Derivatives Foundation puzzles, solved step by step

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

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041Make me a market on the number of heads in 100 flips of a fair coin. How wide do you quote, and how does your quote change when I tell you the first 10 flips produced 8 heads?Market makingCoreDRWNew York · 2026

Try it first

After you hear that the first 10 flips gave 8 heads, where is the new mid?

Show the worked solution

Centre on 50 and quote something like 48 at 52; after 8 heads in 10, move to 51 at 55, around 53. The mid is the expected count, 100 x 1/2. The standard deviation is sqrt(100 x 1/4) = 5, which tells you how much one lot can swing. Nobody can know more than you about a fair coin, so the quote can be tight. Once 8 heads are banked, the 90 flips left average 45, so the mid is 53 and the sd is sqrt(22.5), about 4.74.

Where does the middle of the quote come from?

If a friend asks how many of the next 100 cars past a junction will be white, and one car in two is white, you say 50 and you are not embarrassed by it. The middle of a market is your expected value, and for 100 fair flips that is 100 x 1/2 = 50 heads; everything else in the answer is about the width. The spread of the outcome comes from the binomial: variance n x p x (1 - p) = 25, so the standard deviation is 5. The count lands between 45 and 55 about 72.9% of the time, and outside 40 to 60 only about 3% of the time. Say both numbers out loud; the interviewer is checking that you know the centre and the scale before you name a price.

The relationship

\mu = np = 100 \times \tfrac12 = 50, \qquad \sigma = \sqrt{np(1-p)} = \sqrt{25} = 5, \qquad \mu' = 8 + 90 \times \tfrac12 = 53, \quad \sigma' = \sqrt{90 \times \tfrac14} = 4.74

n	the number of flips still random
p	the chance of heads on each flip, one half
mu, sigma	the mean and standard deviation of the total before any flip
mu prime, sigma prime	the same after the first 10 flips are known: the 8 heads are fixed and only 90 flips remain random

What it says in wordsKnown flips add to the centre one for one, and only the flips still to come feed the standard deviation.

Before any flip the outcome is centred on 50 with standard deviation 5 and the quote sits at 48 bid, 52 offered, and after 8 heads in the first 10 flips the whole curve moves to 53 and narrows to 4.74, so the same four-wide quote moves up to 51 at 55.

How wide should the quote be, and why not plus or minus one standard deviation?

Width is what you charge for two risks: that the person trading with you knows something you do not, and that you have to carry a position whose outcome swings. On a fair coin nobody can know more than you, so the first risk is zero and the quote can be tight, a point or two either side of 50; the standard deviation of 5 sets how many lots you are willing to show, not how far apart your bid and offer are. A quote of 45 at 55 is not wrong, but it says you think the other side may be informed, and an interviewer will ask why. Open at 48 at 52, say you would tighten to 49 at 51 against someone who clearly has no edge, and say what would make you widen: hidden information, a size much bigger than you want to hold, or a coin you have not inspected.

What exactly changes when I tell you 8 of the first 10 were heads?

Two things, and candidates usually get only one. The centre moves to 8 plus the expected heads on the 90 flips left, 45, which is 53. The coin is still fair, so 80% is not the new rate; the 8 are simply banked. The uncertainty also falls, because 10 of the flips are no longer random: the standard deviation goes from 5 to sqrt(90 x 1/4) = 4.74, about 5% lower. Move the whole quote up by 3 to 51 at 55, and notice what the news did to your old offer: a counterparty who had seen those flips would have lifted your 52 and made 1 a lot on average. That is the lesson of the follow-up, and it is why market makers widen or pull quotes when they suspect the other side has seen something they have not. The limitation of the tidy answer is that it trusts the coin; after a run of 8 heads a careful trader at least asks whether the coin is fair.

Where candidates lose it

The common loss is re-centring on 80, as if 8 heads in 10 revealed the coin, or leaving the mid at 50 because the coin has no memory. The flips have no memory, but the total does: 8 heads are already in it, and the right mid is 53.

The second is quoting plus or minus one standard deviation, 45 at 55, without saying why. Width is a statement about information and risk. Give the centre and the standard deviation first, then defend the width you choose.

What the interviewer asks next

I tell you the coin may be biased, with heads anywhere between 40% and 60%. How does your market change?
You are lifted on your 52 offer three times in a row. What do you do?
Make a market on the number of heads in the last 50 flips only, given the same news about the first 10.
Make a market on the square of the number of heads.

Asked at DRW, Quantitative Trading, New York, 2026 (Wall Street Oasis): Make a market on the number of heads out of 100 coin flips.

042An option lets you decide in six months whether it becomes a call or a put, both struck at 100 and expiring in one year. The stock is at 100, pays no dividend, and rates are zero. Express it as a portfolio of plain options.Option pricing intuitionHardExotics tradingStructured products

Try it first

Which portfolio of plain options replicates the chooser?

