Derivatives Foundation puzzles, solved step by step

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

Showing 1–10 of 19 · filtered from 100Clear filters

004Two stocks each have 30% annual volatility and a correlation of 0.5. What is the volatility of a basket that holds half of each? What if the correlation were zero?Volatility and correlationWarm upEquity derivativesRisk management
Try it first
Before the formula: can a 50/50 basket of two 30% stocks ever be more volatile than 30%?
Show the worked solution
25.98% at a correlation of 0.5, and 21.21% at zero. Basket variance is the sum of the two weighted variances plus twice the weighted covariance: 0.25 x 0.09 + 0.25 x 0.09 + 2 x 0.25 x 0.5 x 0.09 = 0.0675, whose square root is 25.98%. At zero correlation the cross term vanishes, leaving 0.045, whose root is 21.21%. The basket is less volatile than either stock because they do not move in step.
Why is the basket calmer than the stocks inside it?
Two commuters who each arrive late by a random ten minutes rarely arrive late together; the average of their lateness swings less than either does alone. Volatilities do not add; variances do, and the cross term that joins them is scaled by the correlation, so anything below a correlation of 1 cuts the basket's swing below its parts. At a correlation of 1 the two stocks are one stock and you get 30% back. At minus 1 they cancel exactly and the basket is flat.
Basket volatility rises with correlation from 0% at minus 1 through 21.21% at zero and 25.98% at 0.5 to the parts' 30% only at a correlation of 1, so the gap below 30% is the diversification and it exists only because the stocks are imperfectly correlated.
The relationship
$\sigma_B^2 = w^2\sigma^2 + w^2\sigma^2 + 2w^2\rho\,\sigma^2 = 2 \times 0.25 \times 0.09\,(1+\rho) \quad\Rightarrow\quad \sigma_B = 0.30\sqrt{\tfrac{1+\rho}{2}}$
w the weight of each stock, 0.5
sigma each stock's volatility, 0.30
rho the correlation between the two stocks
sigma_B the basket's volatility
What it says in wordsWith equal weights and equal volatilities, the basket's volatility is the single-stock volatility times the square root of (1 plus rho) over 2.
How do you do it in your head?
Use the shortcut in the formula: with two equal stocks the basket volatility is 30% times the square root of (1 plus rho) over 2. At rho 0.5 that is 30% times the root of 0.75, about 0.866, giving 26.0%; at rho 0 it is 30% times the root of 0.5, about 0.707, giving 21.2%. Say the structure first, then the number, so a slip in the arithmetic does not look like a slip in the thinking.
What is the limitation you should name?
The formula treats correlation as a fixed number, and it is not. Correlations between stocks tend to rise in a sell-off, which is exactly when a basket holder wants the diversification, so the 25.98% is a fair-weather figure. On a derivatives desk that is why basket options and dispersion trades are priced with a correlation assumption that is marked, stressed and hedged rather than looked up once. Say that the answer depends on the correlation you assume, and that the assumption is the risk.
Where candidates lose it
The fast wrong answer is 30%, from averaging the two volatilities. Volatility is a square root, and square roots do not average. Add the variances and the covariance, then take the root.
The second loss is forgetting the factor of 2 on the cross term. With it, the correlation 0.5 answer is 25.98%; without it, you get 23.72% and an interviewer who knows the number immediately.
What the interviewer asks next
Three stocks at 30% volatility, all pairwise correlations 0.5, equal weights. What is the basket volatility?
As the number of equally correlated stocks grows large, where does the basket volatility settle, and why?
The basket option is quoted at 24% implied volatility. What correlation is the market pricing?
006A trade surveillance system raises an alert on 95% of genuinely suspicious trades and, wrongly, on 2% of normal trades. One trade in a thousand is genuinely suspicious. An alert has just fired on a trade. What is the probability the trade is suspicious?Conditional probability and BayesWarm upCitadelMiami · 2022
Try it first
Gut answer before you count anything.
Show the worked solution
About 4.5%. Count 100,000 trades. One in a thousand is suspicious, so 100 are, and 95 of those alert. The other 99,900 are normal, and 2% of them, 1,998, alert anyway. Alerts total 2,093, of which 95 are genuine, so the probability that an alerted trade is suspicious is 95 over 2,093, about 4.5%. The 95% hit rate is not the answer; the base rate is what decides it.
Why does a 95% accurate system give a 4.5% answer?
A smoke alarm that goes off for 2% of toast is a fine alarm in a house that is never on fire; nearly every ring will be toast. When the thing you are looking for is rare, even a small false-alarm rate applied to the huge normal pile produces more alerts than the true cases produce. Here 2% of 99,900 normal trades is 1,998, twenty times the 95 genuine alerts. The system is not bad; the base rate is low, and that is what the question is testing.
Of 100,000 trades, 100 are suspicious and raise 95 true alerts, while the 99,900 normal trades raise 1,998 false ones, so alerts total 2,093 and a trade that alerts is genuinely suspicious only 4.5% of the time.
The relationship
$P(S \mid A) = \frac{P(A \mid S)\,P(S)}{P(A \mid S)\,P(S) + P(A \mid N)\,P(N)} = \frac{0.95 \times 0.001}{0.95 \times 0.001 + 0.02 \times 0.999} = \frac{95}{2{,}093} \approx 4.5\%$
S, N a suspicious trade, a normal trade
A an alert fires
P(A | S) = 0.95 the hit rate
P(A | N) = 0.02 the false-alarm rate
P(S) = 0.001 the base rate
What it says in wordsTrue alerts divided by all alerts, where all alerts are the true ones plus the false ones from the normal pile.
What is the fastest way to say it in the room?
