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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Showing 51–52 of 52 · filtered from 100Clear filters
  1. 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.
  2. 098One trader in ten has a real edge and wins 60% of days; the rest win 50%. A new hire wins 8 of her first 10 days. What is the probability she has an edge?Conditional probability and BayesCoreBelvedere TradingChicago · 2022

    Try it first

    Roughly how likely is it that she has an edge?

    Show the worked solution

    About 23%. With an edge, 8 wins in 10 has probability 45 x 0.6^8 x 0.4^2 = 12.1%; without, 45/1,024 = 4.4%. Out of 1,000 traders, 100 have an edge and about 12.1 of them go 8 of 10; 900 do not and about 39.6 of them go 8 of 10 by luck. So 12.1 of 51.6, about 23.4%, have an edge. A strong start raises the chance from 10% but leaves luck the likelier story.

    Why is the answer so much lower than her win rate?

    If a test for a rare condition comes back positive, most positives can still be false, simply because there are so many more people without the condition to produce them. Trading records work the same way. Ten days is a weak test and skill is rare, so most traders with a strong start are lucky members of the large unskilled group, not members of the small skilled one. The win rate of 80% is what she did, not what she is; the question is which kind of trader is likelier to produce that record, weighted by how many of each kind there are.

    A strong start mostly picks out the lucky, because skill is rarefirst 4 columns: the 100 with an edge (60% days); the other 900 win 50%highlighted: the traders who happen to go 8 of 10Who goes 8 of 10?edge: 100 x 12.1%12.1no edge: 900 x 4.4%39.6share with an edge12.1 / 51.6 = 23.4%up from 10%: the record is 2.75 timeslikelier with an edge, but most8-of-10 starts are still luck
    Of 1,000 new traders, the 100 with an edge produce about 12.1 who go 8 of 10 and the 900 without produce about 39.6, so only about 23% of traders with that start have an edge, even though the record is 2.75 times likelier with one.

    How do you run the numbers in the room?

    Use natural frequencies. Of 1,000 hires, 100 have an edge. Each goes 8 of 10 with probability C(10,8) x 0.6^8 x 0.4^2 = 45 x 0.01680 x 0.16 = 0.1209, so about 12.1 of them. Of the 900 without, each goes 8 of 10 with probability 45/1,024 = 0.0439, so about 39.6. Among the 52 or so traders with 8 wins, about 12 have an edge, a posterior of 23.4%, because a likelihood ratio of 2.75 cannot overcome prior odds of 1 to 9. In odds form: 1/9 x 2.75 = 0.306, and 0.306/(1 + 0.306) = 23.4%.

    The relationship
    P(edge∣8/10)=0.1×0.12090.1×0.1209+0.9×0.0439≈0.234P(\text{edge} \mid 8/10) = \frac{0.1 \times 0.1209}{0.1 \times 0.1209 + 0.9 \times 0.0439} \approx 0.234
    0.1the share of traders with an edge
    0.1209the chance of exactly 8 wins in 10 at 60% a day
    0.0439the chance of exactly 8 wins in 10 at 50% a day
    What it says in wordsWeight each explanation by how common it is and how well it fits the record, then take the edge's share.

    Two refinements the interviewer may ask for. If the record were 8 or more wins rather than exactly 8, the answer rises a little, to about 25%. And the practical point: a desk that promotes on a ten-day record mostly promotes luck. It takes many more days to separate a 60% trader from a 50% one, because the gap in win rate is small next to the day-to-day noise. The limitation of the model is the two-type world; real skill comes in shades, which spreads the posterior out but does not change the lesson about rare skill and short records.

    Where candidates lose it

    The common loss is answering 80% or 60%, confusing the trader's record with the probability that she is skilled. The interviewer is testing whether you start from the base rate.

    The second loss is computing the likelihoods correctly and forgetting the prior, which gives 12.1/(12.1 + 4.4) = 73%. That is the answer for a world where half of all traders have an edge, which is not the world described.

    What the interviewer asks next

    • How many days would she need at an 80% win rate before you were 90% sure she had an edge?
    • If one trader in three had an edge, what would 8 of 10 imply?
    • Her next 10 days are 5 wins and 5 losses. Update the probability.

    Asked at Belvedere Trading, Capital Markets, Chicago, 2022 (Wall Street Oasis): The technical portion of the interview consisted of probability questions including one questions relating to Bayes' theorem