Show the worked solution

A one-year call struck at 100 plus a six-month put struck at 100. At the decision date you take the larger of the call and the put. Write that as the call plus max(P - C, 0). Put-call parity with zero rates and no dividend says P - C = 100 - S at that date, so the extra piece is max(100 - S, 0) paid at six months: a six-month put. At 20% volatility that is 7.97 + 5.64 = 13.60, against 15.93 for a one-year straddle.

Why is a chooser worth less than a straddle but more than a single option?

Booking a restaurant table for a date six weeks away while keeping the right, at four weeks, to switch it to a different restaurant is worth more than a fixed booking but less than holding two bookings to the end. The chooser is the same: it beats any one plain option because you choose with six months of news in hand, but it is cheaper than a straddle because you must give up one leg at six months. A straddle keeps both the call and the put for the full year; the chooser keeps only whichever looks better at the half-way point, and then lives or dies with it. At 20% volatility the one-year straddle is 15.93, a one-year call alone is 7.97, and the chooser sits in between at 13.60. The question is whether you can find that middle number without a model of the choice itself.

How does put-call parity turn the choice into plain options?

At six months you hold max(C, P), where C and P are the values then of the call and the put with six months left. Split it: max(C, P) = C + max(P - C, 0). Now use parity. With no dividend and zero rates, a put minus a call with the same strike and expiry is worth the strike minus the stock, so P - C = 100 - S at the decision date. The awkward piece max(P - C, 0) becomes max(100 - S, 0), which is exactly the payoff of a put struck at 100 that expires at six months, so the chooser is a one-year call plus a six-month put. The call is always in the portfolio because you either keep it or, by parity, the put you switch into is that call plus a short forward, and the short forward's value is what the six-month put pays you when you switch.

The relationship

V_{t} = \max(C_t, P_t) = C_t + \max(P_t - C_t, 0) = C_t + \max(K - S_t, 0) \quad \Rightarrow \quad \text{Chooser}_0 = C(K, T) + P(K, t)

t	the decision date, six months
T	the final expiry, one year
C t, P t	the values at the decision date of the call and the put struck at K
K - S t	put minus call by put-call parity, with zero rates and no dividend

What it says in wordsThe choice is a call you always own plus a put on the decision date that pays the switch.

The chooser equals a one-year call plus a six-month put, because at the decision date the larger of call and put is the call plus max(100 - S, 0), and at 20% volatility the two pieces add to 13.60, below the 15.93 one-year straddle.

How do you check the answer, and what changes with rates and dividends?

Test the two ends of the decision date. If the choice had to be made today, the chooser is just whichever is dearer now, and at the money with zero rates the call and put cost the same, 7.97; the formula gives a call plus a put expiring today at the money, which is worth zero, so it agrees. If the choice is made at one year, you simply take whichever pays, which is a straddle, and the formula gives a call plus a one-year put, which agrees too. A simulation of 40,000 paths to six months, taking the better of call and put on each, gives 13.61 against 13.60. With rates or dividends the same split works, but parity puts a discounted strike on the put: it is struck at K times the discount factor from six months to a year, adjusted for the dividends paid in that window. The limitation to say: the decomposition relies on European options and on parity holding exactly; an American chooser, or one on a stock with a borrow cost, needs a model.

Where candidates lose it

The common loss is answering a straddle. A straddle keeps both legs for a year; the chooser forces you to give one up at six months, so it must cost less, and the difference is the put's last six months of life.

The second is getting the put's expiry wrong: writing a one-year put alongside the call, or a six-month call alongside the put. The switch happens at six months, so the piece that pays for it expires at six months. Write max(C, P) = C + max(P - C, 0) and parity does the rest.

What the interviewer asks next

What is the chooser worth if the decision date moves to nine months?
Rates are 6% instead of zero. What strike does the six-month put carry?
The stock pays a dividend of 3 at month nine. How does the decomposition change?
The call and the put have different strikes, 95 and 105. Is there still a closed form?

