Hedge Funds puzzles, solved step by step

Puzzles: 100
Traced to a firm: 38
Topics: 14
Hard: 30

Showing 31–40 of 62 · filtered from 100Clear filters

052A family has two children and you learn that at least one of them is a girl. What is the probability that both are girls? How does the answer change if you learn instead that the elder child is a girl?Conditional probability and BayesWarm upQuant and systematic fundsProp and quant trading firms
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At least one child is a girl. What is the chance both are?
Show the worked solution
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.
Of the four equally likely birth orders, at least one girl strikes out only boy-boy and leaves three cells, so both girls is 1 in 3, while the elder is a girl keeps only the two cells of the elder-girl row, so both girls is 1 in 2.
The relationship
$P(GG \mid \text{at least one } G) = \frac{P(GG)}{P(\text{at least one } G)} = \frac{1/4}{3/4} = \frac{1}{3}$
GG both children are girls
1/4 the chance of girl-girl among four equally likely orders
3/4 the chance of at least one girl: every order except boy-boy
What it says in wordsA conditional probability is the chance of both things happening divided by the chance of the thing you were told.
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.
Where candidates lose it
Nearly everyone answers one half for both versions, because the second child feels like a separate coin toss. The interviewer is checking whether you notice that at least one does not say which one.
The other way to lose it is to reach one third and then say one third again for the elder-girl version. Draw the grid, strike the cells each statement rules out, and count what is left.
What the interviewer asks next
You visit the family and a girl, chosen at random from the two children, opens the door. What is the chance both children are girls?
At least one child is a girl born on a Tuesday. What is the chance both are girls?
With three children and at least one girl, what is the chance all three are girls?
053Two independent numbers are drawn uniformly at random between 0 and 1. What is the expected value of the smaller one, and of the larger one?Continuous probability and distributionsCoreQuant and systematic fundsProp and quant trading firms
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What is the expected value of the smaller of the two draws?
Show the worked solution
The smaller draw averages 1/3 and the larger averages 2/3. Two random points cut the stretch from 0 to 1 into three pieces, and nothing distinguishes one piece from another, so each has expected length 1/3. The minimum ends the first piece and the maximum ends the second. The two averages add to 1, as they must, because the minimum plus the maximum is just the sum of the two draws.
What is the one-line argument?
Imagine two friends who each arrive at a random minute within the same hour. The wait until the first arrival, the gap between the two arrivals and the time from the second arrival to the end of the hour are three stretches of time. No stretch is special, so each averages a third of the hour, and the first friend arrives, on average, 20 minutes in. The same symmetry puts the minimum of two uniform draws at 1/3 and the maximum at 2/3.
The diagonal splits the square of possible draws into two mirror-image halves; the density of the smaller draw falls in a straight line from 2 at zero to 0 at one with its average at 1/3, and the density of the larger draw mirrors it with its average at 2/3.
How do you check it with a calculation?
Work from the chance that the minimum sits above some level x. Both draws must be above x, which happens with probability (1 - x) squared. For a quantity that cannot be negative, the average equals the area under its chance of exceeding each level, and the area under (1 - x) squared from 0 to 1 is 1/3. The maximum then follows without a second integral: minimum plus maximum equals X plus Y, which averages 1, so the maximum averages 2/3.
The relationship
$E[\min] = \int_0^1 P(\min > x)\,dx = \int_0^1 (1-x)^2\,dx = \tfrac{1}{3}, \qquad E[\max] = 1 - \tfrac{1}{3} = \tfrac{2}{3}$
P(min > x) the chance both draws exceed x, (1 - x) squared
E[min] the expected value of the smaller draw
E[max] the expected value of the larger draw
What it says in wordsIntegrate the chance that the minimum clears each level to get its average, then use min plus max equals the sum of the draws.
Where does the pattern lead next?
