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. 008You roll a fair die again and again, adding each face to a pot. But if you roll a 1, the whole pot is wiped out and the game ends. You may stop and bank the pot at any time. When should you stop, and why?Expected value and optimal stoppingHardQuant tradingProp trading firms

    Try it first

    First instinct: with 15 in the pot, should you roll once more?

    Show the worked solution

    Roll while the pot is below 20 and stop once it reaches 20 or more. One more roll loses the pot with probability 1/6 and otherwise adds a face of 2, 3, 4, 5 or 6, which sum to 20. The expected change is (20 minus pot) over 6: +3.33 from an empty pot, +0.83 at 15, zero at 20 and negative beyond. The number of rolls so far is irrelevant; only the pot matters. Played this way, the expected bank from an empty pot is about 8.14.

    What does one more roll actually buy you?

    Think of a street game where you can keep picking envelopes that each add a few rupees to your winnings, but one envelope in six says lose everything. Whether to pick again depends on how much you already hold, not on how many envelopes you have opened. The expected change from one more roll is the chance of adding, five sixths, times the average addition, 4, minus the chance of ruin, one sixth, times the pot you would lose. That is (20 minus pot) over 6, and it is positive exactly while the pot is under 20.

    One more roll is worth (20 minus pot) / 6: positive below a pot of 20, negative above0102030pot already held-2+2+4expected gain from one more rollpot = 20: gain is exactly zeroempty pot: +3.33pot 30: -1.67ROLL AGAINBANK ITone more roll adds valueone more roll loses value
    The expected gain from one more roll falls in a straight line from +3.33 with an empty pot to zero at a pot of 20 and to -1.67 at 30, so rolling adds value on the left of 20 and destroys it on the right, and the stopping rule is simply to bank at 20 or above.
    The relationship
    E[Δ∣pot=p]=56×4−16×p=20−p6⇒stop when p≥20E[\Delta \mid \text{pot}=p] = \frac{5}{6}\times 4 - \frac{1}{6}\times p = \frac{20 - p}{6} \quad\Rightarrow\quad \text{stop when } p \ge 20
    pthe pot already held
    5/6 x 4the chance of surviving the roll times the average of the faces 2 to 6
    p/6the pot at risk, times the one-in-six chance of a 1
    What it says in wordsOne more roll is worth the expected addition minus the expected loss, and the two balance when the pot is 20.

    Why is the one-step rule the whole answer here?

    In many stopping problems you cannot trust a one-step look: a roll that loses value today might open a bigger gain tomorrow. Here the gain from one more roll only falls as the pot grows, so once rolling stops being worth it, it never becomes worth it again, and the one-step rule is optimal. You can check it by valuing the whole game for each stopping threshold: stopping at 20 (or 21, which is equivalent because the roll at exactly 20 is worth zero) gives an expected bank of about 8.14, stopping at 15 gives 7.85 and at 25 gives 8.00. Say the word monotone if you know it; say the reason either way.

    What would a desk add to the textbook answer?

    That the rule maximises expected value and nothing else. A player who cannot afford to lose the pot, or who is paid on a target rather than on the average, would stop earlier, and a trader who sizes positions knows that expected value is only the first thing to check. The limitation is the same in the puzzle and on the desk: expected value is right for a game you can play many times, and this game is played once.

    Where candidates lose it

    Candidates who answer by feel either stop far too early, because a one-in-six wipe-out sounds frightening, or never stop, because the pot keeps growing. The question is asking for the point where the two forces balance, and that point is a number: 20.

    The second loss is using 3.5 as the average face. The faces that keep you in the game are 2 to 6, averaging 4, and their total is 20. Using 3.5 gives a threshold of 21 and shows the ruin branch has been counted twice.

    What the interviewer asks next

    • Now rolling a 1 costs you only half the pot. What is the new stopping rule?
    • What is the expected bank from an empty pot under the optimal rule, and how would you compute it?
    • Suppose you must pay 1 for every roll. Does the threshold go up or down, and by how much?
  2. 096A game flips a fair coin until the first tail and pays Rs 2^n if the first tail comes on flip n, but the house can pay at most Rs 1,024. What is the fair price, and why does the uncapped game break the idea of a fair price?Expected value and optimal stoppingCoreMarket makingProp trading firms

    Try it first

    How much does each possible flip contribute to the expected payout, below the cap?