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
044Using only calls, build a payoff that is zero below 90, rises one for one to 10 at 100, falls back to zero at 110, and stays at zero above that.Option payoffs and no-arbitrageCoreEquity derivativesStructured products
Try it first
Which call portfolio gives the tent?
Show the worked solution
Long one 90 call, short two 100 calls, long one 110 call: a call butterfly. Read the slope of the target from left to right: 0, +1, -1, 0. A long call adds +1 to the slope at its strike, so the changes of +1 at 90, -2 at 100 and +1 at 110 give the weights. Check the corners: 0 at 90, 10 at 100, 0 at 110 and above. At 20% volatility and three months it costs about 3.69.
How do you read a payoff picture as a list of calls?
A road that is flat, then climbs, then drops, then is flat again can be described by where the gradient changes and by how much. A payoff made of straight pieces is the same: each call adds one unit of slope from its strike onwards, so the number of calls at a strike is simply the change of slope at that strike. The tent has slope 0 below 90, +1 from 90 to 100, -1 from 100 to 110, and 0 above 110. The changes are +1 at 90, -2 at 100 and +1 at 110. So you buy one 90 call, sell two 100 calls and buy one 110 call. Above 110 the three legs pay (S - 90) - 2(S - 100) + (S - 110) = 0, which confirms the payoff returns to zero and stays there.
The long 90 call, the two short 100 calls and the long 110 call add up to a tent that is zero below 90, peaks at 10 at 100 and returns to zero from 110 onwards, because the slope changes by +1, -2 and +1 at the three strikes.
The relationship
$\text{Payoff}(S) = (S - 90)^+ - 2(S - 100)^+ + (S - 110)^+, \qquad \text{weight at } K = \text{slope just above } K - \text{slope just below } K$
(S - K)+ the payoff of a call struck at K, the larger of S - K and zero
weight at K the number of calls to hold at that strike; negative means sell
S the stock price at expiry
What it says in wordsThe weight on each strike is the jump in slope there, which is how any straight-line payoff is built from calls.
What does the butterfly cost, and what does its price tell you?
Take an illustrative stock at 100 with 20% volatility and three months to expiry. The calls cost 10.71, 3.99 and 0.95, so the butterfly costs 10.71 - 2 x 3.99 + 0.95 = 3.69. The payoff is a tent of height 10 and base 20, and its price is close to the chance of finishing near 100 times the peak. Divide the cost by the peak payoff of 10 and you get 0.369, close to the 0.383 risk-neutral chance that the stock ends between 95 and 105, which is why a butterfly is the market's way of pricing the probability of a narrow range. Shrink the strike gap towards zero and the scaled butterfly becomes the risk-neutral density itself, a result traders use to read the distribution off a strip of call prices.
Where does the no-arbitrage check come in?
The tent never pays less than zero, so it can never cost less than zero. That means C(90) - 2C(100) + C(110) must be at least 0: call prices must be convex in strike. Suppose a screen shows the 90 call at 11.00, the 100 call at 7.00 and the 110 call at 2.50. The butterfly costs 11.00 - 14.00 + 2.50 = -0.50: you are paid 0.50 to hold a payoff that is never below zero. A negative butterfly price is a free lunch, and spotting it on a quote sheet is the reason the question is asked. The limitation in practice is that each leg has a bid and an offer, so the check must use the prices you can actually trade at, buying at offers and selling at bids, and small apparent violations usually vanish once the spread is paid.
Where candidates lose it
The common loss is short one 100 call instead of two. One short call only cancels the slope of the 90 call, which flattens the payoff at 10 forever; it takes two to turn the slope down to -1 and bring it back to zero.
The second is building the tent and stopping. The interviewer usually follows with the price: a butterfly must cost more than zero because it never pays less than zero, and a candidate who connects that to convexity in strike has answered the real question.
What the interviewer asks next
Build the same tent with puts only. Is the cost the same?
Build a payoff that is 0 below 90, rises to 10 at 100 and stays at 10 above.
The 90, 100 and 110 calls trade at 11.00, 7.00 and 2.50. What do you do?
As the gap between the strikes shrinks, what does the scaled butterfly price approach?

045Two players each ante Rs 1 and receive one card from an ace, a king and a queen. Player one may bet Rs 1 or check, and a check goes straight to showdown. Facing a bet, player two may call or fold. How often should player one bluff with the queen, and how often should player two call with the king?Games and logicHardOld Mission CapitalNew York · 2022

Try it first

How often should player one bet the queen?

Show the worked solution

Player one bets the ace always, checks the king, and bluffs the queen one time in three; player two calls with the ace, folds the queen, and calls with the king one time in three. The bluff rate makes one bet in four a bluff, which is the break-even for a call of 1 into a pot of 3. The king's call rate makes player two fold one time in three overall, the break-even for a bluff of 1 into a pot of 2. Player one gains 1/18 of a rupee a hand.

Which decisions are obvious, and which one is the real question?

Start by removing the choices nobody would make. Player one always bets the ace, since it wins every showdown, and player two always calls with the ace and folds with the queen, since those cards win or lose every time; the only real decisions are player one's queen and player two's king. Player one's king should check: if it bets, player two calls only with the ace and folds the queen, so the bet loses 2 half the time and wins 1 half the time, an average of -0.5, worse than the 0 a showdown gives. That leaves two numbers to find, the queen's bluff frequency and the king's calling frequency, and each is set to make the other player's choice a matter of indifference.

How do you find the bluffing frequency?

A goalkeeper who always dives left is easy to beat; a penalty taker mixes his side so the keeper gains nothing by guessing. Poker equilibrium works the same way. Player one bluffs just often enough that player two's king gains nothing by calling over folding, and player two calls just often enough that player one's queen gains nothing by bluffing over checking. Facing a bet with the king, folding loses the ante, -1. Calling wins 2 against a bluff and loses 2 against the ace. With the ace always bet and the queen bet a fraction b of the time, a bet is a bluff with probability b / (1 + b). Setting the call's value equal to -1 gives b = 1/3, so one bet in four is a bluff.

