Derivatives Foundation puzzles, solved step by step

Puzzles: 100
Traced to a firm: 66
Topics: 12
Hard: 29

Showing 91–100 of 100

091I deal a shuffled deck face up one card at a time. At any moment you may say stop; if the next card is red you win, otherwise you lose. What strategy maximises your chance of winning, and what is that chance?Expected value and optimal stoppingHardJump TradingChicago · 2018
Try it first
You decide to wait until more reds than blacks remain, then stop. How does that compare with stopping straight away?
Show the worked solution
Every strategy wins with probability exactly 1/2, so there is nothing to optimise. The chance that the next card is red equals the share of red left in the deck, and that share is a fair game: whatever has been dealt, its expected value one card later is its value now. A neat way to see it: the card after you say stop has the same chance of being red as the last card in the deck, and the last card is red with probability 1/2 whatever you do.
Why can't waiting for a good moment help?
Imagine a jar of 26 red and 26 black sweets that a friend pulls out one at a time, and you bet on the colour of the next one. When more reds are left you feel you have edge, but the jar only gets red-heavy after blacks come out, and you do not control which comes out. The share of red remaining has the same expected value one card from now as it has now, so any rule for when to stop is a bet on a fair game and cannot beat 1/2. The fancy name is a martingale: a quantity whose best forecast of its next value is its current value.
Two simulated deals show the share of red among undealt cards wandering around 1/2, with deck A going red-heavy after 7 cards and deck B never going red-heavy and ending on a black card, and the rule of stopping when reds lead gets its chance in 26 deals out of 27 yet wins exactly 1/2, because the deals where it never comes are certain losses.
What is the one-line proof the interviewer wants?
Whatever rule you use, you say stop at some point and win if the next card is red. Swap that bet for a bet on the last card of the deck. Given everything dealt so far, the next card and the last card are both a random draw from the same set of undealt cards, so they are red with the same probability. Every stopping rule wins with the same probability as a bet on the bottom card of the deck, and the bottom card is red half the time. That holds for any strategy, so 1/2 is both the best and the worst you can do. A full check of every position from 26 red and 26 black confirms that the best achievable chance from r red and b black is exactly r/(r + b), the chance of stopping right there.
The relationship
$E\!\left[\frac{R_{k+1}}{N_{k+1}} \;\middle|\; \frac{R_k}{N_k}\right] = \frac{R_k}{N_k} \;\Rightarrow\; P(\text{win}) = \frac{R_0}{N_0} = \frac{26}{52} = \frac{1}{2}$
R_k red cards still undealt after k cards
N_k all cards still undealt after k cards
R_k / N_k the chance the next card is red if you stop now
What it says in wordsYour winning chance at any moment has the same expected value tomorrow as today, so no rule for when to stop can raise its starting value of a half.
Put a number on why the waiting rule fools people. Starting from an even deck, the first black card dealt leaves 26 red in 51, so reds lead immediately half the time, and over the whole deal the moment arrives in 26 deals out of 27. But it typically arrives at a share only just above 1/2, and in the 1 deal in 27 where it never arrives, the last card is black for certain. The edge in the good deals is paid for exactly by the sure loss in the bad ones. The limitation of the result is the payoff: if you were paid more for winning late, or could bet different amounts, the game would no longer be fair and timing could matter.
Where candidates lose it
The common loss is proposing the wait-for-reds-to-lead rule and claiming a small edge, usually around 51%. The interviewer then asks what happens if reds never lead, and the edge disappears. Count both branches before you claim anything.
The second loss is starting a dynamic programme over 27 by 27 states in the room. It works, but it takes far too long. The last-card argument settles it in one sentence and is what the question is testing.
What the interviewer asks next
Now you win Rs 2 if the next card is red and lose Rs 1 if black, and you must stop at some point. Does timing matter?
With 3 red and 1 black, what is your chance, and can any strategy change it?
What changes if you may skip a card without it being revealed?
Asked at Jump Trading, Research, Chicago, 2018 (Wall Street Oasis): I'm dealing a deck of poker, you can stop me anytime. If the next card is red, you win.
092Screen quotes: the 95 call is 7.00 bid, 7.40 offered, and the 100 call is 4.10 bid, 4.50 offered. You buy the 95 call and sell the 100 call. What do you pay, what is the most you can make and lose, and where do you break even at expiry?Option payoffs and no-arbitrageCoreWolverine Trading, Chicago, ILUSA · 2019
Try it first
What does the call spread cost you to put on?
Show the worked solution
