Quant puzzles, solved step by step

Puzzles: 100
Traced to a firm: 71
Topics: 12
Hard: 30

Showing 1–10 of 100

001Four people must cross a narrow bridge at night with one torch. At most two can be on the bridge at once, anyone crossing must carry the torch, and a pair walks at the slower person's pace. They take 1, 2, 5 and 10 minutes. What is the shortest total time to get everyone across?Logic and algorithmic reasoningCoreBelvedere TradingChicago · 2021
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Before you plan it: what is the fastest time?
Show the worked solution
17 minutes. Send 1 and 2 over (2 minutes), 1 comes back (1), 5 and 10 cross together (10), 2 comes back (2), and 1 and 2 cross again (2). The obvious plan, where the fastest person escorts each of the others, takes 19. The saving comes from putting the two slowest walkers on the bridge at the same time.
Why does the obvious plan lose two minutes?
Think of two slow parcels going by the same courier. If each travels on its own trip you pay for both trips; if they share a van, you pay for the slower one only. Every crossing costs the slower walker's time, so a slow person paired with a fast one wastes the fast one, and two slow people paired together waste nothing. Escorting with the fastest person pays 10 and then 5 as separate crossings. Pairing 5 with 10 pays 10 once and the 5 minutes are free.
Pairing the 5 and 10 minute walkers on one crossing finishes in 17 minutes, while letting the 1 minute walker escort everyone pays for the 10 and the 5 separately and finishes in 19.
What is the price of pairing the slow two?
Somebody has to bring the torch back after the slow pair crosses, and it must not be one of them. So the plan first ferries two fast people over, leaves one on the far side to carry the torch back later, and spends the 2 minute walker's return trip to buy the 5 minute saving. The trade is 1 + 2 extra minutes of shuttling against 5 minutes saved on the slow side, a net gain of 2. With different speeds the trade can flip, which is the real content of the puzzle.
The relationship
$\text{escort: } 2a + b + c + d \qquad \text{pair: } a + 3b + d \qquad \text{pair wins when } 2b < a + c$
a, b the two fastest times, here 1 and 2
c, d the two slowest times, here 5 and 10
What it says in wordsPairing the slow two is better exactly when twice the second fastest time is less than the fastest plus the second slowest.
How do you convince the interviewer 17 cannot be beaten?
There must be at least five crossings, three over and two back, because each trip over moves at most two people and someone must return the torch. The 10 minute walker costs 10 on whatever crossing carries them. If 5 and 10 cross separately you already spend 15 on those two trips, and the three remaining crossings cost at least 1 + 1 + 2, which is 19; if they cross together, the best you can do with the other four crossings is 2 + 1 + 2 + 2. A brute force over every schedule gives the same minimum, 17 minutes.
Where candidates lose it
The strong candidate's trap is a fast answer of 19. Letting the quickest person run every errand feels efficient, and it is the right instinct for returning the torch, but it is the wrong instinct for the slow walkers.
The second loss is getting 17 by trial and error and then being unable to say why. State the principle, that the slow pair shares one crossing, and give the rule for when it wins: when twice the second fastest time is below the fastest plus the second slowest.
What the interviewer asks next
What if the times are 1, 4, 5 and 10?
Six people with times 1, 2, 5, 10, 20 and 25: what is the plan?
Write the general algorithm for n people and say its running time.
Asked at Belvedere Trading, Trading, Chicago, 2021 (Wall Street Oasis): crossing the bridge in the shortest amount of time with one flashlight brainteaser
002You are flying to a city where it rains on 25% of days. You phone three friends who live there. Each tells the truth with probability 2/3, independently of the others, and all three say it is raining. What is the probability that it is actually raining?Conditional probability and BayesCoreJane StreetNew York · 2025
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Pick your answer before working it.
Show the worked solution
8/11, about 72.7%. If it is raining, all three say yes with probability (2/3)^3 = 8/27. If it is dry, all three must be lying, (1/3)^3 = 1/27. Weight each by how often it happens: 1/4 x 8/27 against 3/4 x 1/27, which is 8 parts to 3. Three agreeing witnesses move a 25% prior a long way, but not to certainty.
Why is the answer not simply 8/9?
Picture a clinic where a test is quite reliable but the illness is uncommon. A positive result makes the illness more likely, but how much more depends on how rare it was to begin with. The friends' agreement tells you how much more likely rain makes their answer than dry does, eight times, but it does not erase the fact that dry days are three times as common. 8/9 is the answer you get if rain and dry start level. Here they do not.