Do not write Bayes' formula; count a round number of trades. Say: in 100,000 trades, 100 are suspicious and 95 alert; 99,900 are normal and 1,998 alert; 95 over 2,093 is about 4.5%. Three sentences, no algebra, and every number is checkable by the person listening. The odds form is just as quick: prior odds 1 to 999, likelihood ratio 0.95 over 0.02, about 47.5, so posterior odds 47.5 to 999, roughly 1 to 21.
What does the desk do with a 4.5% answer?
It decides what the alert is for. A 4.5% hit rate is fine for a filter that sends trades to a human for a second look, and useless for an automatic block, because 95% of blocked trades would be legitimate business. That is the trade-off every surveillance, fraud and risk-limit system lives with: a lower threshold catches more of the 100 but drags in more of the 99,900. Say the limitation too: the 2% and 95% are themselves estimates from past data, and a system tuned on last year's patterns can drift.
Where candidates lose it
The whole trap is answering 95%, confusing the probability of an alert given a suspicious trade with the probability of a suspicious trade given an alert. Interviewers ask this precisely because the two sound the same and are twenty times apart.
The second loss is reaching for the formula and tangling the denominator. Count 100,000 trades and the denominator builds itself: 95 plus 1,998.
What the interviewer asks next
The false-alarm rate is cut to 0.5%. What is the probability now?
Two independent systems both alert on the same trade. What is the probability it is suspicious?
What base rate would make an alert a coin flip, and what does that tell you about where surveillance is worth running?
Asked at Citadel, Sales and Trading, Miami, 2022 (Wall Street Oasis): I got a question about Bayes' theorem applied to a practical scenario, which I handled decently
012You are making a market on a contract that settles at the sum of two dice. During the game you sell 5 at 7.5, buy 3 at 6.5 and sell 2 at 8. The dice are rolled and total 9. What is your final position, what is your profit or loss in rupees, and what was your expected profit at the moment you finished trading?Market makingWarm upOptiverAmsterdam · 2023
Try it first
Before the blotter: what is the fair value of the sum of two dice?
Show the worked solution
You finish short 4, you lose Rs 2 on the settlement, and your expected profit when you stopped trading was Rs 6. Sold 5, bought 3, sold 2 is a net short of 4. Cash is +37.5 - 19.5 + 16 = +34. Settling at 9 costs 4 x 9 = 36, so the result is 34 - 36 = -2. Against the fair value of 7 the short would have cost 28, leaving +6: the edge of 2.5 + 1.5 + 2 captured on the three trades.
Why keep three numbers in your head and not one?
A shopkeeper who sells umbrellas at a markup has a profit on each sale, a stock count, and a worry about whether it rains. Three separate things. A market maker tracks the same three: edge per trade against fair value, net position, and the exposure to the final number, and the game checks that you never let one of them slip. Say the fair value, 7, first. Then say each trade's edge as you do it: +2.5 on selling 5 at 7.5, +1.5 on buying 3 at 6.5, +2 on selling 2 at 8. Then say the position: short 4.
Selling 5 at 7.5, buying 3 at 6.5 and selling 2 at 8 leaves a short of 4 and cash of +34 with an edge of 2.5, 1.5 and 2 against fair value 7, so at the settlement of 9 the position costs 36 and the result is Rs 2 lost, while at fair value the same trades were worth Rs 6.
The relationship
$\text{P\&L} = \underbrace{5(7.5) - 3(6.5) + 2(8)}_{\text{cash} = 34} + \underbrace{(-4)}_{\text{position}} \times S, \qquad S = 9 \Rightarrow -2,\quad S = 7 \Rightarrow +6$
cash money received for sales minus money paid for purchases
position contracts bought minus contracts sold, here minus 4
S the settlement value of the contract, the dice total
What it says in wordsProfit is the cash already banked plus the position times the settlement, and replacing the settlement with the fair value gives the expected profit.
Was the loss a mistake?
No. Every trade was done at a better price than fair value, so the trading was right; the dice came in high. Expected profit of +6 is what you controlled, and the realised minus 2 is what the dice did; an interviewer wants to hear you separate the two without being asked. The sum of two dice has a standard deviation of about 2.4, so a short of 4 carries a one-standard-deviation swing of nearly 10, far larger than the 6 of edge. The real question is whether a short of 4 was more risk than you wanted to carry against 6 of edge.
What would you have done differently in the game?
Skewed the quote as the short grew. After selling 5, you are short 5 and should lower both your bid and your offer so that the next trade is more likely to be a buy that cuts the position, which is exactly what buying 3 at 6.5 did. Selling 2 more at 8 added to the short again; at that point a wider or higher quote would have protected you. The limitation of the puzzle: with two interviewers trading against you, their trades carry information about nothing, because the dice are not rolled yet, so here the only reason to skew is inventory, not adverse selection.
Where candidates lose it
Candidates lose the position count under pressure, saying short 6 or short 2 because they forget the buy of 3. State the running position after every trade, aloud, as the sample blotter does.
The second loss is reporting the minus 2 as if the trading was bad. The expected profit was plus 6 and the dice were unkind. Say both numbers and which one you controlled.
What the interviewer asks next
The dice settle at 5 instead. What is your P&L, and does your expected P&L change?
What is the standard deviation of the two-dice total, and how does it size the risk of being short 4?
After the first sale of 5 at 7.5, what market would you show next, and why?