You pay 3.30; the most you can make is 1.70, the most you can lose is 3.30, and you break even at 98.30. You buy the 95 call at the offer, 7.40, and sell the 100 call at the bid, 4.10. At expiry the spread is worth nothing below 95, the stock minus 95 between the strikes, and 5 above 100. So the profit runs from minus 3.30 to 5 - 3.30 = 1.70, and is zero when the stock is 95 + 3.30 = 98.30.
Which side of each quote do you deal on?
At a currency counter at the airport there are two rates on the board: the one they buy at and the one they sell at, and you always get the worse one for you. Option screens are the same. If you want to trade now, you buy at the offer and sell at the bid, so a two-leg trade pays the spread on both legs. Buy the 95 call at 7.40, sell the 100 call at 4.10, net debit 3.30. At the mid prices, 7.20 and 4.30, the same trade would cost 2.90. The 0.40 difference is the cost of crossing two bid-offer spreads of 0.40 each, half of each.
Bought across the bid-offer for 3.30, the 95/100 call spread loses 3.30 below 95, gains one for one between the strikes, makes 1.70 above 100 and breaks even at 98.30, while the same spread at mid prices would cost 2.90 and break even at 97.90.
How do you read off the maximum profit, loss and break-even?
Walk the stock price up. Below 95 both calls expire worthless, so you lose the 3.30 you paid. Between 95 and 100 only the long call pays, one for one, so the profit rises from minus 3.30. Above 100 the short call pays out as fast as the long call pays in, so the value is capped at the strike gap of 5. A call spread can never be worth more than the gap between its strikes, so the most you can make is 5 minus what you paid, 1.70, and you break even where the stock has risen 3.30 above the lower strike, at 98.30.
The relationship
$\text{cost} = 7.40 - 4.10 = 3.30, \qquad \text{max profit} = (100 - 95) - 3.30 = 1.70, \qquad S^{*} = 95 + 3.30 = 98.30$
7.40 the offer on the 95 call, what you pay to buy it
4.10 the bid on the 100 call, what you receive to sell it
S* the break-even stock price at expiry
What it says in wordsPay the offer, receive the bid, and the spread's payoff is boxed between losing the premium and making the strike gap less the premium.
Two checks to say aloud. First, the price passes the no-arbitrage bounds: a 95/100 call spread must cost between 0 and 5, and 3.30 does. Second, the risk-reward: you risk 3.30 to make 1.70, which only makes sense if you think the stock finishes above 98.30 more than 3.30/5 = 66% of the time. The limitation: this is the payoff at expiry. Before expiry the spread's value moves with volatility and time, and you could unwind it, but again only by selling the 95 at its bid and buying the 100 at its offer.
Where candidates lose it
The common loss is using mid prices and answering 2.90, as if the screen would trade with you at the middle. The interviewer gave you two-sided quotes precisely to see whether you know which side you hit.
The second loss is getting the maximum profit wrong by forgetting the cap, and saying it is unlimited because you own a call. You also sold one, and above 100 the two cancel; the most a 5-wide spread can ever be worth is 5.
What the interviewer asks next
What would you pay if you could work both orders at mid?
At what price would you buy the 95 call and sell the 100 call so that you risk exactly as much as you can make?
Build the same view with puts. What are the costs on this screen if the 95 put is 1.80 at 2.10 and the 100 put is 3.70 at 4.00?
Asked at Wolverine Trading, Prop Trading, Chicago, IL, USA, 2019 (Wall Street Oasis): pricing options given an ask and a bid price for options with different strikes if you were to short one and long another
093A stock at 100 will be 120 with probability 70% or 90 with probability 30% in one period, and interest rates are zero. Price a call struck at 100. Why does the 70% not appear in your answer?Option pricing intuitionCoreQuant trading
Try it first
What is the call worth?
Show the worked solution
The call is worth 20/3, about 6.67, and the 70% does not matter because the call can be copied with stock and cash. Hold 2/3 of a share and borrow 60: at 120 the copy is worth 80 - 60 = 20, at 90 it is worth 60 - 60 = 0, matching the call in both states. The copy costs 66.67 - 60 = 6.67, so the call must too. The real-world odds are already in the stock price, which the copy uses.
How do you copy the call?
If a shop sells a gift box of two items for more than the items cost separately, you buy the items and skip the box; the box's price is pinned by what goes in it. Options work the same way. Find a mix of stock and cash that pays exactly what the call pays in every state, and the call must cost what the mix costs, whatever anyone believes about the odds. The call pays 20 or 0, a swing of 20, while the stock swings from 120 to 90, a swing of 30. So hold 20/30 = 2/3 of a share. At 120 that is 80, which is 60 too much, and at 90 it is 60, also 60 too much: borrow 60 today and repay it in either state.
Two thirds of a share financed with 60 of borrowing pays 20 when the stock goes to 120 and 0 when it goes to 90, exactly like the call, so the call costs what that portfolio costs, 2/3 x 100 - 60 = 6.67, and the real-world 70% chance of an up move, which would give an average payoff of 14, never enters.