Rain covers a quarter of days and all three friends say yes on 8/27 of those, while dry days cover three quarters and all three lie on only 1/27 of them, so the shaded areas stand 8 to 3 and the chance of rain given three yeses is 8/11, about 72.7%.
How do you set it up so the arithmetic stays small?
Use odds rather than probabilities. Posterior odds are prior odds times the likelihood ratioHow many times more likely the evidence is if the hypothesis is true than if it is false., and both are easy numbers here. Prior odds of rain are 1 to 3. The likelihood ratio of three yeses is (2/3)^3 over (1/3)^3, which is 2 cubed, 8. So the posterior odds are 8 to 3, and the probability is 8 over 8 plus 3, 8/11. Each additional agreeing friend would double the odds again.
The relationship
$\frac{P(R\mid YYY)}{P(D\mid YYY)} = \frac{P(R)}{P(D)}\cdot\frac{(2/3)^3}{(1/3)^3} = \frac{1}{3}\cdot 8 = \frac{8}{3} \;\Rightarrow\; P(R\mid YYY) = \frac{8}{11}$
R, D rain and dry
YYY all three friends say yes
(2/3)^3 and (1/3)^3 the chance of three yeses when it rains, and when it is dry
What it says in wordsMultiply the prior odds by how much more likely the evidence is under rain, then turn the odds back into a probability.
What assumption is doing the work, and should you say it?
The calculation needs the friends to lie independently. If they could be coordinating a joke, three yeses are really one piece of evidence, and the answer falls back towards the one-friend figure of 2/5. Say the independence assumption out loud, then give 8/11. Interviewers often follow up by making one friend unreliable or by letting them talk to each other.
Where candidates lose it
The most common wrong answer is 8/9: the candidate compares the chance of three truths with the chance of three lies and forgets the weather's own odds. The rain prior is a quarter, and leaving it out quietly assumes it is a coin flip.
The second loss is writing out a full Bayes formula with 27ths and 108ths and losing the thread under time pressure. Odds times likelihood ratio gets 8 to 3 in two lines and is easier to check out loud.
What the interviewer asks next
What if only two of the three friends say yes?
How many agreeing friends would you need before you were 95% sure it is raining?
What changes if the friends can talk to each other before answering?
Asked at Jane Street, Generalist, New York, 2025 (Wall Street Oasis): There was a question about the probability of rain the next day that relied on a very in depth understanding of bayes theorem
003Speed round, ninety seconds: you draw three cards from a well-shuffled 52-card deck without replacement. What is the probability that all three are of different suits?Counting and combinatoricsWarm upQuant tradingProp trading firms
Try it first
Closest answer, fast.
Show the worked solution
About 39.8%. The first card can be anything. The second must come from one of the three other suits: 39 of the 51 cards left. The third must avoid both suits already seen: 26 of the 50 left. Multiply: 39/51 x 26/50 = 1,014/2,550, just under 40%. Drawing with replacement would give 3/4 x 1/2 = 37.5%.
Why does the first card cost nothing?
Picture three guests arriving at a party with four dress colours in the wardrobe, and you want no two to match. The first guest cannot clash with anyone. A condition about the cards differing only bites from the second card on, so the first card contributes a factor of 1 and you start counting from card two. Candidates who write 13/52 for the first card have fixed a particular suit and then have to multiply by the number of suit orders to recover, which is where the slips happen.
The first card is free, the second must avoid one used suit with 39 of 51 cards still good, and the third must avoid two with 26 of 50 still good, so three different suits happen with probability 39.8%.
How do you check it a second way in the time?
Count unordered hands. Choose which three suits appear, 4 ways, then one card from each, 13 cubed, and divide by all three-card hands, 52 choose 3. That is 4 x 2,197 = 8,788 over 22,100, which is the same 0.3976. In a speed round you do not have time for both, but knowing the counting route exists lets you sanity-check the product: 0.765 x 0.52 is a little under 0.40.
The relationship
$P = \frac{39}{51}\cdot\frac{26}{50} = \frac{4\cdot 13^3}{\binom{52}{3}} = \frac{8788}{22100} \approx 0.398$
39/51 cards of a new suit among those left after one draw
26/50 cards of a third suit after two draws
\binom{52}{3} the number of possible three-card hands
What it says in wordsSequential dodging and direct counting give the same 39.8%.
What is a speed round actually testing?
