Derivatives Foundation puzzles, solved step by step

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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.

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?

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
052A bus makes three stops. At each stop half the people on board get off, and then the number on board grows by a third. The bus reaches the end of the route with 16 people. How many were on board at the start?Mental maths and estimationCoreProp trading firms
Try it first
Before working it: each stop multiplies the number on board by what?
Show the worked solution
54 people. Each stop halves the load and then adds a third of what is left, so the load is multiplied by 1/2 x 4/3 = 2/3 at every stop. Undoing three stops means dividing 16 by 2/3 three times: 16 to 24 to 36 to 54. The forward check works: 54 to 27 to 36, then 18 to 24, then 12 to 16.
Why work backwards rather than guess a start?
If someone tells you a shirt was marked down by half, then marked up by a third, and now costs Rs 16, you do not guess the original tag; you undo the two moves in reverse order. Each stop is a pair of multiplications, and multiplications are undone by dividing in the opposite order, so the end number walks back to the start without any trial and error. The growth of a third comes last at each stop, so it is undone first: divide by 4/3, which is multiply by 3/4. Then undo the halving by doubling. From 16: times 3/4 is 12, times 2 is 24. Twice more gives 36 and then 54.
Starting from 16 at the end of the route and undoing each stop, first multiplying by 3/4 then by 2, the count climbs 16 to 24 to 36 to 54, and running the three stops forward from 54 returns exactly 16, because each stop multiplies the load by 2/3.
What is the one-line version an interviewer wants to hear?
Collapse each stop to a single factor. Losing half is x 1/2 and gaining a third is x 4/3, so a stop is x 2/3, and three stops are x 8/27. The start is 16 x 27/8 = 54. Saying that in one breath shows you saw the structure rather than the arithmetic, which is what the question is for. The limitation is that it only works because every stop has the same two moves; a route where the fractions change needs the step-by-step walk back.
The relationship
$N_0 = N_3 \times \left(\tfrac{1}{2}\times\tfrac{4}{3}\right)^{-3} = 16 \times \left(\tfrac{3}{2}\right)^3 = 16 \times \tfrac{27}{8} = 54$
N_0 the number on board at the start
N_3 the number at the end of the route, 16
(3/2)^3 undoing the per-stop factor of 2/3 three times
What it says in wordsThe start is the end count divided by the per-stop factor of two thirds, three times over.
Where do the whole-number checks help you?
The bus carries people, so every intermediate count must be a whole number, and that is a free check. If the backwards walk ever produces a fraction, you have undone the moves in the wrong order. Undoing the halving first from 16 gives 32, then times 3/4 gives 24, which happens to be whole here, but on the next stop it would give 48 then 36, and the order mistake would have cost you nothing visible until the end, where 72 x 3/4 = 54 again by luck. Order matters in general even when the numbers hide it, so say the order out loud.
Where candidates lose it
The common slip is treating the two moves at each stop as a net loss of a sixth, since a half minus a third is a sixth. The third is taken on the smaller number, so the stop is a factor of 2/3, not 5/6.
The second slip is undoing the moves in the wrong order. Growth came last, so it is undone first. Say the per-stop factor, then walk back 16, 24, 36, 54, and finish with the forward check.
What the interviewer asks next
If instead a third get off and then the number on board doubles at each stop, what is the per-stop factor?
How many stops would it take for a bus starting with 54 to get below 5 people?
The bus starts with 54 and the pattern continues. After how many stops is the number no longer a whole number?
053It rains on 20% of days. Two forecasters work independently: each calls rain on 90% of the days it actually rains, and also calls rain on 15% of the dry days. Both call rain for tomorrow. What is the probability it rains? And if one calls rain and the other does not?Conditional probability and BayesCoreJane StreetNew York · 2025
Try it first
Two independent forecasters both say rain. Instinct first: how likely is rain?
Show the worked solution
90% if both call rain, and 15% if they disagree. Over 1,000 days, 200 are rainy and 800 are dry. Both call rain on 200 x 0.9 x 0.9 = 162 rainy days and on 800 x 0.15 x 0.15 = 18 dry days, so rain is 162 of 180. A split call happens on 36 rainy days and 204 dry days, so rain is 36 of 240, which is below the 20% base rate.
Why count days instead of multiplying probabilities?
A doctor reading two test results does the same thing: imagine a thousand patients, split them by who is actually ill, then split each group by what the tests said. Natural frequencies keep the base rate in the picture, where the probability form hides it, so you never confuse the chance of the evidence given rain with the chance of rain given the evidence. The 200 rainy days produce two rain calls 162 times; the 800 dry days, despite each forecaster being wrong only 15% of the time, produce two rain calls 18 times because 800 is a big base. The ratio is 162 to 18, so 90%.
Of 1,000 days, 200 rainy days give 162 double rain calls, 36 split calls and 2 with no call, while 800 dry days give 18, 204 and 578, so two rain calls make rain 90% likely and a split call makes it only 15% likely.
Why does a split call push the chance below the base rate?
