Derivatives Foundation puzzles, solved step by step

Puzzles
100
Traced to a firm
66
Topics
12
Hard
29
Topic
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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Showing 11–13 of 13 · filtered from 100Clear filters
  1. 085You draw two cards without replacement and win if the first is black and the second red. Would you rather draw from one 52-card deck or from a 104-card double deck?Probability and countingWarm upOld Mission CapitalNew York · 2014

    Try it first

    Before you compute: which deck, and why?

    Show the worked solution

    The single deck, 25.49% against 25.24%. The first card is black with probability 1/2 in either deck. The second draw is where they differ: with one deck, 26 reds remain in 51 cards, 50.98%; with two decks, 52 remain in 103, 50.49%. Multiply: 26/52 x 26/51 = 25.49% and 52/104 x 52/103 = 25.24%. Removing a black card tilts a small deck towards red more than it tilts a large one.

    Why does the size of the deck matter when the mix is the same?

    Take one boy out of a class of 20 with 10 boys and 10 girls, and the class is 10 girls in 19, 52.6% girls. Take one boy out of a school of 2,000 split evenly, and it is 1,000 in 1,999, 50.03%. The same removal is a bigger share of a smaller group. Drawing without replacement makes the second draw depend on the first, and the dependence is stronger the smaller the deck, which is exactly what the game rewards. You want a black card to make red more likely next, and a single deck gives you more of that lean.

    Removing one black card tilts a small deck more than a big one1246810number of 52-card decks shuffled together25.0%25.2%25.4%25.6%P(black then red); the axis starts at 25%, not zerolimit with infinite decks: 25.00%25.49%25.24%25.12%25.05%The second drawone deck26 / 51 = 50.98%two decks52 / 103 = 50.49%the first draw is 50%in both; only this differs
    The chance of black then red is 25.49% with one deck, 25.24% with two and 25.12% with four, falling towards the 25% that independent draws would give, because the second draw after a black card is 26 of 51 in a single deck but only 52 of 103 in a double deck.

    What is the general pattern, and what is its limit?

    With n decks shuffled together, the chance is 1/2 x 26n/(52n - 1). As n grows the second factor falls towards 1/2 and the product towards 1/4, which is the answer you would get if the draws were independent. Every finite deck beats 25%, and the smallest deck beats it by the most, because the minus one in the denominator is a larger share of a smaller count. The whole effect is small, a quarter of a percentage point between one deck and two, which is also worth saying: the interviewer wants the direction and the reason more than the third decimal.

    The relationship
    Pn=26n52n⋅26n52n−1=12⋅26n52n−1  →  14P_n = \frac{26n}{52n}\cdot\frac{26n}{52n-1} = \frac{1}{2}\cdot\frac{26n}{52n-1} \;\to\; \frac{1}{4}
    nthe number of 52-card decks combined
    26n/(52n - 1)the chance of red once one black card is gone
    1/4the limit as the deck becomes infinitely large
    What it says in wordsThe first draw is always a half; the second draw's lean towards red shrinks as the deck grows.

    Then show you can flip it. If the game paid on black followed by black, the single deck would be worse: 26/52 x 25/51 = 24.51% against 52/104 x 51/103 = 24.76%, and the big deck wins. Same mechanism, opposite sign. Saying that unprompted is what turns a one-line puzzle into evidence that you understand sampling without replacement rather than remembering an answer.

    Where candidates lose it

    The fast wrong answer is that the decks are the same because both are half red. That is true of the first draw and false of the second; the question is about the pair.

    The second loss is computing 25.49% and 25.24% and then picking the double deck because there are more reds in it. More reds and more blacks in the same ratio is not more red; only the removal effect differs, and it favours the small deck.

    What the interviewer asks next

    • Now you win on black then black. Which deck?
    • What is the chance of black then red if you draw with replacement?
    • Three cards: black, red, black. Which deck, and does the answer still favour the smaller one?

    Asked at Old Mission Capital, Quantitative Research, New York, 2014 (Wall Street Oasis): If your goal is to draw a black card followed by a red card, which deck would you choose?

  2. 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?

  3. 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.