Asked at Optiver, Prop Trading, Amsterdam, 2023 (Wall Street Oasis): some difficult trading games where you had to profit making a market whilst remembering your position and the position of two interviewers
018A stock trades at Rs 1,000 and its options are priced at 16% implied volatility. You buy an at-the-money option and delta-hedge it every day. Roughly how large a daily move does the stock need to make for you to break even?Option pricing intuitionWarm upVolatility tradingMarket making
Try it first
Answer in your head before reading on: the breakeven daily move is about
Show the worked solution
About 1% a day, Rs 10. With roughly 256 trading days in a year and volatility growing with the square root of time, daily volatility is annual volatility divided by 16, so 16% a year is 1% a day. A delta-hedged long option earns half its gamma times the square of each day's move and pays theta every day; the two cancel when the move equals the implied daily move. Using 252 days gives 1.008%, still Rs 10.
Where does dividing by 16 come from?
Walk randomly on a straight road, one step forward or back each second, and after 100 seconds you are typically about 10 steps from where you began, not 100, because most steps undo each other. Price moves add up the same way. Volatility scales with the square root of time, and the square root of 256 trading days is 16, so an annual volatility divided by 16 is the standard deviation of one day's move. Traders call this the rule of 16. It makes 16% implied volatility the cleanest number on the screen: 1% a day, Rs 10 on this stock.
The relationship
$\sigma_{\text{day}} = \frac{\sigma_{\text{year}}}{\sqrt{256}} = \frac{16\%}{16} = 1\%,\qquad 1\% \times 1{,}000 = \text{Rs } 10$
sigma year the implied volatility, quoted per year
256 trading days in a year, rounded so the square root is a whole number
sigma day the standard deviation of one day's percentage move
What it says in wordsOne day's typical move is the annual volatility divided by the square root of the number of trading days.
Why does that daily move decide whether the hedged option makes money?
Once the delta is hedged, the option's daily P&L is two pieces. Gamma pays you half gamma times the square of the move, in either direction; theta charges you a fixed amount for the day passing. For a one-month at-the-money option here, gamma is 0.0087 per rupee and theta is 0.435 a day. A Rs 10 move earns 0.5 x 0.0087 x 100 = 0.435, exactly the theta, because the option's price was built so that theta pays for a move of one implied standard deviation. A flat day loses 0.435; a Rs 20 day makes 1.31.
A delta-hedged long at-the-money option loses its theta of 0.435 on a flat day, breaks even when the stock moves Rs 10 either way, and makes 1.31 on a Rs 20 move, because the gain grows with the square of the move while theta is fixed, and Rs 10 is 16% divided by 16.
What does the quick answer leave out?
Three things, and naming one earns the follow-up. First, the breakeven is on the average squared move, not the average move. If the stock's moves are normal with a standard deviation of Rs 10, its average absolute move is only about Rs 7.98, so a stock that typically moves Rs 8 a day is already moving enough to break even. Second, gamma changes as the stock drifts away from the strike and as expiry nears, so the Rs 10 holds for an at-the-money option on the day you measure it. Third, hedging once a day adds noise to the P&L even when realised volatility exactly matches implied. The rule of 16 is a desk shortcut, not a pricing model.
Where candidates lose it
The common slip is dividing 16% by the number of trading days, or by 365, and quoting a breakeven of a few paise. Volatility adds in squares, so time enters under a square root; dividing by days is the mistake the question is built to catch.
The second loss is quoting Rs 10 as an average move to expect every day. It is a standard deviation: plenty of days will move Rs 2 and a few will move Rs 25, and the hedged option breaks even only if the average of the squared moves matches 100.
What the interviewer asks next
The same stock's options are priced at 32% volatility. What is the breakeven move, and what is the theta in terms of gamma?
Over a week the stock moves 5, minus 12, 3, 15 and minus 9. Did a delta-hedged long option make or lose money, roughly?
Why might a trader quote 252 days rather than 256, and when does the difference matter?
020One glass holds 100 ml of wine and another holds 100 ml of water. You take a spoonful of wine, tip it into the water and stir. Then you take a spoonful of the mixture and tip it back into the wine glass. Is there now more wine in the water glass, or more water in the wine glass?Games and logicWarm upProp trading firms
Try it first
Decide before any arithmetic: after the two spoonfuls,
Show the worked solution
Exactly the same. Each glass ends with 100 ml, so whatever wine is missing from the wine glass has been replaced, millilitre for millilitre, by water, and the missing wine can only be in the water glass. With a 10 ml spoon and a thorough stir, the return spoon carries back 0.91 ml of wine and 9.09 ml of water, leaving 9.09 ml of water in the wine and 9.09 ml of wine in the water.
Why does the first spoon feel like it settles the question?
Because it is pure wine going one way and a diluted mixture coming back, so it feels as though more wine travelled. Think instead of two cricket teams of eleven who swap some players and still field eleven each. Each glass ends with exactly 100 ml, so every millilitre of wine that left the wine glass and did not come back has been replaced by a millilitre of water: the two foreign amounts must be equal. The number of team A players now in team B is the number of team B players now in team A, however the swaps were done.