With n draws, the points cut the interval into n + 1 pieces, so the smallest averages 1/(n + 1) and the largest n/(n + 1). The ordered values of uniform draws are evenly spaced on average, and that single fact answers a whole family of follow-ups. With ten bids drawn at random across a range, the highest averages 10/11 of the way up. The quick route here is also the one to say first: the order statisticsThe draws sorted from smallest to largest: the minimum is the first order statistic, the maximum the last. of uniform draws split the interval evenly.
Where candidates lose it
The common wrong answer is 1/4, reached by reasoning that the smaller draw lies in the bottom half and so sits at the middle of it. The smaller draw is not confined to the bottom half; it only has to be below the other draw, and it lands above 0.5 a quarter of the time.
The second loss is grinding through the integral in silence and never giving the symmetry argument. Say the three-pieces picture first, then offer the integral as the check.
What the interviewer asks next
What is the expected distance between the two draws?
With three draws, what is the expected value of the middle one?
What is the probability that the smaller draw is below 0.5?
054You flip a fair coin five times. What is the probability of seeing a run of at least three heads in a row somewhere in the five flips?Counting and combinatoricsCoreQuant and systematic fundsProp and quant trading firms
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What is the chance of at least three heads in a row in five flips?
Show the worked solution
The chance is 8 in 32, exactly 1/4. Sort the qualifying sequences by where their first run of three heads starts. Starting at flip 1: HHH then anything, 4 sequences. Starting at flip 2: flip 1 must be tails, then HHH, then anything, 2 sequences. Starting at flip 3: flip 2 must be tails and flip 1 is free, 2 sequences. Four plus two plus two is eight.
Why does the obvious count go wrong?
Suppose you count guests at a party by asking each room how many people it holds, while some guests stand in doorways. They get counted twice. A run of three can start in three places, but one sequence can contain a run starting at more than one of them, so three places times four fillings, 12, counts HHHHH three times and two other sequences twice. The fix is to give each sequence exactly one home: the position where its first run starts.
The eight qualifying sequences sort into four whose first run starts at flip 1, two at flip 2 and two at flip 3, so the chance is 8 of 32, one quarter; the naive count of 12 counts overlapping runs more than once.
How do you make every sequence count once?
Force the flip just before the run to be tails. If the first run starts at flip 2, flip 1 must be tails; if it starts at flip 3, flip 2 must be tails, and flip 1 can be anything. That gives 4, then 2, then 2. Then check a second way by counting the complement: sequences with no run of three follow a Tribonacci-style rule, each term the sum of the three before, giving 1, 2, 4, 7, 13, 24 for 0 to 5 flips. And 32 minus 24 is 8.
The relationship
$P = \frac{2^2 + 2 + 2}{2^5} = \frac{8}{32} = \frac{1}{4} \qquad \text{check: } 1 - \frac{24}{32} = \frac14$
2^2 first run at flip 1: the last two flips are free
2 first run at flip 2 or 3: one free flip each
24 sequences with no run of three heads
What it says in wordsCount each qualifying sequence once, by the start of its first run, and confirm by counting the sequences that fail.
Why would a quant fund ask this?
Streaks are what people notice in trading records and backtests. A run of three heads turns up in a quarter of all five-flip sequences of a fair coin, so a short winning streak is weak evidence of skill. The counting habit matters as much as the number: giving each outcome one home is the same discipline that stops you double counting overlapping signals when you tally how often a strategy fired.
Where candidates lose it
The trap is 12 out of 32: three places a run can start times four fillings of the other flips. It sounds rigorous, and it is wrong because a sequence with a long run is counted once for every place a run of three fits inside it.
The second loss is giving 1/4 with no check. Offer the complement count, 24 sequences with no run of three, as the second route to the same answer.
What the interviewer asks next
What is the chance of at least three heads in a row in ten flips?
In five flips, what is the chance of a run of exactly three heads, no longer?
How many flips do you expect to make before you first see three heads in a row?
055Your average winning trade makes 1.5 times what your average losing trade loses. What hit rate do you need just to break even?Estimation and mental mathsWarm upMulti-manager platformsLong-short equity funds
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Winners are 1.5 times the size of losers. What hit rate breaks even?