Where did the 70% go?
It is in the stock price. A stock that goes up 70% of the time to 120 and is still priced at 100 today is one the market demands a return on, because it is risky. The copy buys the stock at that price, so it inherits whatever the market thinks of the odds and the risk. The option is priced relative to the stock, not relative to anyone's forecast, so the real-world probability cancels out of the answer. What does appear is a different probability, q, the one that makes the stock earn the risk-free rate: 100 = q x 120 + (1 - q) x 90, so q = 1/3. Discounting the call's payoff at q gives 1/3 x 20 = 6.67, the same answer by another route.
The relationship
$\Delta = \frac{20 - 0}{120 - 90} = \frac{2}{3}, \qquad C = \Delta S - B = \tfrac{2}{3}(100) - 60 = 6.67, \qquad q = \frac{100 - 90}{120 - 90} = \tfrac{1}{3}$
Delta shares held in the copy, the call's swing over the stock's swing
B cash borrowed, 60, so the copy pays nothing in the down state
q the risk-neutral probability of the up move, not a forecast
What it says in wordsHold two thirds of a share, borrow sixty, and you have built the call for 6.67; the risk-neutral probability of a third gives the same number.
Then show the arbitrage, since that is what makes the answer binding. If the call traded at 8, sell it and buy the copy for 6.67: you pocket 1.33 today and the two positions cancel in both states. If it traded at 5, do the reverse. The limitation is the one-step world: real prices take many values, so the copy has to be rebalanced as the stock moves, which is where the Black-Scholes model comes from, and where trading costs and jumps make the copy imperfect.
Where candidates lose it
The common loss is answering 14, the call's expected payoff under the stated odds. It is the natural first move and the one the question is built to catch. Expected payoff under real-world odds is not a price unless everyone is indifferent to risk.
The second loss is getting q = 1/3 and then calling it the true chance of an up move. It is not a forecast. Say what it is: the probability that makes the stock's expected return equal the risk-free rate.
What the interviewer asks next
Price the put struck at 100 in the same tree, and check put-call parity.
If interest rates were 5% for the period, what is q and what is the call worth?
The stock's up probability rises to 90% but its price stays at 100. What happens to the call price, and why?
094A broad equity index has returned an average of 11% a year over the past 30 years, with annual volatility of 16%. Give a 95% confidence interval for its true expected annual return.Distributions and statisticsCoreOld Mission CapitalChicago · 2025
Try it first
How wide is the 95% interval for the true expected return, either side of 11%?
Show the worked solution
About 5.3% to 16.7%. The standard error of a 30-year average is the yearly volatility over the square root of 30: 16/5.48 = 2.92 points. A 95% interval is 1.96 standard errors either side, about 5.7 points, so 11% plus or minus 5.7%. Thirty years of data leave the true expected return anywhere from modest to spectacular, and narrowing it to plus or minus one point would take nearly a thousand years.
Why is the interval so wide after 30 years?
Weigh yourself on a scale that is off by up to two kilos each time. One reading tells you little; the average of four readings is better, but only twice as good, not four times, because errors cancel in proportion to the square root of the count. The uncertainty in an average shrinks with the square root of the number of observations, so 30 noisy years cut a 16-point yearly spread only to about 2.9 points. That is the standard error. Multiply by 1.96 for 95% and you get roughly 5.7 points either side of 11%.
The 95% interval for the true expected return narrows only with the square root of the years of data, so with an 11% average and 16% volatility it runs from about -3% to 25% after 5 years, from 5.3% to 16.7% after 30 years, and would need about 983 years to shrink to plus or minus one point.
What does the interviewer do with the answer?
Usually they push on what it means. A 30-year history cannot tell a 6% market from a 16% market with any confidence, which is why long-run return assumptions are judgements, not measurements. To get the interval down to plus or minus 2 points needs (1.96 x 16/2)^2 = 246 years, and to plus or minus 1 point, 983 years. Volatility, by contrast, is estimated far better from the same data, because daily or monthly returns give thousands of observations of spread, while there is only one 30-year path for the mean. Sampling more often does not help the mean: the average depends only on the first and last levels.
The relationship
$\bar{r} \pm 1.96\,\frac{\sigma}{\sqrt{n}} = 11\% \pm 1.96 \times \frac{16\%}{\sqrt{30}} = 11\% \pm 5.73\% = [5.27\%,\; 16.73\%]$
r bar the sample average yearly return, 11%
sigma the volatility of one year's return, 16%
n the number of years, 30
1.96 the multiplier for a 95% normal interval
What it says in wordsThe average of 30 noisy years is itself noisy, with an error of the volatility over the square root of 30, so the interval is about 5.7 points either side.