The relationship

\text{King indifferent: } \tfrac{b}{1+b}(+2) + \tfrac{1}{1+b}(-2) = -1 \Rightarrow b = \tfrac13, \qquad \text{Queen indifferent: } \tfrac12(-2) + \tfrac12\big(c(-2) + (1-c)(+1)\big) = -1 \Rightarrow c = \tfrac13

b	how often player one bets the queen
c	how often player two calls with the king
-1	the value of the alternative: folding the king, or checking the queen, each loses the ante
1/2	given the queen, player two holds the ace or the king with equal chance

What it says in wordsEach player's frequency is chosen so that the other player's two choices are worth the same.

Player one bets the ace, checks the king and bluffs the queen one time in three, while player two calls with the ace, folds the queen and calls with the king one time in three, because those frequencies leave each player's marginal choice worth exactly the same either way.

How do you sanity check the frequencies, and who wins the game?

Use pot odds as the check. Player two calls 1 to win a pot of 3, two antes and the bet, so a call breaks even when bluffs are 1 / (3 + 1) of bets, a quarter, which is what b = 1/3 delivers. Player one bluffs 1 to win the pot of 2, so a bluff breaks even when player two folds 1 / (2 + 1) of the time; he folds the queen always and the king two times in three, which averages to one in three. Average over the six deals and player one gains 1/18 of a rupee a hand, because only he can bet; neither player can improve on that by changing his own frequency. Against a player two who never calls with the king, bluffing every queen would earn 1/6 of a rupee a hand instead of 1/18. The limitation is that equilibrium is a defence, not the most profitable play against a predictable opponent; a desk uses it as the baseline and then leans towards the other side's mistakes.

Where candidates lose it

The common loss is saying player one should never bluff because the queen cannot win a showdown. Without bluffs player two simply folds his king to every bet, and player one's ace stops getting paid.

The second is giving the bluff rate as a quarter. A quarter is the share of bets that are bluffs; because the ace is always bet, that needs the queen bet one time in three. Keep the two fractions apart and say which one you mean.

What the interviewer asks next

Player two may also bet after player one checks. How does the equilibrium change?
The bet size doubles to Rs 2. What are the new bluffing and calling frequencies?
Player two never calls with the king. What is player one's best response, and what does it earn?
Why does player one never want to bet the king?

Asked at Old Mission Capital, Quantitative Research, New York, 2022 (Wall Street Oasis): Asking to find the game theory optimal strategy in a simplified poker game

046Give the next term in each sequence: 2, 6, 12, 20, 30, ... ; 1, 1, 2, 6, 24, ... ; 3, 5, 9, 17, 33, ...Mental maths and estimationCoreProp trading firmsSell-side sales and trading

Try it first

What comes after 3, 5, 9, 17, 33?

Show the worked solution

42, 120 and 65. In the first, the differences 4, 6, 8, 10 rise by 2, so the next difference is 12 and the term is 42; the terms are n(n + 1). In the second, each term is the previous one times 1, 2, 3, 4, so the next multiplier is 5 and the term is 120; these are the factorials. In the third, the differences 2, 4, 8, 16 double, so add 32 to get 65; the terms are 2 to the n plus 1.

What order should you test patterns in?

A mechanic with an unknown rattle checks the cheap, common causes first and the exotic ones last. Sequence questions reward the same discipline: write the first differences, then the second differences, and only if neither settles try ratios and then rules that mix the two, such as double and subtract one. Each test takes a few seconds and most interview sequences give way to one of the first three. The order matters because the interviewer is timing you; the candidate who stares at the numbers hoping to recognise them is slower than the one who writes a row of differences under them without thinking.

Writing a row under each sequence exposes the rule: the first has constant second differences of 2 and continues to 42, the second has ratios climbing by one and continues to 120, and the third has doubling differences and continues to 65.

How does each of the three give way?

For 2, 6, 12, 20, 30, the first differences are 4, 6, 8, 10 and the second differences are all 2. Constant second differences mean the sequence is a quadratic in n, so the next difference is 12 and the next term 42, and the closed form is n(n + 1): 1 x 2, 2 x 3, up to 6 x 7. For 1, 1, 2, 6, 24, the differences 0, 1, 4, 18 tell you nothing, so try ratios: 1, 2, 3, 4. The next ratio is 5 and the term is 120; the terms are 0!, 1!, 2!, 3!, 4!, and 5! is 120. For 3, 5, 9, 17, 33, the differences are 2, 4, 8, 16, a doubling, so the next difference is 32 and the term is 65. Spot the rule in one line too: each term is twice the previous minus 1, and every term is a power of two plus 1, so the next is 2 to the 6 plus 1.

The relationship

a_n = n(n+1) \Rightarrow a_6 = 42, \qquad b_n = (n-1)! \Rightarrow b_6 = 5! = 120, \qquad c_n = 2^n + 1 \Rightarrow c_6 = 65

a n	the n-th term of the first sequence, a quadratic, so its second differences are constant
b n	the factorials, each term the previous one times the next whole number
c n	a power of two plus one, so its differences are powers of two

What it says in wordsEach sequence has a one-line rule, and the difference or ratio row is how you find it in seconds.