Show the worked solution
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%.
The relationship
$p \times 1.5 = (1-p) \times 1 \;\Rightarrow\; p = \frac{1}{1 + 1.5} = 40\%$
p the hit rate, the share of trades that win
1.5 the average win as a multiple of the average loss
1 - p the share of trades that lose
What it says in wordsThe book breaks even when expected winnings per trade equal expected losses per trade.
At a 30% hit rate expected wins of 0.45 units fall short of expected losses of 0.70; at 40% both are 0.60 and the book breaks even; at 45% wins of 0.675 beat losses of 0.55, a profit of 0.125 units a trade.
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.
Where candidates lose it
The instinct is to say 50% or more, because being right more often than wrong sounds like the definition of a good trader. The interviewer wants you to weigh each outcome by its size rather than count outcomes.
The other slip is inverting the ratio and answering 60%, which is the breakeven if losers were 1.5 times winners. Write the one equation, 1.5p = 1 - p, before you say a number.
What the interviewer asks next
Your hit rate is 55% and your winners are 0.8 times your losers. Are you making money?
Costs are 0.1 units a trade. What hit rate do you need now?
Why might a manager's win-loss ratio fall as the fund grows?
058A market maker knows that 30% of the orders she receives come from traders who know the true value is exactly Rs 1 above or below the mid. The other 70% are uninformed and buy or sell at random. How wide must her bid-ask spread be just to break even?Market making and trading gamesCoreProp and quant trading firmsVolatility and relative value funds
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How wide must the full spread be?
Show the worked solution
The spread must be Rs 0.60: a bid Rs 0.30 below the mid and an offer Rs 0.30 above it. Call the half-spread s. Each uninformed order earns her s on average. Each informed order trades on the side where the value really is Rs 1 away, so she loses 1 minus s. Breaking even needs 0.7s = 0.3(1 - s), which gives s = 0.30.
Why do some of the people who trade with you cost you money?
A shop that buys second-hand phones pays the same price to every seller, but the sellers who know their phone is faulty are the keenest to sell. A quote is an offer made to everyone, and the people who know more take it only when it hurts you. The spread is the fee charged to the whole crowd to pay for the few who know more, which is called adverse selectionThe tendency for the counterparties who choose to trade with you to be the ones who know the trade is good for them..
How do you set up the break-even?
Charge a half-spread s on each side. An uninformed order is as likely to be a buy as a sell and carries no information, so on average it earns you s; an informed order buys at your offer only when the value is Rs 1 above the mid, so it costs you 1 minus s. Weight each by how often it arrives and set the total to zero: 0.7s - 0.3(1 - s) = 0, so s = 0.30 and the spread is Rs 0.60.
The relationship
$0.7\,s = 0.3\,(1 - s) \;\Rightarrow\; s = 0.30, \qquad \text{spread} = 2s = 0.60$
s the half-spread, the distance from the mid to the bid or the offer
0.7 the share of orders from uninformed traders
1 - s what an informed trade takes: the Rs 1 value gap less the half-spread crossed
What it says in wordsThe spread earned from uninformed flow must pay for the expected loss to informed flow.
At a half-spread of Rs 0.20 she earns Rs 0.14 an order from uninformed flow but loses Rs 0.24 to informed flow, a loss of Rs 0.10; at Rs 0.30 she earns and loses Rs 0.21 each, the break-even spread of Rs 0.60.
What moves the spread?
Two things: how much of the flow is informed and how much the informed know. In general the half-spread is s = aV, where a is the informed share and V is how far the true value sits from the mid, so doubling either doubles the spread. Here 0.3 x 1 = 0.3. That is why quotes widen ahead of a results announcement, when more of the flow may know something, and in small stocks, where one piece of news moves the value a long way. The model leaves out inventory risk and competing quoters, both of which move a real spread; say so.