Say the assumptions, because the interviewer will. The interval treats the 30 yearly returns as independent draws from one unchanging distribution, which real markets are not; regime shifts make the true uncertainty wider. The 11% and 16% are the inputs given for this exercise, so confirm any real index's figures before using them. And an arithmetic average of yearly returns is not the compound growth rate; with 16% volatility the compound rate is roughly 1.3 points lower, which is a separate question worth flagging rather than answering.
Where candidates lose it
The common loss is using 16% as the error, giving 11% plus or minus 31 points, which confuses the spread of one year with the uncertainty of an average. The opposite loss is a tiny interval from forgetting that only 30 independent years exist.
The second loss is reaching for daily data to shrink the interval. More frequent sampling sharpens the volatility estimate, not the mean: the total return over 30 years depends only on where the index started and where it ended.
What the interviewer asks next
How many years of data would you need to tell a 6% expected return from an 8% one?
Why does more frequent data help estimate volatility but not the expected return?
If returns were autocorrelated, would the interval be wider or narrower?
Asked at Old Mission Capital, Prop Trading, Chicago, 2025 (Wall Street Oasis): Confidence interval on S&P 500 return past 30 years
095Three people stand in a line facing forward, wearing hats drawn from 3 red and 2 blue. The back person sees the two hats ahead, the middle person sees only the front hat, and the front person sees none. The back says "I don't know my colour", then the middle says "I don't know my colour". What colour is the front person's hat?Games and logicCoreWolverine Trading, Chicago, ILUSA · 2019
Try it first
What does the back person's "I don't know" rule out?
Show the worked solution
Red. The back person would know his hat if he saw two blues, since only two exist; his silence rules that out. The middle person now knows the two front hats are not both blue. If he saw blue on the front person, he would know his own was red, and he would say so. His silence means the front hat is not blue. So the front person, who sees nothing, deduces red from the two silences alone.
What does a silence tell everyone else?
If a friend who can see the scoreboard says she cannot tell who is winning, you learn that the scores are close. Her not knowing is information, because you know what she would have said if they were not. Each "I don't know" rules out every world in which that person would have known, and everyone behind and in front can use it. The back person would know only in one world: two blue hats ahead, which forces his own to be red. So his silence removes that world, and the middle and front people both hear it.
Of the four hat pairs the back person might see on the middle and front people, his silence crosses out blue and blue, the middle person's silence then crosses out a red middle with a blue front, and both rows that survive have a red hat on the front person, which is why the front person knows the answer without seeing anything.
How does the middle person's silence finish it?
The middle person now knows that he and the front person are not both blue. He looks at the front hat. If it is blue, he cannot be blue too, so he is red, and he would say so. He does not. The middle person's silence can only mean he sees a red hat in front, because a blue one would have told him his own colour. The front person runs the same reasoning, does not need to see anything, and says red. The full check enumerates the seven possible hat triples, removes the ones where the back person would know, then the ones where the middle person would, and every survivor has red in front.
The relationship
$\{\text{back silent}\} \Rightarrow \neg(M = B \wedge F = B), \qquad \{\text{middle silent}\} \Rightarrow \neg(F = B) \;\Rightarrow\; F = R$
M, F the middle and front hats
B, R blue and red
the arrow what each silence lets everyone conclude
What it says in wordsThe first silence removes the case of two blues ahead, and the second removes a blue in front, which leaves only red.
Then say what it rests on, because that is the interview point. The puzzle needs common knowledge: everyone knows the hat counts, everyone reasons perfectly, and everyone knows the others do too. If the middle person might simply be slow, his silence carries no information and the front person learns nothing. On a trading floor the same logic runs in the other direction: a counterparty who could have traded and chose not to has told you something, and reading those non-events is part of the job.
Where candidates lose it
The common loss is saying the front person cannot know anything because they see nothing. That ignores that the two silences are data. The question is built to see whether you treat a non-answer as information.
The second loss is running the logic from the front. Start with the person who has the most information, the back, ask what would have let him know, and strike that case. Then move forward one person at a time.
What the interviewer asks next