With a 10 ml spoon, the wine glass goes from 100 ml of wine to 90 ml and then back to 100 ml holding 9.09 ml of water, while the water glass goes to 110 ml and back to 100 ml holding 9.09 ml of wine, so the two foreign amounts are equal.
What do the millilitres actually look like?
Take a 10 ml spoon. After the first transfer the water glass holds 100 ml of water and 10 ml of wine, 110 ml in all, so a stirred spoonful from it is 10/110 wine. The return spoon carries 0.91 ml of wine and 9.09 ml of water, so 9.09 ml of wine stays behind in the water glass and 9.09 ml of water arrives in the wine glass. The arithmetic confirms the argument, but the argument came first and did not need the spoon size, the stirring or any division.
The relationship
$\text{water in wine} = 10 \times \frac{100}{110} = 9.09,\qquad \text{wine in water} = 10 - 10 \times \frac{10}{110} = 9.09$
10 the spoon, in millilitres
100/110 the share of water in the stirred water glass after the first transfer
10/110 the share of wine in that glass
What it says in wordsThe water carried into the wine glass equals the wine left behind in the water glass, both 9.09 ml for a 10 ml spoon.
Why do the interviewer's variations not change the answer?
Interviewers vary the story: no stirring, five spoonfuls back and forth, a ladle instead of a spoon. As long as both glasses end at their starting volume, the answer is equal, because the argument uses only the totals. The limitation to say out loud: if the return spoon is a different size from the first, the glasses end at different volumes and the amounts differ, so check the volumes before using the shortcut. The desk lesson is the bookkeeper's: in a closed system, look at the totals before tracking every transfer, the same way a net position check catches a booking error faster than replaying every ticket.
Stage Wine glass Water glass
Start 100 wine 100 water
After spoon 1 90 wine 100 water + 10 wine
After spoon 2 90.91 wine + 9.09 water 90.91 water + 9.09 wine
Tracking a 10 ml spoon through both transfers leaves each glass at 100 ml with 9.09 ml of the other liquid, which is what the conservation argument predicted without any arithmetic.
Where candidates lose it
The common answer is more wine in the water, because the first spoon was undiluted. It anchors on one transfer and forgets that the second spoon also took some of that wine back.
The second loss is reaching the right answer by long arithmetic and then failing the follow-up, such as an unstirred glass or several transfers, because there was no argument underneath. Give the volume argument first and use the numbers only as a check.
What the interviewer asks next
The return spoon is 5 ml instead of 10 ml. Which glass now holds more of the other liquid, and by how much?
You repeat the two-spoon swap many times. What do both glasses converge to?
Where on a trading desk does checking a total first save you from tracking every transfer?
023A ticket pays Rs 100 if a card drawn from a well-shuffled standard deck is a picture card (jack, queen or king), and nothing otherwise. I quote the ticket at 20 bid, 22 offered. Would you trade, and on which side?Betting and sizingWarm upAkuna CapitalChicago · 2026
Try it first
Before you work out the fraction: the right action against a 20 bid, 22 offered quote is
Show the worked solution
Yes: buy at 22, because the ticket is worth 23.08. A standard deck has 12 picture cards out of 52, a 23.1% chance, so the fair value of a Rs 100 payout is 12/52 x 100 = 23.08. You buy at the offer, 22, which is below fair, and expect to make about Rs 1.08 a ticket. Selling at 20 would give away about Rs 3.08. The edge is small next to the outcome, so it pays only over many tickets.
How do you value the ticket?
If a friend offers you Rs 100 when a coin lands heads, you would pay up to Rs 50 and no more; the coin's odds set the price. A ticket that pays a fixed amount on an event is worth the payout times the probability of the event, before any talk of the quote. A standard deck has four suits, each with a jack, a queen and a king, so 12 picture cards in 52. The ticket is worth 12/52 x 100 = 3/13 x 100 = 23.08. Notice that the ace is not a picture card; counting it is the quickest way to get 16/52 and the wrong side.
The relationship
$V = \frac{12}{52} \times 100 = \frac{3}{13} \times 100 \approx 23.08$
12 picture cards: jack, queen and king in four suits
52 cards in the deck
100 the payout in rupees if a picture card is drawn
What it says in wordsThe ticket is worth its payout times the chance of being paid.
Which side of the quote do you compare with?
A quote has two prices, and you can only use one of each: you buy at the offer and sell at the bid. Trade only when fair value sits outside the quote: buy when it is above the offer, sell when it is below the bid, and pass when it falls between them. Here fair is 23.08 against a 22 offer, so you buy and expect 1.08. The figure runs two other quotes for contrast: at 22 bid, 24 offered, fair is inside and there is no trade; at 24 bid, 26 offered, you sell at 24 and expect 0.92.
Against a fair value of 23.08, the quote of 20 bid, 22 offered lets you buy below fair for an expected 1.08 a ticket, a quote of 22 at 24 straddles fair and gives no trade, and a quote of 24 at 26 lets you sell above fair for an expected 0.92.
How much would you buy, and what could change your mind?
Each ticket pays 0 or 100, so its standard deviation is 100 x the square root of 0.231 x 0.769, about Rs 42, against an edge of Rs 1.08. The edge is real but small next to the noise, so it is worth taking in size over many independent draws and worth very little on a single ticket. Before trading, ask the two questions a desk would: is the deck standard and well shuffled, and does the person quoting know something you do not, such as a card already removed? The limitation of the clean answer is that it trusts the setup; in a game where the quoter controls the deck, a quote this generous is itself a warning.