Where candidates lose it
The common slip is answering Rs 0.30, which is the half-spread, or quoting a width without saying where the bid and offer sit. Name both sides: bid at the mid minus 0.30, offer at the mid plus 0.30.
The deeper loss is charging each informed trade the full Rs 1 and forgetting that the informed trader also crosses the spread, which gives 0.7s = 0.3 and s of about 0.43. Her loss to an informed order is the value gap less the half-spread.
What the interviewer asks next
Half the flow is now informed. What spread breaks even?
You receive three buy orders in a row. Should you move your mid, and which way?
A rival quotes Rs 0.40 wide. What happens to the mix of flow you receive?
059A stock trades at Rs 1,000 and pays no dividend. A six-month European call struck at Rs 1,000 costs Rs 60 and the matching put costs Rs 45. Six-month simple interest is 3%. Is there an arbitrage, and how do you lock it in?Options and payoffsCoreVolatility and relative value fundsProp and quant trading firms
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What should the call minus the put be worth?
Show the worked solution
Yes: buy the call, sell the put, short the stock and lend Rs 970.87, which collects about Rs 14.13 today with nothing owed at expiry. Put-call parity says the call less the put should equal the stock less the present value of the strike, 1,000 minus 1,000/1.03, about Rs 29.13. The market prices that gap at Rs 15, so the call is cheap against the put by about Rs 14.
What does a call minus a put actually pay?
Imagine agreeing today to buy a used car in six months at a fixed price. If car prices rise you gain the difference; if they fall you lose it. A long call with a short put at the same strike does exactly that: at expiry it pays the stock price minus the strike, whichever way the stock moved. So it must cost the same as buying the stock today with money borrowed against the strike, 1,000 minus 970.87, which is Rs 29.13. That identity is put-call parityFor European options on the same stock, strike and expiry, call minus put equals the stock price minus the present value of the strike, less the value of any dividends..
The relationship
$C - P = S - \frac{K}{1 + r} = 1{,}000 - \frac{1{,}000}{1.03} \approx 29.13$
C, P the call and put prices, Rs 60 and Rs 45
S the stock price today, Rs 1,000
K the strike, Rs 1,000
r six-month simple interest, 3%
What it says in wordsA call less a put is worth the stock less the strike discounted to today.
How do you build the trade?
Buy what is cheap and sell what is dear. The option version of the forward, long call and short put, costs Rs 15; the cash version, long stock bought with borrowed money, is worth Rs 29.13, so buy the option version and sell the cash version. Buy the call for 60, sell the put for 45, short the stock for 1,000 and lend 970.87 at 3%. The cash left over today is Rs 14.13, worth about Rs 14.55 at expiry.
At expiry the long call and short put pay the stock price minus Rs 1,000 while the short stock and the maturing Rs 1,000 loan pay Rs 1,000 minus the stock price, so the package nets to zero at every price and the Rs 14.13 collected at the start is kept.
What could stop the arbitrage working?
Say what you assumed. The lock holds only if the options are European, the stock pays no dividend before expiry, and you can short the stock without a borrowing fee that eats the Rs 14. A dividend lowers the forward and so the fair gap; an American put can be exercised against you early; a hard-to-borrow stock charges a fee. A gap this size in a liquid name usually means one of those is present, so your first question back is which one.
Where candidates lose it
The common loss is comparing the call and the put directly, seeing the call is dearer and deciding it must be the one to sell. Parity says the call should be dearer, by about Rs 29; the trade depends on whether the actual gap is too wide or too narrow.
The second is forgetting to discount the strike and calling zero fair, or using 3% of the stock, Rs 30, which is close but wrong. Say the formula before the numbers, then build the four legs.
What the interviewer asks next
The stock will pay a Rs 20 dividend before expiry. Is there still an arbitrage, and which way?
The call is Rs 80 and the put Rs 45. What trade do you put on now?
Why does parity hold exactly for European options but only as a band for American ones?
062A fund makes 1.5% every month for a year. What is its return for the year? Another fund made 40% in total over three years. What is its annual rate of return?Returns, compounding and feesWarm upFund of funds and allocatorsMulti-manager platforms
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What does 1.5% a month for twelve months come to?