Is the answer really unique, and what should you say if pushed?

Strictly, no. Any five numbers can be continued by any sixth, because a polynomial of degree five can be passed through all six points. For example n(n + 1) + (n - 1)(n - 2)(n - 3)(n - 4)(n - 5) matches 2, 6, 12, 20, 30 exactly and then gives 162. The expected answer is the simplest rule that fits, and the way to show you know that is to name the rule, not just the number. That habit carries over to the desk: a pattern in five data points is a hypothesis, and the trader who states the rule can test it on the sixth point, while the one who only extrapolates cannot tell when the pattern has broken. If the interviewer offers a sequence that resists all three tests, try alternating terms, or interleaved sequences, before guessing.

Where candidates lose it

The common loss is 66 for the third sequence: doubling 33 and forgetting that the rule doubles and then subtracts 1. Check the rule on an earlier pair, 17 to 33, before you say the answer.

The second is hunting the factorials through differences, which give 0, 1, 4, 18 and lead nowhere. When differences grow faster than the terms, switch to ratios immediately.

What the interviewer asks next

What comes next: 1, 4, 9, 16, 25, 36, and what are its second differences?
Next term: 1, 2, 6, 15, 31, ...
Next term: 2, 3, 5, 7, 11, 13, ...
Find a rule for 1, 3, 7, 15, 31 and give its closed form.

