Hedge Funds puzzles, solved step by step

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  1. 006You owe exactly Rs pi, that is Rs 3.14159..., and can only pay in whole paise. How do you pay a fair amount on average, and what is the chance you end up paying Rs 3.15?Expected value and dice gamesCoreMillennium ManagementSheung Wan · 2025

    Try it first

    Under the fair scheme, what is the chance you pay Rs 3.15?

    Show the worked solution

    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.

    Weight each paisa by how close pi is to it, and the beam balances at pi84.1%15.9%pay Rs 3.14pay Rs 3.15Rs 3.14Rs 3.15pi = 3.14159...0.159 paise0.841 paise: the gap to 3.15Expected payment = 3.14 x 0.8407 + 3.15 x 0.1593= 3.14159265..., exactly pi
    Pi sits 0.159 paise above Rs 3.14 and 0.841 paise below Rs 3.15, so paying Rs 3.15 with probability 15.9% and Rs 3.14 otherwise balances exactly at pi, which makes the expected payment fair.
    The relationship
    E[pay]=3.14 (1−p)+3.15 p=π  ⟺  p=π−3.140.01≈0.1593E[\text{pay}] = 3.14\,(1-p) + 3.15\,p = \pi \iff p = \frac{\pi - 3.14}{0.01} \approx 0.1593
    pthe probability of paying Rs 3.15
    \pi - 3.14how far pi sits above the lower whole-paisa amount
    What it says in wordsThe chance of paying the higher amount equals how far along the gap pi lies.

    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.

    Where candidates lose it

    Candidates round to Rs 3.14 and argue the gap is too small to matter. The interviewer is not asking about a sixth of a paisa; the question is whether you see that a fair expected value can be built from amounts that are each individually wrong.

    The second loss is the coin flip. Fifty-fifty between 3.14 and 3.15 feels even-handed but averages 3.145, overpaying by nearly half a paisa every time. The probability has to match where pi sits in the gap.

    What the interviewer asks next

    • How would you hit the probability exactly using only a fair coin?
    • How many coin tosses does that take on average?
    • What if you owe Rs e, 2.71828...?

    Asked at Millennium Management, Quantitative Research, Sheung Wan, 2025 (Wall Street Oasis): How to pay the restaurant fairly if I owe pi dollars. Need to pay with usual dollars and cents.

  2. 056Game A: roll two fair dice and be paid the product of the faces in rupees. Game B: roll one fair die and be paid the square of its face. Which game is worth more, and by exactly how much?Expected value and dice gamesCoreCitadelMiami · 2022

    Try it first

    Which game would you rather play, and by roughly how much?

    Show the worked solution

    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.

    The relationship
    E[X2]=Var(X)+E[X]2=3512+12.25=916≈15.17,E[XY]=E[X] E[Y]=12.25E[X^2] = \mathrm{Var}(X) + E[X]^2 = \tfrac{35}{12} + 12.25 = \tfrac{91}{6} \approx 15.17, \qquad E[XY] = E[X]\,E[Y] = 12.25
    X, Ythe faces of two independent dice
    Var(X)the variance of one die, 35/12
    E[X]the average face, 3.5
    What it says in wordsThe average square is the square of the average plus the variance; the average product of independent dice has no variance term.
    Squaring one die pays for its spread; independent dice do notOne die squared: the six equally likely payouts1face 14face 29face 316face 425face 536face 6mean of squares 15.173.5 x 3.5 = 12.25Game A: two dice, paid the product12.25Game B: one die, paid its square15.17+2.92The gap is the variance of one die:15.17 - 12.25 = 35/12 = 2.92
    The six squares from 1 to 36 average 15.17, above the 12.25 that squaring the average roll gives; the two-dice product game is worth 12.25, so the one-die square game is worth 2.92 more, exactly the variance of one die.

    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.

    Where candidates lose it

    The quick wrong answer is that the games are worth the same, because both feel like three and a half times three and a half. That is true only for independent dice. Squaring one die ties the factors together.

    The second loss is getting 15.17 and 12.25 and not naming the gap. Say that 2.92 is the variance of a die; that one sentence is what the interviewer is waiting for.

    What the interviewer asks next

    • What is game A worth if you are paid the sum of the two dice instead of the product?
    • The second die always shows 7 minus the first. What is the average product now?
    • What is the expected value of the larger of two dice?

    Asked at Citadel, Sales and Trading, Miami, 2022 (Wall Street Oasis): No, pretty typical interview questions, dice questions, etc.