Where candidates lose it
The common slip is to sell at 20 on the feeling that the ticket usually loses. That is true and already priced: losing most of the time is why the ticket is worth 23, not 50. Value it first, then compare.
The second loss is comparing fair value with the middle of the quote, 21, and saying buy without naming the price. You buy at the offer. Saying buy at 20 tells the interviewer you do not know which side of a quote you can trade on.
What the interviewer asks next
The ticket now pays on a picture card or an ace. Where would you trade against the same quote?
I draw a card, look at it without showing you, and then quote 20 at 22. What do you do now?
Make me your own market on the original ticket, and say why you chose that width.
Asked at Akuna Capital, Junior Trader Interview, Chicago, 2026 (Wall Street Oasis): if you win you get 1$. how much money would be a fair bet.
026I will pay you if at least 60% of the flips of a fair coin come up heads. Do you want 10 flips or 100 flips?Distributions and statisticsWarm upHudson River TradingNew York · 2020
Try it first
Before you count anything: which do you take?
Show the worked solution
Take 10 flips. You are betting on luck, and luck averages out as the trials pile up. With 10 flips, 6 or more heads happens 386 times in 1,024, about 37.7%. With 100 flips, 60 or more heads happens about 2.8% of the time. The share of heads settles towards 50% at the rate of one over the square root of n, so the 60% line gets harder to reach with every extra flip.
Why does the number of flips change the odds at all?
Think of a school with two cricket teams, one of eleven and one of a hundred and ten. If someone offers a prize for a team whose average height is 10 cm above the national average, the small team is the one that can win: one or two tall players move its average, while the big team's average is pinned down by sheer numbers. The same coin, flipped more often, produces a share of heads that sits ever closer to a half, so a payout that needs an unusual share wants the fewest flips you can get. This is the law of large numbers working against you, and the question is testing whether you know which side of it you are on.
With 10 flips the bars at 6 heads and above hold 37.7% of the probability, but with 100 flips the bars at 60 heads and above hold only 2.8%, because the share of heads tightens around a half as the flips increase.
How do you put a number on it without a table?
Count the small case exactly: 6 or more heads in 10 flips means adding the ways to get 6, 7, 8, 9 and 10 heads, which are 210, 120, 45, 10 and 1, a total of 386 out of 1,024, so 37.7%. For 100 flips use the normal approximation. The standard deviation of the share of heads is 0.5 over the square root of n: 15.8% for 10 flips, 5% for 100. The 60% line is 0.6 standard deviations out in the first case and 2.0 in the second, and two standard deviations in one tail is about 2.3%. The exact binomial answer is 2.8%; the approximation gets you to the right decision in one breath.
The relationship
$\sigma_{\bar{X}} = \frac{0.5}{\sqrt{n}} \qquad z_{10} = \frac{0.1}{0.158} = 0.63, \quad z_{100} = \frac{0.1}{0.05} = 2.0$
sigma of X bar the standard deviation of the share of heads
n the number of flips
z how many standard deviations the 60% line sits from the mean of 50%
What it says in wordsThe 60% line gets further from the centre, measured in standard deviations, as the flips increase, so it becomes rarer to cross.
What is the interviewer listening for after the answer?
Say the general rule and then the exception. The rule: whenever a payout needs the sample to look unlike the population, choose the smallest sample. The exception: if the payout were for landing between 40% and 60%, you would want the most flips, for exactly the same reason. Read the sign of the bet before you choose the sample size: a bet on luck wants few trials and a bet on the average wants many. If the interviewer changes the wording to more than 60%, 7 or more heads in 10 is 17.2%, still far above the 100-flip figure. And if the choice is between 100 and 1,000 flips, 600 or more heads happens about 1.36e-10 of the time, which is as close to never as a desk needs.
Where candidates lose it
The fast wrong answer is 100 flips, because more flips feel like more chances. They are more chances for the average to assert itself, not for luck. Candidates who say it have the law of large numbers backwards, and the interviewer hears it immediately.
The second loss is saying 10 without a number. Have the 386 out of 1,024 ready, then the standard deviation argument for 100, so the answer sounds reasoned rather than remembered.
What the interviewer asks next
I pay you if the share of heads is between 45% and 55%. Now which do you want?
What if the coin has a 60% bias towards heads? Does the answer flip?
Roughly how many flips make the 60% line a three standard deviation event?
Asked at Hudson River Trading, Prop Trading, New York, 2020 (Wall Street Oasis): Questions on EV for coin tosses, law of large numbers, Bayes theorem
031A stock is at 1,000, volatility is 20% and rates are near zero. Estimate the three-month at-the-money call in your head, and the straddle.Option pricing intuitionWarm upMarket makingVolatility trading
Try it first
Say the call price before you reach for a formula.
Show the worked solution
Call about 40, straddle about 80. The rule is 0.4 x S x sigma x sqrt T. Three months is a quarter of a year, so sqrt T is 0.5 and the volatility over the period is 10%; 0.4 x 1,000 x 0.1 = 40. With rates at zero the at-the-money put is worth the same, so the straddle is 80. The full model gives 39.88 for the call, because the exact constant is 1 over sqrt(2 pi), 0.3989, not 0.4.
Where does the 0.4 come from?