Suppose the back person says "I know". What can the other two conclude?
With 2 red and 3 blue hats, does the same chain of silences tell the front person anything?
If only the back person speaks and says "I don't know", what can the middle person conclude about his own hat when he sees red in front?
Asked at Wolverine Trading, Prop Trading, Chicago, IL, USA, 2019 (Wall Street Oasis): a brain teaser about the hat problem where 3 people go into a room with I think 3 red and 2 blue hats
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.
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
$E = \sum_{n=1}^{10} 2^n \cdot \frac{1}{2^n} \;+\; 1024 \cdot P(\text{first tail on flip } 11 \text{ or later}) = 10 + 1024 \cdot \frac{1}{1024} = 11$
2^n the payout if the first tail comes on flip n
1/2^n the chance the first tail comes on flip n
1/1024 the 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.
097I pick a whole number from 1 to 100 and know it; you do not. You quote 45 at 55 and I buy at 55. You requote 58 at 68 and I buy again at 68. What have you learned, and where should your next quote be?Market makingHardJane StreetNew York · 2025
Try it first
After the first lift at 55, where is the middle of what the number could still be?
Show the worked solution
The number is between 69 and 100, so centre the next quote on 84.5, say 80 at 89. Someone who knows the number buys only when your offer is too low. The lift at 55 put it in 56 to 100, midpoint 78, and your requote at 58 to 68 was far too low, so it was lifted too. Now it is in 69 to 100. You are short two at 55 and 68, worth about -46 against a fair value of 84.5, and the next quote should sit on what is left, not drift up from the last one.
What does a trade tell you when the other side knows the answer?
If a friend who has already peeked at the exam paper offers to bet you that the pass mark is above 55, you do not need to ask why. A counterparty who knows the number only trades when your price is wrong in their favour, so every lift of your offer is proof that the number is above it, and your fair value should jump to the middle of what is left. After the lift at 55 the number lies in 56 to 100, centred on 78. The requote at 58 to 68 treated the lift as noise and moved only 13 points. A quote centred near 78, say 73 at 83, would have used the information.
The first lift at 55 leaves 56 to 100 with a midpoint of 78, the requote of 58 at 68 sits 15 below that midpoint and is lifted again, which leaves 69 to 100 with a midpoint of 84.5, so the next quote of 80 at 89 is centred on what can still be true.
Where should the next quote sit, and what have the trades cost?
After the second lift the number is in 69 to 100: 32 values, mean 84.5. Centre the next quote on the middle of the remaining range, 84.5, so that whichever side is hit, the range halves, which is the fastest way to stop losing to someone who knows more than you. A quote of 80 at 89 does that. The damage so far: you sold at 55 and 68 something now worth about 84.5, an expected loss of 29.5 + 16.5 = 46. Most of it came from the timid requote; a quote centred on 78 would have been lifted only if the number was above 83.
The relationship
$E[X \mid X > 55] = \frac{56 + 100}{2} = 78, \qquad E[X \mid X > 68] = \frac{69 + 100}{2} = 84.5$
X the number I picked, uniform on 1 to 100 before any trade
X > 55 what my buying at 55 reveals
84.5 the fair value after both lifts, where the next quote should be centred
What it says in wordsEach lift cuts the range to everything above your offer, and fair value moves to the middle of what is left.
Then say how the answer changes with the counterparty, because that is the judgement being marked. If I might be guessing rather than knowing, a lift is weaker evidence and you should move less, perhaps halfway towards 78. If I know the number exactly, any tight quote inside the range loses whenever it trades, and the only quote that cannot lose is one that spans the whole range, 69 bid, 100 offered, which is no market at all. Real desks live between the two: they widen against flow they think is informed and tighten against flow they think is not.
Where candidates lose it
The common loss is moving the quote up a few points after each lift, as the 58 at 68 requote did. It treats an informed trade as a random one and pays for the lesson again on the next trade.
The second loss is the opposite overreaction: refusing to quote, or quoting 1 at 100. The interviewer wants a market that uses the information and still trades, centred on the conditional mean of what is left.
What the interviewer asks next
Instead I sell to you at 45 on the first quote. What is your next quote?
Suppose I know the number only half the time and guess otherwise. How far should the first lift move you?
How many lifts or hits does it take, at most, to find the number if you always quote around the middle of the range?
Asked at Jane Street, Generalist, New York, 2025 (Wall Street Oasis): It was a probability theory based quant trading style market making questions which were intense
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.
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
$P(\text{edge} \mid 8/10) = \frac{0.1 \times 0.1209}{0.1 \times 0.1209 + 0.9 \times 0.0439} \approx 0.234$