Thirty questions in forty-five minutes cannot all be worked in full. The skill being tested is choosing the shortest correct route and estimating the product well enough to pick from the options. Here, 39/51 is about 0.76 and 26/50 is 0.52; 0.76 x 0.52 is about 0.40, which eliminates every other option before you finish the exact fraction.
Where candidates lose it
The fast wrong answer is 37.5%, from treating the draws as if cards go back in the deck. It feels close enough, and in a multiple choice round it sits right next to the correct option on purpose.
The other slip is starting with 13/52 for the first card, which silently fixes that card's suit. You then need to multiply by 4 for the suit choice, and under time pressure most people forget.
What the interviewer asks next
What is the probability that four cards are all of different suits?
What is the probability that three cards share a suit?
Draw until you have seen all four suits. What is the expected number of cards?
004You keep drawing independent random numbers, each uniform on 0 to 1, until their running total exceeds 1. What is the expected number of draws?Continuous and geometric probabilityHardCitadel SecuritiesChicago · 2025
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What is your instinct for the answer?
Show the worked solution
e, about 2.718. The chance that n uniforms still add to at most 1 is 1/n!, the volume of a corner of the n-dimensional cube. The number of draws N exceeds n exactly when that happens, and an expected count is the sum of the chances of exceeding each n. So E[N] is 1 + 1 + 1/2 + 1/6 + 1/24 and so on, which is e.
Why is two draws the wrong answer?
Fill a one litre jug with cups of random size, each somewhere between empty and full. On average two cups make a litre, but you stop at the first cup that overflows, and some pairs of cups fall short. Averages of the draws do not tell you the average stopping time: you need the chance that you are still short after each draw. After two draws you are still at or below 1 exactly half the time, so a third draw is needed often, and occasionally a fourth.
The chance of still being at or below 1 after n draws is 1/n!, so the bars run 1, 1, 1/2, 1/6, 1/24, and their running sum, the expected number of draws, closes in on e, about 2.718.
Where does 1/n! come from?
For two draws, the pairs with a total at or below 1 fill the triangle under the line x + y = 1 in the unit square, area 1/2. For three, they fill a corner of the unit cube, volume 1/6. In general the region where n uniforms add to at most 1 is a corner of the n-dimensional cube with volume 1/n!, because the n! orderings of the coordinates carve the cube into equal pieces. You can also build it by convolutionThe density of a sum of independent variables, found by combining every way the parts can add up to the same total.: the density of the sum below 1 is s to the power n-1 over (n-1)!, and integrating from 0 to 1 gives 1/n!.
The relationship
$E[N] = \sum_{n\ge 0} P(N > n) = \sum_{n\ge 0} P(U_1+\dots+U_n \le 1) = \sum_{n\ge 0}\frac{1}{n!} = e$
N the number of draws needed
P(N > n) the chance that n draws were not enough
U_i the uniform draws
What it says in wordsThe expected count equals the sum over n of the chance that n draws were still not enough, and those chances are 1/n!.
How do you check an answer this surprising?
Check the pieces. N is at least 2 always, since one draw never exceeds 1, so the answer must be above 2; the bars for n = 0 and n = 1 are both 1 for that reason. A simulation of 200,000 runs gives an average of 2.721 draws, within a whisker of 2.718. Saying that you would simulate it, and roughly what you expect to see, is a good close in a research interview.
Where candidates lose it
The instinctive answer is 2, because two draws average exactly 1. It confuses the average of the draws with the average stopping time, and it ignores that the stopping rule waits for the total to pass 1, not reach it on average.
The second loss is knowing the answer is e without being able to say why. The tail-sum formula for an expected count, plus the 1/n! volume, is the whole argument, and it takes three sentences.
What the interviewer asks next
What is the expected number of draws to exceed 2?
What is the expected value of the total at the moment it first exceeds 1?
What is the probability that exactly two draws are needed?
Asked at Citadel Securities, Quant Research Interview, Chicago, 2025 (Wall Street Oasis): He was asking some questions about the probability, especially on the convolution.
005Construct two random variables that are uncorrelated but clearly dependent, and show that their covariance is zero.Correlation, regression and linear algebraWarm upTwo SigmaNew York · 2025
Try it first
Which pair works?
Show the worked solution
Take X equal to -1, 0 or 1 with probability 1/3 each, and Y = X squared. Y is fixed by X, so they are as dependent as variables can be. But E[X] = 0 and E[XY] = E[X cubed] = (-1 + 0 + 1)/3 = 0, so the covariance E[XY] - E[X]E[Y] is zero. Correlation only measures straight-line association, and this relationship is a V.