047Your fair value on a contract is 50 and you quote 49 at 51. You have been lifted until you are short 20 lots against a risk limit of 25. Where do you quote now, and why not simply widen?Market makingCoreMarket makingProp trading firms
Try it first
Short 20 of a 25-lot limit, with fair value still 50. What do you do with the quote?
Show the worked solution
Skew the whole quote up, to about 50 at 52, keeping the width of 2. The short is a risk you want to shed, so make your bid attractive to sellers and your offer less attractive to buyers. A simple rule moves the mid 0.05 per lot of inventory, so 20 lots short moves it up 1. Widening to 48 at 52 also stops the buying, but it pushes sellers away too, so the short stays on your book and you earn less while you wait.
What is the position telling you to do?
A fruit seller who has run short of mangoes by mid-morning does not shut the stall; she raises the price she pays suppliers and nudges up the price to customers, so more mangoes arrive and fewer leave. Inventory changes what you want to trade next, not what the contract is worth: with fair value still 50, a short of 20 lots means your next trade should be a buy, so the quote should lean towards buying. You have used 80% of the risk limit, and five more lifts would put you at it. The question is testing whether you separate the two numbers a market maker carries: the fair value, which has not moved, and the price at which you want to trade, which has.
Moving both sides up 0.05 per lot of short keeps the width at 2 while the quote climbs from 49 at 51 to 50 at 52 at 20 lots short, so sellers find your bid at fair value and buyers find a dearer offer, whereas widening to 48 at 52 pushes both sides away.
How far should you skew, and what does it cost?
A common rule moves the mid in proportion to the position: here 0.05 per lot, so 20 lots short lifts the mid by 1 and the quote becomes 50 at 52. The bid now sits at fair value, so you buy back with no edge, and the offer is 2 above fair, so a buyer who still lifts it pays you well for adding to the risk. Skewing trades some edge for risk reduction, and the closer the position is to the limit, the more edge you should be willing to give up. In a toy model where a quote d away from fair trades with probability 0.6 x e to the minus d each period, the unchanged quote earns 0.44 a period and never reduces the short; the full skew earns 0.16 but buys back a net 0.52 lots a period, about 39 periods to flatten; the half skew sits between at 0.38 and 0.23. The numbers are illustrative; the direction is not.
Quote Bid fill Offer fill Net lots bought Edge a period
49 at 51, unchanged 0.22 0.22 +0.00 0.44
49.5 at 51.5, half skew 0.36 0.13 +0.23 0.38
50 at 52, full skew 0.60 0.08 +0.52 0.16
48 at 52, widened 0.08 0.08 +0.00 0.32
In the toy fill model only a skewed quote buys back the short; the widened quote earns more per fill but leaves the position where it is.
Why not simply widen?
Widening to 48 at 52 makes the offer as unattractive as the skew does, but it also moves the bid two points below fair, so sellers go elsewhere. In the toy model both sides fill 0.08 of the time and the expected change in the position is zero: a wider quote stops new risk arriving but does nothing about the risk you already hold, and the 20-lot short keeps moving with the market while you wait. Widening has its place, when you think fair value is uncertain or that the people lifting you know something, because then both sides are dangerous. Say that distinction, and then say the last resort: if skewing does not bring sellers fast enough, hedge the short in a related market, or cross the spread and buy, rather than drift up to the limit.
Where candidates lose it
The common loss is widening, because it feels cautious. It protects against new risk, but the 20 lots you already hold are the problem, and a wide quote does not bring sellers to buy them back.
The second is moving fair value. Nothing about the contract has changed; only your position has. Keep 50 as fair, move the quote, and say why the two are now different numbers.
What the interviewer asks next
You suspect the buyers lifting you know something. Does that change skew into widen?
You reach the 25-lot limit. What do you do with the offer?
How would you choose the skew per lot of inventory?
A related contract is liquid and moves with yours. How does that change your quote?
048You may roll a fair die up to three times, stopping whenever you like, and you are paid the value of the last roll. What is the optimal stopping rule and the value of the game?Expected value and optimal stoppingCoreRBC Capital MarketsToronto · 2025
Try it first
On the first roll you get a 4. Do you stop?
Show the worked solution
Stop on 5 or 6 on the first roll, on 4 or more on the second, and take the third; the game is worth 14/3, about 4.67. Work backwards. One roll left is worth 3.5. With two left, keep anything above 3.5, so 4, 5 or 6, which makes the game worth 15/6 + 3/6 x 3.5 = 4.25. With three left, keep anything above 4.25, so 5 or 6, which gives 11/6 + 4/6 x 4.25 = 14/3.
Why start from the last roll?
If you are flat-hunting and can see three flats, one a week, you take the first only if it beats what you expect from the remaining two, and you can only know that by thinking about the last week first. Every stop-or-continue decision compares the number in hand with the value of continuing, and the value of continuing is only known once you have solved the rounds after it, so you solve from the end. With one roll left there is no choice: you take whatever comes, worth 3.5 on average. That number becomes the bar for the roll before it, and the value of that roll becomes the bar for the one before that.
Read from right to left, the last roll is worth 3.5, so with two rolls left you keep 4 or more and the game is worth 4.25, so with three rolls left you keep only 5 or 6 and the game is worth 14/3, about 4.67.
How do the two thresholds come out?
With two rolls left, you keep the first of them if it beats 3.5, so 4, 5 or 6 are kept, each with probability 1/6, and on 1, 2 or 3 you roll once more for 3.5. The value is (4 + 5 + 6)/6 + (3/6) x 3.5 = 2.5 + 1.75 = 4.25. With three rolls left, the bar rises to 4.25, so a 4 is no longer good enough: keep only 5 or 6, and the value is (5 + 6)/6 + (4/6) x 4.25 = 11/6 + 17/6 = 14/3. The thresholds rise as more rolls remain, because the option to keep rolling is worth more when there are more chances left. Notice that the threshold is a value, not a face: you keep a face only if it is strictly greater than the value of carrying on.
The relationship
$V_1 = 3.5, \qquad V_{n} = \frac{1}{6}\sum_{f=1}^{6} \max(f, V_{n-1}) \quad\Rightarrow\quad V_2 = 4.25, \quad V_3 = \tfrac{14}{3} \approx 4.67$
V n the value of the game with n rolls left, played optimally
max(f, V n-1) on rolling face f you keep it or carry on, whichever is worth more
the sum over f the average over the six equally likely faces
What it says in wordsThe value with n rolls left is the average, over the faces, of the better of keeping the face and rolling on.
What do the follow-ups test?
They test whether you can re-run the recursion. With more rolls the value climbs towards 6 but slowly: 4 rolls give 4.944, 6 give 5.275, 10 give 5.650. A cost per roll lowers each continuation value and drops the thresholds. Keeping a 4 on the first roll costs you: that rule is worth 4.625 against 4.667, a small gap that an interviewer will still ask you to explain. A simulation of 300,000 games of the optimal rule gives 4.666. The desk link is direct: an American option is a stopping problem of exactly this shape, exercise when the value in hand beats the value of holding on, and a binomial tree solves it by the same backward pass.
Where candidates lose it
The common loss is keeping a 4 on the first roll because it beats 3.5. The right comparison is with the value of continuing, which with two rolls left is 4.25, not 3.5.
The second is solving forwards, trying to guess the first-roll threshold before knowing what the later rolls are worth. Say the last roll is worth 3.5, then build up; the answer arrives in two lines.
What the interviewer asks next
What is the game worth with four rolls, and what are the thresholds?
Each roll after the first costs 0.25. How do the thresholds change?
You are paid the square of the final roll. Does the stopping rule change?
How is this related to exercising an American option?
Asked at RBC Capital Markets, Quantitative Trading, Toronto, 2025 (Wall Street Oasis): Best way to maximize EV across 3 chosen dice rolls (can choose to continue or not).
049A 3 x 3 x 3 cube is painted on the outside and cut into 27 small cubes. You pick one small cube at random and roll it like a die. What is the probability the top face is painted?Probability and countingCoreJane StreetNew York · 2026