0.1 the share of traders with an edge
0.1209 the chance of exactly 8 wins in 10 at 60% a day
0.0439 the 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
099Daily returns come from a normal with 1% volatility on 90% of days and a normal with 3% volatility on the other 10%, both with mean zero. What are the overall volatility and kurtosis, and what does this do to out-of-the-money option prices?Distributions and statisticsHardTwo SigmaNew York · 2025
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What is the overall daily volatility?
Show the worked solution
Volatility 1.34% a day and kurtosis 8.33, against 3 for a normal, which makes far out-of-the-money options worth much more than a single-volatility model says. Variance is 0.9 x 1 + 0.1 x 9 = 1.8. The fourth moment is 3 x (0.9 x 1 + 0.1 x 81) = 27, so kurtosis is 27/1.8^2 = 8.33. With the same volatility, a 4% move is 6.4 times as likely as the normal says, and a one-day put 4% out of the money is worth about 23 times as much.
How do the moments of a mixture combine?
Picture a road with quiet days and occasional storms. Most days the traffic varies a little; on storm days it varies a lot. Averaged over a year, the variability you see is dominated by the storm days, out of proportion to how rare they are. For a mixture with mean zero, each even moment is the weighted average of the components' moments: variances add up by weight, and fourth moments, which grow with the fourth power of volatility, are dominated by the rare wild days. Variance: 0.9 x 1 + 0.1 x 9 = 1.8, a volatility of 1.34%. Fourth moment: each normal's is 3 sigma^4, so 3 x (0.9 x 1 + 0.1 x 81) = 27. The 3% days are a tenth of the time but supply 90% of the fourth moment.
Against a normal with the same 1.34% daily volatility, the mixture of 1% and 3% days is taller in the middle and lies above the normal in the tails beyond about 3.7%, so its kurtosis is 8.33 instead of 3 and a move beyond 4% either way is 6.4 times as likely.
What does the fat tail do to option prices?
Kurtosis is the fourth moment over the square of the variance: 27/3.24 = 8.33. A single normal fitted to the same data has the same volatility, 1.34%, but kurtosis 3, so it puts far too little weight on big moves. An option far out of the money pays only on a big move, so its value is driven by the tail, and a mixture that keeps the volatility but fattens the tail makes it worth many times what the matched normal says. A one-day put 4% out of the money is worth 0.0127% of spot under the mixture against 0.00055% under the normal, about 23 times as much. To match the mixture's price, a single-volatility model needs 1.91% for that put but only 1.20% at the money: the smile.
The relationship
$\sigma^2 = 0.9(1)^2 + 0.1(3)^2 = 1.8, \qquad \kappa = \frac{3\,[\,0.9(1)^4 + 0.1(3)^4\,]}{(1.8)^2} = \frac{27}{3.24} \approx 8.33$
sigma^2 the variance of the daily return, in square percentage points
kappa the kurtosis, 3 for any single normal
3 sigma^4 the fourth moment of a normal with mean zero
What it says in wordsVariances average by weight, fourth moments average by weight, and the ratio comes out near 8.3 because the rare 3% days dominate the fourth power.
Say what it means on a desk. This mixture is the simplest model of regime-switching volatility, and it reproduces two things a single normal cannot: a peak that is too tall and tails that are too fat, the shape seen in daily returns of most liquid assets. It is also why at-the-money options can look rich and far wings cheap if you price both off one historical volatility. The limitation: the mixture here is symmetric and draws each day independently. Real markets cluster their wild days and fall harder than they rise, which adds skew to the smile, not just curvature.
Where candidates lose it
The common loss is averaging the volatilities, 0.9 x 1 + 0.1 x 3 = 1.2%, and then getting the kurtosis wrong as a result. Volatilities never average in a mixture; variances do.
The second loss is computing the moments correctly and stopping, without connecting kurtosis to option prices. The question asks what the fat tail does to the wings, and the answer is a smile: more implied volatility the further out of the money you go.
What the interviewer asks next
What is the kurtosis if the wild days are 5% volatility but only 4% of days?
Which is more mispriced by a single-volatility model, an at-the-money option or a far out-of-the-money one, and in which direction?
How would you add skew to this model?
Asked at Two Sigma, Quantitative Research, New York, 2025 (Wall Street Oasis): They asked a couple questions involving Mixture Gaussians (e.g., probability density and moments).
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
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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.
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
$E[\max(D_1, D_2)] = \sum_{k=1}^{6} k \cdot \frac{2k - 1}{36} = \frac{161}{36} \approx 4.47, \qquad 4.47 - 1 = 3.47 < 3.50 < 4$
max(D1, D2) the higher face of two dice
(2k - 1)/36 the chance the higher face is exactly k
1 the 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.

