Derivatives Foundation puzzles, solved step by step

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100
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29
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All topicsMental maths and estimation9Random walks and Markov chains7Conditional probability and Bayes7Volatility and correlation7Option pricing intuition7Expected value and optimal stopping10Market making11Option payoffs and no-arbitrage10Probability and counting11Distributions and statistics8Games and logic8Betting and sizing5
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  1. 087What is the expected value of one roll of a fair die? What is the most you would pay to play a game that pays the roll in rupees, and what changes if the game pays the square of the roll instead?Expected value and optimal stoppingWarm upOld Mission CapitalChicago · 2020Wolverine TradingChicago · 2024Tower Research CapitalNew York · 2012

    Try it first

    The game pays the square of the roll. What is it worth?

    Show the worked solution

    3.5, so pay at most Rs 3.50 for the plain game, and Rs 15.17 for the squared one, not Rs 12.25. The plain game averages 1 to 6: 21/6 = 3.5. The squared game averages 1, 4, 9, 16, 25 and 36: 91/6, about 15.17. Squaring the average roll gives only 12.25 and misses 2.92, which is the variance of the roll. A payoff that bends upward is worth more than its value at the average outcome.

    What does fair price mean here?

    If a friend offers to pay you whatever a die shows, and you play it a thousand times, you collect close to 3,500 rupees. Paying more than 3.50 a go loses money over the thousand games; paying less makes it. The fair price of a game is the average of what it pays, weighted by how likely each payout is, and the most a market maker would pay is a little under it, to leave room for edge. For one roll each face has weight 1/6, so the expected value is (1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5. Say both numbers: fair value Rs 3.50, and a bid of perhaps Rs 3.25 if you are asked to trade it.

    The average of the squares beats the square of the average1roll 12roll 23roll 34roll 45roll 56roll 6average 3.5Game 1: pays the rollfair price Rs 3.501roll 14roll 29roll 316roll 425roll 536roll 6average of squares 15.173.5 squared = 12.25Game 2: pays the roll squaredfair price Rs 15.17, not Rs 12.25the gap, 15.17 - 12.25 = 2.92, is the variance of one roll
    The plain game pays 1 to 6 and averages 3.5, while the squared game pays 1, 4, 9, 16, 25 and 36 and averages 15.17, which sits 2.92 above 3.5 squared, and that gap is the variance of a single roll.

    Why is the squared game worth more than 3.5 squared?

    The trap answer squares the average roll and says 12.25. But the game does not pay the square of the average; it pays the square of whatever is rolled. A 6 pays 36, which is 23.75 above 12.25, and a 1 pays 1, only 11.25 below it. Squaring stretches the high outcomes more than it squeezes the low ones, so the average of the squares, 15.17, sits above the square of the average, 12.25. The gap has a name: E[X^2] - (E[X])^2 is the variance of the roll, 35/12, about 2.92.

    The relationship
    E[X]=216=3.5,E[X2]=916≈15.17,E[X2]−(E[X])2=3512≈2.92E[X] = \frac{21}{6} = 3.5, \qquad E[X^2] = \frac{91}{6} \approx 15.17, \qquad E[X^2] - (E[X])^2 = \frac{35}{12} \approx 2.92
    Xthe number rolled
    E[X^2]the fair value of the squared game
    35/12the variance of one roll, the extra value the curve adds
    What it says in wordsThe squared game is worth the plain value squared plus the variance, so more spread means more value.

    Then say why a derivatives desk asks it. An option is a payoff that bends upward, like the square, so it is worth more than its payoff at the expected price, and the extra is driven by the spread of outcomes. That is the whole reason volatility has a price. The limitation: fair value assumes you are indifferent to risk and can play many times. Offered one roll of the squared game for Rs 15, a person who cannot afford to lose Rs 14 might sensibly refuse, and an interviewer will sometimes push on that.

    Where candidates lose it

    On the plain game, candidates sometimes answer 3 or 4 because a die cannot show 3.5. The expected value need not be a possible outcome; it is the long-run average per roll.

    On the squared game the loss is 12.25, squaring the average instead of averaging the squares. It is the exact mistake that leads people to value an option at its payoff on the expected price, which is why the interviewer asks it straight after the easy one.

    What the interviewer asks next

    • What would you pay for a game that pays the cube of the roll?
    • You may reroll once after seeing the first roll. What is the game worth now?
    • Make me a market on the sum of two dice.

    Asked at Old Mission Capital, Trading, Chicago, 2020 (Wall Street Oasis): What is the expected value of a dice roll?
    Asked at Wolverine Trading, Sales and Trading, Chicago, 2024 (Wall Street Oasis): Question about dice, if I rolled a 6 sided dice what is the EV of each roll.
    Asked at Tower Research Capital, Intern Interview -, New York, 2012 (Wall Street Oasis): What is the expected value for rolling a dice?