Because a forecaster saying dry is itself evidence, and it is strong evidence. Each forecaster calls dry on 85% of dry days but only 10% of rainy days, so a dry call divides the odds of rain by 8.5, while a rain call multiplies them by only 6. Put together, one rain call and one dry call multiply the odds by 6 x (1/8.5), about 0.71, so the odds fall from 1 to 4 to about 1 to 5.67, which is 15%. The common feeling that the two calls cancel is wrong because the two kinds of error are not symmetric.
The relationship
$\frac{P(R\mid\text{both})}{P(D\mid\text{both})} = \frac{0.2}{0.8}\times\frac{0.9}{0.15}\times\frac{0.9}{0.15} = \frac{1}{4}\times 36 = 9 \;\Rightarrow\; P(R\mid\text{both}) = \tfrac{9}{10}$
0.2/0.8 the prior odds of rain, 1 to 4
0.9/0.15 the likelihood ratio of one rain call, 6
9 the posterior odds of rain after two independent calls
What it says in wordsPrior odds of one to four, multiplied by six for each independent rain call, give odds of nine to one, which is ninety per cent.
What assumption does the answer lean on, and when would it fail?
Independence given the weather. If both forecasters read the same satellite feed, two rain calls are closer to one piece of evidence than two, and the odds should be multiplied by something well short of 36. On a desk the same mistake appears when three models trained on the same data all flag a trade: the signals agree because they share inputs, not because the evidence is three times as strong. The answer also treats the 90% and 15% rates as known; with rates estimated from a short record, the posterior is softer than it looks.
Where candidates lose it
The fast wrong answer is 81%, from 0.9 times 0.9, which is the likelihood of the evidence rather than the probability of rain. Candidates who write Bayes' formula from memory often lose the base rate and give the same number.
The second trap is the split call: most people say about 50% or go back to 20%. Say that a dry call is strong evidence, divide the odds by 8.5 and multiply by 6, and the 15% follows.
What the interviewer asks next
Both forecasters say dry. What is the chance of rain now?
The two forecasters share the same data and always agree. What does two rain calls tell you?
A third independent forecaster with the same accuracy calls rain. Update the 90%.
How accurate would one forecaster need to be for a single rain call to make rain more likely than not?
Asked at Jane Street, Generalist, New York, 2025 (Wall Street Oasis): a question about the probability of rain the next day that relied on a very in depth understanding of bayes theorem
054Give me two random variables that are uncorrelated but clearly dependent, and show me that the correlation is zero.Distributions and statisticsCoreTwo SigmaNew York · 2025Tower Research CapitalNew York · 2014
Try it first
X is a standard normal and Y = X squared. What is the correlation of X and Y?
Show the worked solution
Take X standard normal and Y = X squared. Y is completely determined by X, the strongest dependence there is, yet Cov(X, Y) = E[X cubed] - E[X] E[X squared] = 0 - 0 x 1 = 0, because a symmetric distribution has zero third moment. A discrete version is X uniform on minus 1, 0 and 1 with Y = X squared, where E[XY] = (minus 1 + 0 + 1)/3 = 0 and E[X] = 0.
What does correlation actually measure?
Think of a shop whose sales rise when it is very cold and when it is very hot, and sag in mild weather. Plot sales against temperature and you get a U. Ask a straight line to summarise the U and it comes out flat: the hot days pull it up on the right exactly as much as the cold days pull it up on the left. Correlation is the slope of the best straight line through the cloud, scaled to lie between minus 1 and 1, so it can only detect the part of a relationship that runs in one direction. A relationship that turns around is invisible to it.
Forty-one points on the parabola Y = X squared, with X spread like a standard normal, have a best-fit line that is exactly flat at height 0.97, so their correlation is zero while every Y is fixed by its X.
How do you prove the zero in two lines?
Write the covariance as E[XY] minus E[X] E[Y]. With Y = X squared, E[XY] is E[X cubed], and every distribution symmetric about zero has E[X cubed] = 0, because each positive value x cubed is cancelled by the equally likely minus x cubed. E[X] is also zero, so both terms vanish and the correlation, which is the covariance divided by two positive standard deviations, is zero. The same proof works for any even function of a symmetric X, such as the absolute value, and for any symmetric X, not just the normal.
The relationship
$\operatorname{Cov}(X, X^2) = E[X^3] - E[X]\,E[X^2] = 0 - 0 \times 1 = 0 \qquad \text{while } Y = X^2 \text{ exactly}$
E[X^3] the third moment, zero for any distribution symmetric about zero
E[X] the mean of X, zero
E[X^2] the variance of a standard normal, one
What it says in wordsThe covariance of a symmetric variable with its own square is zero, so correlation is zero even though the square is a function of the variable.
Why does a derivatives desk care about this distinction?
Because a hedge built on correlation only neutralises straight-line exposure. A stock's return and its squared return are close to uncorrelated, yet the squared return is exactly what a long option position pays for, so a delta hedge that zeroes the linear exposure leaves the gamma exposure untouched. The limitation runs the other way too: independence does imply zero correlation, so a measured correlation near zero rules nothing out, and a desk that reads zero correlation as no relationship will be surprised in both tails.
Where candidates lose it
Candidates reach for a definition, saying independence implies zero correlation but not the reverse, and then cannot produce an example. The interviewer wants the example first: X and X squared, said inside ten seconds.
The second loss is giving the example without the proof. Write the covariance as E[X cubed] minus E[X] E[X squared] and say why the third moment of a symmetric variable is zero.