R_k	red cards still undealt after k cards
N_k	all cards still undealt after k cards
R_k / N_k	the chance the next card is red if you stop now

7.40	the offer on the 95 call, what you pay to buy it
4.10	the bid on the 100 call, what you receive to sell it
S*	the break-even stock price at expiry

Delta	shares held in the copy, the call's swing over the stock's swing
B	cash borrowed, 60, so the copy pays nothing in the down state
q	the risk-neutral probability of the up move, not a forecast

r bar	the sample average yearly return, 11%
sigma	the volatility of one year's return, 16%
n	the number of years, 30
1.96	the multiplier for a 95% normal interval

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Why can't waiting for a good moment help?

What is the one-line proof the interviewer wants?

Where candidates lose it

What the interviewer asks next

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Which side of each quote do you deal on?

How do you read off the maximum profit, loss and break-even?

Where candidates lose it

What the interviewer asks next

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How do you copy the call?

Where did the 70% go?

Where candidates lose it

What the interviewer asks next

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Why is the interval so wide after 30 years?

What does the interviewer do with the answer?

Where candidates lose it

What the interviewer asks next

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What does a silence tell everyone else?

How does the middle person's silence finish it?

Where candidates lose it

What the interviewer asks next

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Why does every flip add exactly one rupee?

What goes wrong without the cap?

Where candidates lose it

What the interviewer asks next

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What does a trade tell you when the other side knows the answer?

Where should the next quote sit, and what have the trades cost?

Where candidates lose it

What the interviewer asks next

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

How do you run the numbers in the room?

Where candidates lose it

What the interviewer asks next

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How do the moments of a mixture combine?

What does the fat tail do to option prices?

Where candidates lose it

What the interviewer asks next

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How much is the second die really worth?

Does playing 1,000 times change the choice?

Where candidates lose it

What the interviewer asks next