  2. 100Pick one: (a) roll a die and take the face in rupees, (b) pay Rs 1 to roll two dice and take the higher face, or (c) take a sure Rs 4. Which do you choose, and would the answer change if you played 1,000 times?Expected value and optimal stoppingWarm upJane Streetlondon · 2025

    Try it first

    What is the expected value of the higher of two dice, before the fee?

    Show the worked solution

    Take the sure Rs 4, once or a thousand times. One die is worth Rs 3.50. The higher of two dice averages 161/36 = Rs 4.47, so after the Rs 1 fee it is worth Rs 3.47, the worst of the three. The sure Rs 4 beats both. Over 1,000 plays the choice only gets clearer: the games total about Rs 3,500 and Rs 3,472, each with a spread of about Rs 50, against exactly Rs 4,000.

    How much is the second die really worth?

    If a shop lets you take the better of two mangoes for one rupee extra, the question is how much better the better one usually is, not how good a mango can be. Two dice give you the higher face, which is often a 5 or 6, so it feels valuable. The higher of two dice is k with probability (2k - 1)/36, so its average is 161/36, about 4.47: the second die adds about Rs 0.97, a little less than the Rs 1 it costs. After the fee, option (b) is worth Rs 3.47, below the plain die at Rs 3.50 and well below the sure Rs 4.

    The sure Rs 4 beats both games on average and on a thousand playsRs 0Rs 1Rs 2Rs 3Rs 4Rs 5Rs 6expected payout per play, with one standard deviation3.50(a) one die3.47(b) higher of two4.00(c) sure Rs 4dice, minus Rs 1before fee 4.47Over 1,000 plays(a)Rs 3,500 +/- 54(b)Rs 3,472 +/- 44(c)Rs 4,000 exactlythe gap is about 9 standarddeviations: (c) wins withnear certaintyone play: (a) beats Rs 4 with chance 33.3%, (b) with 30.6%; the sure Rs 4 never loses to itself
    One die is worth Rs 3.50 with a spread of 1.71, the higher of two dice after the Rs 1 fee is worth Rs 3.47 with a spread of 1.40, and the sure Rs 4 has no spread, so over 1,000 plays the games total about Rs 3,500 and Rs 3,472, each give or take about Rs 50, against exactly Rs 4,000.

    Does playing 1,000 times change the choice?

    Played once, a gambler might take (a) for the one-in-three chance of a 5 or 6, or (b) for its 30.6% chance of a 6. Both lose to Rs 4 on average and carry risk, so nobody who dislikes risk should prefer them, and nobody who is neutral to risk should either. Over 1,000 plays the spread of each game's total grows only with the square root of the plays, about Rs 54 for (a) and Rs 44 for (b), while the gap to the sure Rs 4,000 grows with the plays themselves, to Rs 500 and Rs 528. The gap is about nine standard deviations, so the sure option wins with near certainty. Repetition turns expected value into the outcome.

    The relationship
    E[max⁡(D1,D2)]=∑k=16k⋅2k−136=16136≈4.47,4.47−1=3.47<3.50<4E[\max(D_1, D_2)] = \sum_{k=1}^{6} k \cdot \frac{2k - 1}{36} = \frac{161}{36} \approx 4.47, \qquad 4.47 - 1 = 3.47 < 3.50 < 4
    max(D1, D2)the higher face of two dice
    (2k - 1)/36the chance the higher face is exactly k
    1the fee for option (b)
    What it says in wordsThe second die adds just under one rupee of value and costs exactly one, so it loses to the plain die and both lose to the sure four.

    The interviewer usually follows with a market: what would you pay for option (b) without the fee? Fair value is Rs 4.47, so a bid around Rs 4.25 leaves edge. Then they change the numbers: if the sure amount were Rs 3.40, the plain die would become the best choice on average, and the decision for a single play would depend on how much you care about risk. The limitation is the usual one: expected value is the right yardstick only when the stakes are small relative to what you can afford to lose.

    Where candidates lose it

    The common loss is picking (b) because the higher of two dice sounds strong, without subtracting the fee or working out that the second die adds only about Rs 0.97.

    The second loss is saying the answer flips with 1,000 plays, as if repetition favoured the gamble. Repetition shrinks the relative spread of a game's total; it makes the higher expected value more certain to win, and here that is the sure Rs 4.

    What the interviewer asks next

    • What would you pay for option (b) if there were no fee?
    • What is the expected value of the higher of three dice?
    • If the sure amount were Rs 3.40, which would you pick once, and which a thousand times?

    Asked at Jane Street, Trading, london, 2025 (Wall Street Oasis): Given this game, what is the expected value of winning given 3 different strategies, which one would you choose.