Show the worked solution
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.
The relationship
$1.015^{12} - 1 = 19.6\%, \qquad 1.40^{1/3} - 1 = 11.9\%$
1.015 the monthly growth factor, 1 plus 1.5%
1.40 the three-year growth factor, 1 plus 40%
1/3 the cube root, which undoes three years of compounding
What it says in wordsRaise the growth factor to the number of periods to go forward, and take the matching root to go back.
Twelve months at 1.5% compound to 19.6% rather than 18.0%, and a 40% three-year gain is 11.9% a year rather than 13.3%, because 13.3% compounded for three years would give 45.6%.
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.
Where candidates lose it
The trap is simple arithmetic: 12 x 1.5% = 18% and 40 / 3 = 13.3%. Both come out fast and both are wrong, in opposite directions, which is why the interviewer asks the pair together.
The second loss is getting 19.6% and then dividing on the second half out of habit. State the rule once, multiply going forward and take the root going back, and apply it to both halves.
What the interviewer asks next
A fund loses 1.5% every month for a year. What is its annual return?
Which pays more: 1% a month, or 12.5% paid once a year?
A manager reports a three-year return of 40% and an average annual return of 13.3%. What is wrong with the second number?
064An exporter earns Rs 100 of revenue, 60% of it billed in dollars, and has Rs 85 of costs, all paid in rupees. The rupee weakens from 80 to 88 to the dollar. What happens to its operating margin?Valuation, accounting and macro riddlesCoreLong-short equity fundsGlobal macro funds
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What is the operating margin after the move?
Show the worked solution
The margin rises from 15% to about 19.8%, and operating profit rises 40%. Each dollar now buys 10% more rupees, which turns Rs 60 of dollar revenue into Rs 66, so revenue reaches Rs 106. Costs are all in rupees and stay at Rs 85, so profit climbs from Rs 15 to Rs 21, and 21 over 106 is 19.8%. A 6% rise in revenue becomes a 40% rise in profit.
Why does a weaker rupee help this company so much?
Think of a tutor in Pune who teaches some students abroad for fees in dollars and pays rent and a salary in rupees. When the rupee weakens, each dollar fee converts into more rupees while the rent does not move. A currency move lands on the revenue billed in the foreign currency and on nothing else, so when every cost is local the whole gain drops into profit. Here the gain is Rs 6, on a profit that started at only Rs 15.
Before the move, Rs 40 of rupee revenue and Rs 60 of dollar revenue less Rs 85 of rupee costs leave Rs 15 of profit, a 15% margin; at 88 to the dollar the dollar revenue is worth Rs 66, costs stay at Rs 85 and profit rises to Rs 21, a 19.8% margin.
How do you size the profit move quickly?
Chain three numbers. The dollar buys 10% more rupees, 60% of revenue is exposed, so revenue rises 6%; a thin 15% margin turns that Rs 6 into a 40% rise in profit. The ratio of the profit change to the revenue change, 40 over 6, almost 7 times, is the company's operating leverageHow much operating profit moves for a given move in revenue, high when most costs are fixed or unaffected by the change. to the currency. An analyst on a long-short book uses that multiplier to judge how much of a stock's move a currency view is really driving.
The relationship
$\text{margin} = \frac{40 + 60 \times \tfrac{88}{80} - 85}{40 + 60 \times \tfrac{88}{80}} = \frac{21}{106} \approx 19.8\%$
40 revenue billed in rupees, unchanged
60 x 88/80 dollar revenue restated at the new rate, Rs 66
85 costs, all in rupees, unchanged
What it says in wordsRestate only the dollar-billed revenue at the new rate, hold rupee costs fixed, and divide profit by the new revenue.
What would you check before trusting the new margin?
