Derivatives Foundation puzzles, solved step by step

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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?