A tailor who knows a customer's height is normally distributed around 170 cm with a spread of 10 cm can say how far above 170 the average tall customer stands: about 0.4 of the spread, which is 4 cm, because the mean of the positive half of a normal is sigma over sqrt(2 pi). An at-the-money call pays the positive half of the stock's move, and the average of the positive half of a normal is 0.4 of its standard deviation, so the call is worth 0.4 times the standard deviation of the move over its life. The standard deviation of the move is S x sigma x sqrt T, which is 1,000 x 0.2 x 0.5 = 100 here, and 0.4 of 100 is 40. The exact constant is 1 / sqrt(2 pi) = 0.3989, so the rule gives 40 where the precise version gives 39.89, and the model 39.88.
The relationship
$C_{ATM} \approx 0.4\, S\, \sigma \sqrt{T} = 0.4 \times 1000 \times 0.20 \times 0.5 = 40 \qquad C_{exact} = S\left(2N\!\left(\tfrac{\sigma\sqrt{T}}{2}\right) - 1\right)$
S the stock price, 1,000
sigma sqrt T the volatility scaled to the option's life: 20% x 0.5 = 10%
0.4 the approximation to 1 over root 2 pi, which is 0.3989
N the standard normal distribution function
What it says in wordsAn at-the-money call is about four tenths of one standard deviation of the stock's move over its life.
Across volatilities from 0 to 100% the rule 0.4 x S x sigma x sqrt T sits almost on top of the full model price for a three-month at-the-money call, giving 40 against 39.88 at 20% volatility, and bends below it only at high volatility where the model's price curves.
Why is the straddle just double, and when is it not?
With rates at zero and no dividends, the forward equals the spot, so an at-the-money call and put have the same value by put-call parity, and the straddle is simply two calls, about 80. The straddle costs 8% of the stock for a three-month bet, which is the whole quarter's one-standard-deviation move of 10% times 0.8: that is the number a volatility trader carries in their head. With rates or dividends the forward moves away from spot, at-the-money means at-the-forward, and the call and put split the straddle unevenly, though their sum barely changes. The rule also assumes the volatility is the right one for this strike, which on a real surface with skew it may not be.
How far can you push the rule?
The rule is linear in volatility and time, the model is not. For three months the gap is 0.12 on 40 at 20% volatility, and even at 60% volatility the rule gives 120 against 119.2. Over one year at 20% the rule gives 80 against the model's 79.66. Up to a total move of about 50% the rule is within a couple of percent, which is every interview case and most of the real book; beyond that the model price flattens because a call can never be worth more than the stock. Say the limitation and then use the rule anyway: on a desk the question is never whether 40 is exactly right, it is whether 44 on the screen is rich or cheap.
Where candidates lose it
The common loss is forgetting to scale the volatility to the horizon and quoting 0.4 x 1,000 x 0.2 = 80 for the call, which is the one-year number. Say sqrt T out loud: three months is a half.
The second is giving the call and then stalling on the straddle, or doubling the call without saying why. The put equals the call only because rates are zero and there is no dividend; name parity and the interviewer knows you understand what at the money means.
What the interviewer asks next
Rates are now 8%. Which is worth more at the money, the call or the put, and by roughly how much?
The stock pays a 2% dividend before expiry. What changes?
Quote me the one-month straddle on the same stock, then the one-year.
The market is paying 44 for the call. What volatility is it implying, roughly?
035There are 100 coins on the table. Players take turns removing 1 to 10 coins, and whoever takes the last coin wins. Do you want to go first, and what is your first move?Games and logicWarm upQuant trading
Try it first
Go first or second, and what is the opening?
Show the worked solution
Go first and take 1, leaving 99. Work backwards: whoever faces 11 coins loses, because any take of 1 to 10 leaves 1 to 10 for the other player to finish. The same holds for 22, 33 and every multiple of 11. From 100, taking 1 leaves 99, a multiple of 11; after that, whatever the opponent takes, you take 11 minus it, stepping down 88, 77, 66 and so on to 0, where you take the last coin.
Why work backwards from the last coin?
If you are climbing stairs with a friend and the rule is that the person who steps onto the top stair wins, you do not plan from the bottom; you ask which stair you must leave your friend on so that they cannot reach the top in one go. Games with a fixed last move are solved from the end: find the positions where the player to move loses, then find the positions from which you can push your opponent onto one of them. With 1 to 10 coins allowed, facing 1 to 10 coins is a win, you take them all. Facing 11 is a loss, because every move leaves between 1 and 10. Facing 12 to 21 is a win, since you can reduce to 11. Facing 22 is a loss again. The losing positions repeat every 11.
Every multiple of 11 from 0 to 99 is a losing position for the player who must move, so the first player takes 1 to leave 99 and then answers every take of t with 11 minus t, stepping down through 88, 77 and 66 until the last coin.
How do you find the period without listing every position?
The period is the largest take plus one, 11, because that is the one total a pair of moves can always be made to add up to: whatever your opponent takes between 1 and 10, you can take the balance of 11. The losing positions are the multiples of the largest take plus one, and the winning opening move is the remainder when the pile is divided by that number. 100 divided by 11 is 9 remainder 1, so take 1. If the rule allowed 1 to 7 coins, the period would be 8 and the opening would be 100 mod 8, which is 4. If the pile had been 99 to start with, you would want to go second, because the first player cannot leave a multiple of 11.