Three things that often shrink the gain. Hedges, pricing and imported inputs: a company that sold its dollars forward keeps the old rate until the hedges roll off, buyers abroad may push for lower dollar prices, and any imported materials get dearer too. Each one eats into the Rs 6. The 19.8% is the unhedged, first-round answer; naming those three tells the interviewer you would not stop at it.
Where candidates lose it
The first slip is saying the margin is unchanged because revenue and costs both grow. Only the dollar-billed revenue moves; rupee costs stay where they were, and that asymmetry is the whole point.
The second is applying the 10% to all Rs 100 of revenue, which gives Rs 110 and a 22.7% margin. Only 60% of revenue is billed in dollars, so revenue rises 6%, not 10%.
What the interviewer asks next
Half of the costs are imported and paid in dollars. What is the new margin?
The company hedged half its dollar revenue at 80. What margin does it report this year?
What does the same move do to an importer with the mirror-image structure?
065A stock rises 50% or falls 40% each year with equal probability. Its expected return is positive, but what happens to a buy-and-hold investor over time? And what fraction of wealth should sit in the stock if the rest is held in cash and the mix is rebalanced every year?Betting and sizingHardMulti-manager platformsProp and quant trading firms
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Over many years, what happens to the typical buy-and-hold investor?
Show the worked solution
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.
The relationship
$g(f) = \tfrac12\ln(1 + 0.5f) + \tfrac12\ln(1 - 0.4f), \qquad g'(f) = 0 \;\Rightarrow\; f^{*} = 0.25$
f the fraction of wealth held in the stock, the rest in cash
g(f) the expected log growth per year of the rebalanced mix
f* the stake that maximises it
What it says in wordsPick the stake that makes the average log return per year as large as possible; here that is a quarter of your wealth.
Over 20 alternating years the all-stock investor falls to about 0.35 of the starting wealth while the average across all paths climbs to 2.65, and a 25% stake rebalanced every year grows to about 1.13, because growth depends on the log return, not the average return.
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..
Where candidates lose it
Most candidates answer that the investor earns 5% a year, because that is the average. The interviewer built the numbers so the average and the typical outcome point in opposite directions, and wants to see you notice.
The second loss is concluding the stock is simply bad and putting nothing in it. Zero earns nothing; the point is that a small, rebalanced stake turns the same gamble into positive growth.
What the interviewer asks next
Cash now earns 3% a year. How does the best stake change?
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069You have 12 coins that look identical. One is either heavier or lighter than the others, and you do not know which. Using a two-pan balance only three times, find the odd coin and say whether it is heavy or light.Logic and brainteasersHardProp and quant trading firmsLong-short equity funds
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Why is three weighings enough, in principle?
Show the worked solution
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.
Weighing four against four splits the 24 cases into three groups of 8; the second weighing mixes suspects with known good coins to split each 8 into 3, 2 and 3; the third weighing then separates what is left, so all 24 cases fit inside the 27 outcomes.
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.
Where candidates lose it
The usual loss is weighing six against six first. It wastes the balance outcome, because the odd coin is always in one of the pans, and it leaves 12 cases on a branch that only two weighings, 9 outcomes, must resolve.
The second is forgetting that heavy or light is part of the answer. Candidates find the coin and stop; the counting argument, 24 cases in 27 outcomes, is the proof that you have not left anything to luck.
What the interviewer asks next
What is the largest number of coins you can handle with three weighings if you must also say heavy or light?
How does the problem change if you have one extra coin known to be genuine?
Can you design all three weighings in advance, without looking at the earlier results?

GG	both children are girls
1/4	the chance of girl-girl among four equally likely orders
3/4	the chance of at least one girl: every order except boy-boy

P(min > x)	the chance both draws exceed x, (1 - x) squared
E[min]	the expected value of the smaller draw
E[max]	the expected value of the larger draw

2^2	first run at flip 1: the last two flips are free
2	first run at flip 2 or 3: one free flip each
24	sequences with no run of three heads

p	the hit rate, the share of trades that win
1.5	the average win as a multiple of the average loss
1 - p	the share of trades that lose

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