    Show the worked solution

    Rs 11. A first tail on flip n pays 2^n with probability 1/2^n, so each of flips 1 to 10 contributes exactly Rs 1, Rs 10 in all. Reaching flip 11 or later has probability 1/1,024 and pays the cap of Rs 1,024, adding one more rupee. Without the cap every flip keeps adding Rs 1 and the expected value is infinite, yet nobody would pay much to play, which shows expected value alone cannot price a bet on outcomes the payer cannot honour.

    Why does every flip add exactly one rupee?

    Think of a raffle where each ticket in the next bundle is half as likely to win but the prize is twice as large. Every bundle is worth the same to you. Here the first tail on flip 1 pays Rs 2 half the time, on flip 2 pays Rs 4 a quarter of the time, on flip 3 pays Rs 8 an eighth of the time. The payout doubles exactly as the probability halves, so each flip contributes 2^n x 1/2^n = Rs 1 to the expected value, and the fair price is simply the number of flips before the cap bites, plus the capped tail. Ten flips reach Rs 1,024; anything later is paid at the cap, worth 1,024 x 1/1,024 = Rs 1.

    Every flip adds exactly Rs 1, so the cap is the whole priceRs 11121/21241/41381/814161/1615321/3216641/64171281/128182561/256195121/5121101,0241/1,024111+1,0241/1,02411uncapped: Rs 1 morefor every flip, foreverflippaysprobfirst tail on flip n: payout x probability = Rs 1 for every n up to the capcapped tail10 bars of Rs 1 + the capped tail Rs 1 = fair price Rs 11, though 7 times in 8 the game pays Rs 8 or less
    Each flip from 1 to 10 contributes payout times probability of exactly Rs 1, and reaching flip 11 or later adds the capped Rs 1,024 times a 1 in 1,024 chance, another Rs 1, for a fair price of Rs 11, while without the cap the Rs 1 contributions would carry on forever.

    What goes wrong without the cap?

    Remove the cap and the sum is 1 + 1 + 1 + ... with no end, an infinite expected value. Yet the game pays Rs 8 or less seven times in eight, and more than Rs 11 only one time in eight. The infinite value lives entirely in outcomes so rare and so large that no house could pay them, so the cap is not a detail; it is the price. Raising the cap barely moves it: a cap of Rs 1 crore makes the game worth only about Rs 24.19, because each doubling of the cap adds one rupee. That is the market maker's answer to the paradox: price what the counterparty can actually pay.

    The relationship
    E=∑n=1102n⋅12n  +  1024⋅P(first tail on flip 11 or later)=10+1024⋅11024=11E = \sum_{n=1}^{10} 2^n \cdot \frac{1}{2^n} \;+\; 1024 \cdot P(\text{first tail on flip } 11 \text{ or later}) = 10 + 1024 \cdot \frac{1}{1024} = 11
    2^nthe payout if the first tail comes on flip n
    1/2^nthe chance the first tail comes on flip n
    1/1024the chance the first ten flips are all heads
    What it says in wordsTen rupees from the flips the cap does not touch, and one more from the capped tail.

    Mention the other classic resolution, then put it in its place. Economists answer the paradox with diminishing utility: a doubling of wealth is worth less to you than the first rupee, so a risk-averse player pays little. True, but on a desk the binding constraint is usually the counterparty's balance sheet, not your utility. The limitation of the Rs 11 is the same as any fair value: it is an average over many plays. Played once, the price feels steep because 87.5% of the time you get Rs 8 or less.

    Where candidates lose it

    The common loss is computing an infinite expected value and stopping there, or saying the cap makes the game worth Rs 1,024. The cap removes the infinity but adds only one rupee for the capped tail.

    The second loss is getting the boundary wrong: counting 11 flips below the cap, or forgetting the capped tail entirely and saying Rs 10. Write the n = 10 payout, Rs 1,024, next to the cap and the count is obvious.

    What the interviewer asks next

    • What is the fair price if the house can pay at most Rs 1 crore?
    • What is the median payout of the uncapped game?
    • Make me a market on this capped game, and say which side you would rather be on if the house's credit were in doubt.