What the interviewer asks next
Give a second example where neither variable is a function of the other.
X is uniform on 0 to 1 rather than symmetric. Is X still uncorrelated with X squared?
What single number would catch the dependence between X and X squared that correlation misses?
Asked at Two Sigma, Generalist, New York, 2025 (Wall Street Oasis): Come up with two uncorrelated but dependent variables. Lots of questions regarding regression
Asked at Tower Research Capital, Quantitative Research, New York, 2014 (Wall Street Oasis): He also asked a question about independence and correlation. I did not respond fast enough
055A token sits on one corner of a square. Every second it moves to one of the two neighbouring corners, chosen by a fair coin. What is the expected number of seconds until it first reaches the opposite corner?Random walks and Markov chainsCoreQuant trading
Try it first
The opposite corner is two steps away. Expected time to get there?
Show the worked solution
4 seconds. By symmetry, the two corners next to the start are the same state, call it adjacent. From the start you always move to adjacent in one step. From adjacent, half the time you reach the target and half the time you return to the start. So E(start) = 1 + E(adjacent) and E(adjacent) = 1 + E(start)/2, giving E(adjacent) = 3 and E(start) = 4.
Why collapse four corners into three states?
If you are lost in a town with a river on one side, what matters is how far you are from the river, not which street you are on. The square is the same: standing at either corner next to the start, the token's future looks identical, one coin flip from the target and one from the start. Grouping corners by their distance from the target turns a four-state chain into a three-state line, and a line is solved with one equation per unknown. You have two unknowns, the expected time from the start and from an adjacent corner, because the target itself takes zero time.
Grouping the two corners next to the start into one state gives a three-state chain in which the start always moves to adjacent, and adjacent finishes or returns to the start with equal chance, so E(adjacent) = 3 and E(start) = 4 seconds.
How do you set up and solve the equations in the room?
Each equation says the same sentence: one step, plus the average of what is left from where you land. From the start, every step lands on an adjacent corner, so E(start) = 1 + E(adjacent); from an adjacent corner, half the steps finish and half return, so E(adjacent) = 1 + (1/2) x 0 + (1/2) x E(start). Substitute the first into the second: E(adjacent) = 1 + (1/2)(1 + E(adjacent)), so E(adjacent)/2 = 3/2, E(adjacent) = 3, and E(start) = 4. Saying the sentence before the symbols is what keeps the equations honest.
The relationship
$E_0 = 1 + E_1, \qquad E_1 = 1 + \tfrac{1}{2}E_0 \;\Rightarrow\; E_1 = 3,\; E_0 = 4$
E_0 the expected steps to the target from the starting corner
E_1 the expected steps from either corner next to the start
1/2 the chance a step from an adjacent corner lands on the target
What it says in wordsFrom the start you always move one step closer; from there a coin flip either finishes or sends you back, and the two equations give four steps on average.
What is the check, and what does the general case look like?
Check it with the geometric picture. After the first step you are adjacent, and from there each attempt either finishes in one step or costs two steps, back to the start and out again, before you are adjacent once more. Each attempt succeeds with chance 1/2, so E(adjacent) = (1/2) x 1 + (1/2) x (2 + E(adjacent)), which again gives 3, and the first step makes it 4. On a cube the same method with four distance classes gives 10 steps to the opposite vertex; on a general graph the method is the same, but the number of distance classes grows and the arithmetic stops being mental.
Where candidates lose it
The fast wrong answer is 2, the length of the shortest path. The interviewer is checking whether you see that the token can bounce back, and whether you reach for the first-step equations rather than trying to sum a series.
The second loss is writing four equations, one per corner. Say the symmetry out loud, collapse to three states, and the whole thing is two lines.
What the interviewer asks next
What is the expected time to return to the starting corner for the first time?
Same walk on the eight corners of a cube. Expected time to the opposite vertex?
What is the probability the token reaches the opposite corner within 4 steps?
The coin is biased: it moves clockwise with chance 0.7. Does the expected time change?

(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

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

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

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

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

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 work backwards rather than guess a start?

What is the one-line version an interviewer wants to hear?

Where do the whole-number checks help you?

Where candidates lose it

What the interviewer asks next

Try it first

Why count days instead of multiplying probabilities?

Why does a split call push the chance below the base rate?

What assumption does the answer lean on, and when would it fail?

Where candidates lose it

What the interviewer asks next

Try it first

What does correlation actually measure?

How do you prove the zero in two lines?

Why does a derivatives desk care about this distinction?

Where candidates lose it

What the interviewer asks next

Try it first

Why collapse four corners into three states?

How do you set up and solve the equations in the room?

What is the check, and what does the general case look like?

Where candidates lose it

What the interviewer asks next