The relationship
$\text{losing positions} = \{0, 11, 22, \ldots, 99\}, \qquad \text{opening} = 100 \bmod 11 = 1$
11 the largest allowed take plus one, the amount you can always complete in a pair of moves
100 mod 11 the remainder when 100 is divided by 11; take exactly this many
What it says in wordsTake the remainder on your first move, then keep each pair of moves summing to 11.
What changes if the last coin loses instead of wins?
Then you want to hand your opponent the last coin, so the position you avoid facing is 1 coin, and the losing positions shift up by one: 1, 12, 23 and so on up to 100. Facing 100 in that version you are already lost, so you would want to go second, which shows the interviewer that you re-derive the pattern rather than remember it. The method is the same in every variant: name the terminal position, step back one move at a time to find the first losing position, then find the period. The limitation of the trick is that it needs a game with perfect information and no chance; add a die that sets each turn's maximum and the clean period disappears.
Where candidates lose it
The common loss is taking 10, because a bigger move feels like a stronger start. It leaves 90, which is not a multiple of 11, and a prepared opponent takes 2 to leave 88 and wins from there.
The second is knowing the answer and not the reason. Say why 11 is the period: any take of 1 to 10 can be completed to 11. Without that sentence the interviewer will change the numbers and watch you stall.
What the interviewer asks next
Players may take 1 to 7 coins instead. Do you go first, and what is the opening?
The player who takes the last coin loses. Do you go first?
There are two piles, 100 and 60, and you may take from either pile. Who wins?
Each turn a die sets the maximum take. Is there still a strategy, and what is it?
043In your head, no paper: convert 3/32 to a decimal, work out 38 x 42, and give 1/7 to four decimal places.Mental maths and estimationWarm upBelvedere TradingChicago · 2021
Try it first
What is 38 x 42?
Show the worked solution
0.09375, 1,596 and 0.1429. For 3/32, halve 1 five times to get 1/32 = 0.03125 and multiply by 3. For 38 x 42, both sit 2 away from 40, so the product is 40 squared minus 2 squared, 1,600 - 4. For 1/7, the repeating block is 142857, because 7 x 142857 = 999,999, so 1/7 = 0.142857... which rounds to 0.1429. Each one is a known anchor plus one step.
Why does the interviewer ask three small sums in a row?
A shopkeeper who totals a bill in his head is not doing long addition; he rounds to the nearest hundred and fixes the difference. Trading desks ask quick sums to see whether you reach for an anchor you already know, because that is how prices get checked in the two seconds before someone else trades. The three questions here each have a short route: fractions with a power of two below them are halvings, products of numbers either side of a round number are a difference of squares, and sevenths are one repeating block of six digits. The speed comes from knowing which route fits, and the interviewer listens to the route as much as to the number.
Halving 1 five times gives 1/32 = 0.03125 and three of those make 0.09375, a 40 by 40 square with a 2 by 2 corner removed shows 38 x 42 = 1,596, and the six-digit cycle 142857 gives 1/7 = 0.1429 to four places.
What is the route for each one?
For 3/32, keep halving from a half: 0.5, 0.25, 0.125, 0.0625, 0.03125. Every halving adds a digit, and 32 is five halvings, so 1/32 is 0.03125 and three of them are 0.09375. Bond traders do this all day, because US Treasury prices are quoted in 32nds. For 38 x 42, notice that both numbers sit 2 from 40, so the product is 40 squared minus 2 squared, 1,596, a trick that works for any pair placed evenly around a round number. The same trick gives 47 x 53 = 2,500 - 9 = 2,491. For 1/7, remember that 7 x 142857 = 999,999. So 1/7 is 142857 divided by 999,999, which is 0.142857 repeating, and the fifth decimal is 5, so it rounds up to 0.1429.
The relationship
$\frac{3}{32} = 3 \times \frac{1}{2^5} = 3 \times 0.03125, \qquad (a - b)(a + b) = a^2 - b^2, \qquad \frac17 = \frac{142857}{999999} = 0.\overline{142857}$
2 to the 5 32, five halvings of 1
a, b the round midpoint, 40, and the distance to each number, 2
the bar the block of six digits that repeats forever
What it says in wordsTurn each sum into a fact you already know plus one small step.
What does a strong candidate add after the answers?
Add the sanity check and the neighbours. A product of two numbers around a centre is always a little below the centre squared, so 1,596 must be under 1,600. Three 32nds must be just under a tenth, since 3.2 of them would make exactly 0.1. The sevenths share the same six digits in rotation, 2/7 = 0.285714 and 3/7 = 0.428571, so knowing one seventh gives you all six. The limitation is honest too: anchors cover the common cases, and for an awkward product like 37 x 46 you fall back on splitting, 37 x 46 = 37 x 50 - 37 x 4 = 1,850 - 148 = 1,702. Saying when you switch method sounds better than pretending one trick does everything.
Where candidates lose it
The common loss is starting long multiplication for 38 x 42 aloud, carrying digits and losing track. The interviewer gives you a pair either side of 40 on purpose; a candidate who does not spot it looks slow even if the answer is right.
The second is 1/7 as 0.1428, truncating instead of rounding. The next digit is 5, so the fourth decimal rounds up. Say the repeating block first, then round, and the slip disappears.
What the interviewer asks next
What is 5/16 as a decimal, and 7/64?
Work out 67 x 73 and 96 x 104 the same way.
What is 5/7 to four places?
Estimate 1/13 to three decimal places.
Asked at Belvedere Trading, Trading, Chicago, 2021 (Wall Street Oasis): 3/32 mental math, 38*42, crossing the bridge in the shortest amount of time

