a queue where two friends always stand together. Pick someone at random who is not one of the friends and ask whether the person behind them is a friend. Only one of the four has a friend behind them: the one standing just in front of the pair. The empty click is information: it tells you which chambers you could be on, and using it is the whole puzzle. The cylinder has no memory, but you do.

Why does the answer flip if the bullets are not adjacent?

Separate the bullets, say into chambers 1 and 4. The four empties are now 2, 3, 5 and 6, and the chambers after them are 3, 4, 6 and 1: two empties and two bullets. Not spinning now survives only 2 times in 4, 50%, so spinning, at 66.7%, becomes the better choice. Adjacency is what bunches both bullets behind a single empty chamber. Ask where the bullets sit before you answer, and say that the answer depends on it.

Where does this reasoning show up on a desk?

The same move, updating on what you have just observed instead of resetting to the base rate, is how a trader reads a fill. Getting filled on your bid tells you something about who was selling, just as the empty click tells you which chamber you are on. Ignoring it is the equivalent of spinning the cylinder: it feels neutral, but it throws away an edge you were handed for free.

The probability is 7/8, and the density of X + Y is a triangle, a tent peaking at 1. Because X and Y are independent and uniform, every point of the unit square is equally likely, so probability is area. The line x + y = 1.5 slices off a corner triangle with legs of 0.5, area 1/8. The sum's density rises in a straight line from 0 to 1 and falls back to 0 at 2.

Why does probability become area here?

Throw a dart at a square board so that every point is equally likely to be hit. The chance it lands in a region is that region's share of the board. Two independent uniforms are exactly that dart: the pair (X, Y) lands evenly on the unit square, so any question about X and Y becomes a question about an area. The condition X + Y below 1.5 is everything under the line x + y = 1.5, which is the whole square except one corner.

How do you get the shape of the sum's density?

Slide the line x + y = s across the square and watch how long it is inside. Near s = 0 it barely clips the corner; at s = 1 it runs corner to corner, the longest it gets; past 1 it shortens again. The density of the sum at s is proportional to the length of that line inside the square, which gives a triangle rising from 0 to a peak at 1 and falling to 2. This is the convolutionThe density of a sum of independent variables, found by adding up every way the two parts can combine to the same total. of two flat densities, and the same reason two dice most often total 7.

Check the first answer with the tent. The area beyond 1.5 is a triangle with base 0.5 and height 0.5, which is 1/8 again. Two routes that agree is the check an interviewer wants to hear before you commit. Add a third uniform and the density becomes three joined curved pieces; add many and the sum looks normal, which is the central limit theorem arriving in slow motion.

Randomise: pay Rs 3.15 with probability 0.159 and Rs 3.14 otherwise. Pi is 3.14159..., which sits 0.159 of the way from 3.14 to 3.15. Paying the higher amount with exactly that probability makes the expected payment 3.14 + 0.01 x 0.1593..., which is pi. So the chance you pay Rs 3.15 is about 15.9%, and over many meals nobody is short-changed.

Why can no fixed amount be fair?

Always round to Rs 3.14 and the restaurant loses 0.159 paise every time; always pay 3.15 and you overpay 0.841 paise. Any fixed amount is unfair to one side, so the only way to be exactly fair is to be fair on average. Two friends who split a Rs 101 bill by taking turns to pay the odd rupee are doing the same thing: neither is exact on any one night, both are exact over time.

How do you actually draw a probability of 0.159?

Use any randomness you can split finely. Draw a uniform number between 0 and 1 and pay Rs 3.15 if it falls below 0.1593. With only a die, paying 3.15 on a six gives 1/6, an expected payment of Rs 3.141667: close, not exact. With only a fair coin you can be exact: toss it to generate the binary digits of a uniform number one at a time and stop as soon as the digits so far settle which side of 0.1593 it falls. Each toss settles it with probability one half, so on average it takes two tosses.

Where does randomised rounding show up in a fund?

Whenever a quantity has to be split in whole units. A fund allocating 1,003 shares across three accounts cannot give each 334.33; handing the odd share out by lottery, or in rotation, keeps each account fair on average. The principle is the same: when the exact amount is impossible, make the expected amount exact and keep the error unbiased. The limitation is that fair on average is not fair every time, which is why allocation policies also cap how far any account can drift.

The price falls by about 2.94%. Spread duration of 6 says a 0.50 percentage point widening costs 6 x 0.50% = 3.00%. Convexity of 50 adds back one half x 50 x 0.005 squared, about 0.06%, because the price curve bends upwards. On a bond priced at 100 that is a move to about 97.06. At 50 basis points the convexity term is small; at 300 or 500 it is not.

What do duration and convexity each measure?

Picture a playground slide that curves and flattens towards the bottom. Judge the drop from the steepness at the top and you overstate it, because the slide levels off as you go. Spread durationThe percentage change in a bond price for a one percentage point change in its credit spread, holding the risk-free rate fixed. is the steepness at today's spread; convexity is the flattening, so the straight-line estimate always overstates the loss when spreads widen. Duration gives the first-order move, 6 x 0.50% = 3.00% down; convexity corrects it by a term that depends on the square of the move.

When does the convexity term start to matter?

It grows with the square of the move. At 50 basis points convexity is worth 0.06% against a 3.00% duration loss; at 300 basis points duration says -18% and convexity adds back 2.25%, which is no longer small. That is why a credit desk can run duration-only risk for everyday moves but needs convexity for stress scenarios. One more distinction marks a strong answer: for a fixed-coupon bond spread duration and rate duration are close, but a floating-rate note has almost no rate duration and still carries several years of spread duration.