Try it first
What is the chance the top face is painted?
Show the worked solution
1/3. Picking a cube at random and then a face at random makes every one of the 27 x 6 = 162 small faces equally likely to end up on top. The painted small faces are exactly the squares on the big cube's surface, 6 faces of 9 each, 54 in all. So the chance is 54/162 = 1/3. The breakdown by cube agrees: corners give 8 x 3, edges 12 x 2, face centres 6 x 1, the core 0, total 54.
Why count faces rather than cubes?
If a bag holds sweets of different sizes and you want the chance a random bite is chocolate, you count chocolate bites, not chocolate sweets. Every small cube is equally likely and every face of it is equally likely to land on top, so every one of the 162 small faces has the same chance, 1/162, and the answer is just the share of small faces that are painted. That share is easy, because the painted small faces are exactly the visible squares of the big cube: 6 faces with 9 squares each, 54. The answer, 54/162 = 1/3, comes in one line without classifying a single cube, which is what the interviewer hopes to see.
The 8 corner cubes carry 24 painted faces, the 12 edge cubes 24, the 6 face centres 6 and the core none, so 54 of the 162 small faces are painted and a random top face is painted with probability one third.
How does the cube-by-cube count confirm it?
Classify the 27 cubes by position. The 8 corners have 3 painted faces each, the 12 edge cubes have 2, the 6 face centres have 1, and the single core cube has none: 8 + 12 + 6 + 1 = 27. Weight each type by how often you pick it and by the chance its top is painted: 8/27 x 3/6 + 12/27 x 2/6 + 6/27 x 1/6 + 1/27 x 0 = (24 + 24 + 6) / 162 = 1/3, the same answer by the long road. The two methods are the law of total probability written two ways, once by cube and once by face, and saying that out loud shows you know why they must agree. For an n x n x n cube the face count gives 6n squared painted faces out of 6n cubed, so the answer is 1/n: 1/2 for a 2 x 2 x 2 cube, 1/10 for a 10 x 10 x 10.
The relationship
$P(\text{top painted}) = \frac{6 \times 3^2}{27 \times 6} = \frac{54}{162} = \frac13, \qquad \text{in general } \frac{6n^2}{6n^3} = \frac1n$
6 x 3 squared the painted small faces, the 9 squares on each of the 6 outer faces
27 x 6 all small faces, each equally likely to end on top
n the number of cuts along each edge
What it says in wordsThe chance is the painted share of all small faces, which for an n-cube is one over n.
What is the natural follow-up, and how do you answer it?
Turn it round: the top face is painted; what is the chance you picked a corner? That is Bayes on the same count. Of the 54 painted faces, 24 belong to corners, so the chance is 24/54 = 4/9, far above the 8/27 a corner has before you look. Seeing paint is evidence for the cubes with more paint, and the face count gives the posterior directly without a formula. A second follow-up asks for the chance that the picked cube has any paint at all, which is 26/27, and the gap between 26/27 and 1/3 is exactly the trap in the original question. The limitation is that the face count relies on every face being equally likely to land on top; a weighted cube, or a rule that picks cubes by size, would need the long route.
Where candidates lose it
The common loss is answering 26/27, the chance the cube has some paint. The question asks about the top face, and most painted cubes are painted on only a few of their six faces.
The second is miscounting the cube types, often 6 edges instead of 12, and then forcing the total to 27 with the core. Skip the classification: count the 54 visible squares, divide by 162, and use the breakdown only as a check.
What the interviewer asks next
The top face is painted. What is the probability the cube is a corner?
Do the same for a 4 x 4 x 4 cube. Is there a general formula?
You roll the chosen cube twice. What is the chance both tops are painted?
How many of the 27 cubes have exactly two painted faces, and for an n-cube?
Asked at Jane Street, Engineering, New York, 2026 (Wall Street Oasis): How you got to the answer matters even if you got the question right. Strawberry question + 3x3 cube question
050You can buy the 90 call, sell the 110 call, buy the 110 put and sell the 90 put, all European and expiring in one year, for a net 19.20. What does the position pay at expiry, what interest rate does it imply, and when is that attractive?Option payoffs and no-arbitrageHardMarket makingRates derivatives
Try it first
What does the four-leg position pay at expiry?
Show the worked solution
It pays exactly 20 at any stock price, so 19.20 today implies a one-year rate of 20/19.20 - 1, about 4.17%. The 90 call and short 90 put make a long forward at 90; the short 110 call and long 110 put make a short forward at 110. The stock cancels and 110 - 90 = 20 is left. Buying the box is lending at 4.17% and selling it is borrowing at that rate, so it suits a lender whose other return is lower, or a borrower whose funding costs more.
Why does the payoff not depend on the stock?
Agree to buy a scooter from one friend for 90 and to sell it to another for 110, both next year, and you will make 20 whatever scooters cost by then. The box is the same pair of agreements: long the 90 call and short the 90 put is a promise to buy at 90, and short the 110 call and long the 110 put is a promise to sell at 110, so the stock comes in and goes out and only the gap between the strikes, 20, is left. Seen as spreads, the call spread pays from 0 below 90 up to 20 above 110, and the put spread pays the mirror image, 20 below 90 down to 0 above 110. Wherever the stock ends, the two add to 20.
The long call spread rises from 0 to 20 between 90 and 110 while the long put spread falls from 20 to 0 over the same range, so their sum is a flat 20 at every stock price, and paying 19.20 for it is lending at 4.17%.
How do you turn the price into a rate?
You pay 19.20 today and receive 20.00 in a year with no market risk, so it is a deposit. The simple rate is 20 / 19.20 - 1 = 4.17%, and the continuously compounded rate is ln(20 / 19.20) = 4.08%. A box is a zero-coupon bond built from options, and its price is the strike gap times the discount factor, so any box that trades away from the market's interest rate is mispriced. Prices consistent with this rate, for an illustrative stock at 100 with 20% volatility, are a 90 call at 16.11, a 110 call at 5.69, a 110 put at 11.29 and a 90 put at 2.51: 16.11 - 5.69 + 11.29 - 2.51 = 19.20. Volatility does not enter the box price at all, because every volatility effect in the calls is cancelled by the puts.
The relationship
$\underbrace{(S - 90)^+ - (90 - S)^+}_{S - 90} \; - \; \underbrace{\big[(S - 110)^+ - (110 - S)^+\big]}_{S - 110} = 20, \qquad r = \frac{20}{19.20} - 1 = 4.17\%$
(S - K)+ a call payoff struck at K
(K - S)+ a put payoff struck at K
call minus put at the same strike a forward to buy at that strike, paying S - K
r the simple one-year rate implied by paying 19.20 for 20
What it says in wordsA long forward at 90 and a short forward at 110 leave a fixed 20, and its price gives the interest rate.
When is it attractive, and what can go wrong?
Compare 4.17% with your own rates. If cash would otherwise earn 3.5%, buying the box lends at a better rate: 19.20 at 3.5% grows to only 19.87, against 20 from the box. If your funding costs 5%, selling the box borrows more cheaply: you receive 19.20 today and owe 20, while 20 owed at 5% would have raised only 19.05. Whether a box is cheap or dear is never a property of the box alone; it depends on the rate you can otherwise lend or borrow at, which is why boxes are a funding trade. The risks to name are practical. American options can be exercised early, which breaks the box, so use European index options. Four bid-offer spreads and fees can eat the 0.80 of interest. And the counterparty, or the clearing house, must still be there in a year.
Where candidates lose it
The common loss is analysing the four legs as a view on the stock, describing a bull call spread and a bear put spread and forgetting to add them. Add them: the stock cancels and the position is a loan.
The second is calling a box at 19.20 an arbitrage on its own. It is only cheap or dear against a rate, so say which rate you are comparing with, your deposit rate if you buy and your funding rate if you sell.
What the interviewer asks next
The same box trades at 19.80. What rate does that imply, and who would sell it?
Why does an American-style box carry early-exercise risk, and which leg is the danger?
Build a box with strikes 95 and 105. What should it cost at the same rate?
How is a box related to put-call parity at each strike?