w	the weight of each stock, 0.5
sigma	each stock's volatility, 0.30
rho	the correlation between the two stocks
sigma_B	the basket's volatility

S, N	a suspicious trade, a normal trade
A	an alert fires
P(A \| S) = 0.95	the hit rate
P(A \| N) = 0.02	the false-alarm rate
P(S) = 0.001	the base rate

cash	money received for sales minus money paid for purchases
position	contracts bought minus contracts sold, here minus 4
S	the settlement value of the contract, the dice total

sigma year	the implied volatility, quoted per year
256	trading days in a year, rounded so the square root is a whole number
sigma day	the standard deviation of one day's percentage move

Try it first

Why is the basket calmer than the stocks inside it?

How do you do it in your head?

What is the limitation you should name?

Where candidates lose it

What the interviewer asks next

Try it first

Why does a 95% accurate system give a 4.5% answer?

What is the fastest way to say it in the room?

What does the desk do with a 4.5% answer?

Where candidates lose it

What the interviewer asks next

Try it first

Why keep three numbers in your head and not one?

Was the loss a mistake?

What would you have done differently in the game?

Where candidates lose it

What the interviewer asks next

Try it first

Where does dividing by 16 come from?

Why does that daily move decide whether the hedged option makes money?

What does the quick answer leave out?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the first spoon feel like it settles the question?

What do the millilitres actually look like?

Why do the interviewer's variations not change the answer?

Where candidates lose it

What the interviewer asks next

Try it first

How do you value the ticket?

Which side of the quote do you compare with?

How much would you buy, and what could change your mind?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the number of flips change the odds at all?

How do you put a number on it without a table?

What is the interviewer listening for after the answer?

Where candidates lose it

What the interviewer asks next

Try it first

Where does the 0.4 come from?

Why is the straddle just double, and when is it not?

How far can you push the rule?

Where candidates lose it

What the interviewer asks next

Try it first

Why work backwards from the last coin?

How do you find the period without listing every position?

What changes if the last coin loses instead of wins?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the interviewer ask three small sums in a row?

What is the route for each one?

What does a strong candidate add after the answers?

Where candidates lose it

What the interviewer asks next