Say the limitation plainly. Both numbers are local, measured at today's spread, and a distressed bond stops behaving like this long before default, when its price starts tracking the expected recovery instead. For a bond trading near par, as here, the two-term estimate is good to a few hundredths of a per cent for moves of this size.

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.

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 4 years for a Sharpe of 1.0 and about 16 years for a Sharpe of 0.5. The t-statistic of a mean return is the mean over its standard error, which works out to the annual Sharpe ratio times the square root of the number of years, whatever the data frequency. Setting Sharpe x root(years) = 2 gives years = (2 / Sharpe) squared: 4 for 1.0, 16 for 0.5 and just 1 for 2.0.

Why does the t-statistic grow with the square root of time?

A coin that lands heads 55% of the time looks fair after 20 tosses; you need hundreds before the bias shows through the noise. The average return grows in proportion to time, but the noise around it grows only with the square root of time, so the signal-to-noise ratio, the t-statistic, grows with root time. With monthly data, the t-statistic is the monthly Sharpe times root(12 x years), and the monthly Sharpe is the annual Sharpe divided by root 12, so the twelves cancel: t = annual Sharpe x root(years).

Why does monthly data not shorten the wait?

More frequent data gives more observations but each is noisier relative to its mean. Sampling the same years more often does not add information about the mean return; only more years do. This is why a {term('t-statistic', 'An estimate divided by its standard error; a value around 2 is the usual threshold for saying an effect is unlikely to be pure noise.')} on the average return depends on the span of the data, not the number of rows. Frequency helps you estimate volatility, not the mean.

Say the practical point. Most real strategies have Sharpe ratios well below 1, so their track records are too short to separate skill from luck with any confidence. An allocator looking at a three-year record with a Sharpe of 0.8 sees a t-statistic of about 1.4. The limitation of the rule: it assumes returns are independent and stable over the whole sample, and fat tails or regime changes make the real uncertainty larger.

Bet 20% of your current money each time; bet 40% and the long-run growth is gone. For an even-money bet the Kelly fraction is the win probability minus the loss probability, 0.6 minus 0.4. At 20% the typical path grows about 2.0% a bet. At double that, the average log growth is about -0.2% a bet, so the same edge now earns nothing over time.

Why not bet everything when the odds are in your favour?

Think of a shopkeeper who puts all of today's takings into tomorrow's stock. Six days in ten the stock sells and she doubles her money; on the other four she is back to zero, and one zero ends the shop. Wealth compounds, so a single wipe-out is permanent, and the stake that maximises the average payoff of one round is not the stake that maximises what you end with after many. Betting 100% has the highest expected value per round and a certainty of eventual ruin.

How do you find the best fraction?

Maximise the growth rate, which is the average of the log of what each bet does to your money. Bet a fraction f and a win multiplies your wealth by 1 + f, a loss by 1 - f. The growth rate 0.6 ln(1 + f) + 0.4 ln(1 - f) peaks where f equals the edge, 0.6 - 0.4 = 20%, at about 2.01% a bet. Over 100 bets that turns Rs 1,000 into about Rs 7,490 on the typical path. This rule is the Kelly criterionA sizing rule that picks the bet size maximising the long-run growth rate of wealth, the expected log return per bet..

What happens at twice the Kelly bet?

Near the peak the growth curve is close to a parabola centred on 20%. Double the Kelly fraction and you sit as far down the far side as not betting at all sits on the near side: growth of about zero, -0.24% a bet. Over 100 bets the typical path ends near Rs 783, below where you started, although every single bet had a positive expected value. The extra swings cost more in compounding than the extra stake earns.

Say the limitation too. Kelly assumes you know the 60% exactly. On a desk the edge is an estimate, and overestimating it pushes you to the right of the peak, where the curve falls fastest. That is why many traders bet half Kelly: 10% here keeps about 75% of the growth with half the swings.

The one-die game is worth more: Rs 15.17 against Rs 12.25, a gap of 35/12, about Rs 2.92. With two independent dice the average product is the product of the averages, 3.5 x 3.5 = 12.25. With one die the average square is (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6. The gap between the two is exactly the variance of one die.

Why is averaging a square not the same as squaring an average?

Take two students who score 2 and 8 in a test. Their average is 5, and 5 squared is 25. Square each score first and then average, and you get (4 + 64)/2 = 34. Squaring rewards the high value more than it penalises the low one, so the average of the squares is always at least the square of the average, and the gap is the spread. Here the gap, 9, is exactly the variance of the two scores.

How do the two games come out?

In game A the dice are independent, so the average of the product is the product of the averages. Two independent dice give 3.5 x 3.5 = Rs 12.25, because a high roll on one die is as likely to meet a low roll on the other as a high one. In game B one number is multiplied by itself, so a 6 always meets a 6 and a 1 always meets a 1. The average of the six squares is 91/6, about Rs 15.17.

What is the desk lesson?

A payoff that is the square of a move gains from dispersion; one built from two independent pieces does not. Correlation is what turns a product into a square: if the second die always copied the first, game A would be game B. Flip it and make the second die show 7 minus the first, and the average product falls to 9.33. Say that range out loud and the interviewer knows you see the payoff as a bet on varianceThe average squared distance of an outcome from its mean; for one fair die it is 35/12, about 2.92. as well as on the average.