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

Quote	Bid fill	Offer fill	Net lots bought	Edge a period
49 at 51, unchanged	0.22	0.22	+0.00	0.44
49.5 at 51.5, half skew	0.36	0.13	+0.23	0.38
50 at 52, full skew	0.60	0.08	+0.52	0.16
48 at 52, widened	0.08	0.08	+0.00	0.32

6 x 3 squared	the painted small faces, the 9 squares on each of the 6 outer faces
27 x 6	all small faces, each equally likely to end on top
n	the number of cuts along each edge

Try it first

Where does the middle of the quote come from?

How wide should the quote be, and why not plus or minus one standard deviation?

What exactly changes when I tell you 8 of the first 10 were heads?

Where candidates lose it

What the interviewer asks next

Try it first

Why is a chooser worth less than a straddle but more than a single option?

How does put-call parity turn the choice into plain options?

How do you check the answer, and what changes with rates and dividends?

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

Try it first

How do you read a payoff picture as a list of calls?

What does the butterfly cost, and what does its price tell you?

Where does the no-arbitrage check come in?

Where candidates lose it

What the interviewer asks next

Try it first

Which decisions are obvious, and which one is the real question?

How do you find the bluffing frequency?

How do you sanity check the frequencies, and who wins the game?

Where candidates lose it

What the interviewer asks next

Try it first

What order should you test patterns in?

How does each of the three give way?

Is the answer really unique, and what should you say if pushed?

Where candidates lose it

What the interviewer asks next

Try it first

What is the position telling you to do?

How far should you skew, and what does it cost?

Why not simply widen?

Where candidates lose it

What the interviewer asks next

Try it first

Why start from the last roll?

How do the two thresholds come out?

What do the follow-ups test?

Where candidates lose it

What the interviewer asks next

Try it first

Why count faces rather than cubes?

How does the cube-by-cube count confirm it?

What is the natural follow-up, and how do you answer it?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the payoff not depend on the stock?

How do you turn the price into a rate?

When is it attractive, and what can go wrong?

Where candidates lose it

What the interviewer asks next