What does correlation actually measure?
Think of a thermostat that runs the air conditioner hard on very hot days and the heater hard on very cold days. Energy use is clearly driven by temperature, but a straight line through the data is flat: high use at both ends, low in the middle. Correlation measures only how well a straight line summarises the relationship, so any symmetric U or V shape can have zero correlation while being completely determined. Independence is the stronger claim that knowing X tells you nothing about Y at all.
With X symmetric about zero and Y equal to X squared, Y is fixed exactly by X, yet the best straight line through the points is flat, so the covariance and the correlation are both zero.
How do you show the covariance is zero in one line?
Write the definition and let symmetry do the work. Cov(X, Y) = E[XY] - E[X]E[Y], and with Y = X squared the first term is E[X cubed], which is zero for any X symmetric about zero; the second term has E[X] = 0 in it. With the three-point version you can even list the products: -1 x 1, 0 x 0 and 1 x 1 add to zero. Yet P(Y = 0 given X = 0) is 1 while P(Y = 0) is 1/3, which is dependence in plain sight.
The relationship
$\operatorname{Cov}(X, X^2) = E[X^3] - E[X]\,E[X^2] = 0 - 0\cdot\tfrac{2}{3} = 0 \qquad P(Y=0\mid X=0) = 1 \ne P(Y=0) = \tfrac{1}{3}$
E[X^3] zero because the values of X are symmetric about zero
P(Y = 0 | X = 0) knowing X changes the odds on Y, so they are dependent
What it says in wordsThe covariance cancels by symmetry, while a single conditional probability proves the dependence.
Why does a quant interviewer care?
Because models quietly substitute zero correlation for no relationship. A delta-hedged option book gains or loses roughly with the square of the underlying's move, so its daily P and L can show near-zero correlation with the market while being entirely driven by it. The same holds for a volatility strategy or any payoff with a kink. The one case where zero correlation does mean independence is when the pair is jointly normal, which is worth adding before the interviewer asks.
Where candidates lose it
Candidates reach for two independent variables, which are uncorrelated but not dependent, or for X and -X, which are dependent but perfectly correlated. Both show the definitions are fuzzy.
The quieter trap is choosing X uniform on 0 to 1 and Y = X squared. Without symmetry about zero the covariance is positive, 1/12, and the example fails. Centre X first.
What the interviewer asks next
When does zero correlation imply independence?
Give an example with zero correlation where Y is not a function of X.
If you regress Y on X in the example, what do the fitted line and R squared look like?
Asked at Two Sigma, Generalist, New York, 2025 (Wall Street Oasis): Come up with two uncorrelated but dependent variables.
006A ticket pays Rs 1 if at least one six appears when three fair dice are rolled, and nothing otherwise. What is the fair price of the ticket?Market making, betting and sizingWarm upAkuna CapitalChicago · 2026
Try it first
Your price, to the nearest paisa band?
Show the worked solution
91/216 of a rupee, about 42 paise. A fair price for a ticket paying Rs 1 is the probability of winning. The fastest route is the complement: the chance of no six on three dice is 5/6 x 5/6 x 5/6 = 125/216, so the chance of at least one six is 1 - 125/216 = 91/216, or 0.421. Adding 1/6 three times gives 50 paise and overcounts.
Why is the price just a probability?
If a raffle pays Rs 100 and you win one time in four, playing many times earns you Rs 25 a ticket on average, so Rs 25 is the break-even price. A ticket paying Rs 1 on some event is worth exactly the probability of that event, because that is its average payout. Trading firms phrase probability questions as prices on purpose: it makes you answer in the units a desk uses, and it sets up the next question, which is where you would quote a bid and an offer.
Of the 216 equally likely rolls of three dice, 125 contain no six, so 91 contain at least one and the ticket's fair price is 91/216 of a rupee, about 42 paise, not the 50 paise that adding 1/6 three times suggests.
Why is at least one a signal to use the complement?
At least one six covers exactly one six, exactly two, or three, and each needs its own count. The opposite event, no six at all, is a single clean case: every die avoids six, and independent dice multiply. So the complement takes one line. Adding 1/6 + 1/6 + 1/6 fails because the three events overlap: a roll of 6, 6, 2 is counted once for the first die and again for the second. With ten dice the same mistake would give a probability above 1.
The relationship
$P(\text{at least one six}) = 1 - \left(\tfrac{5}{6}\right)^3 = 1 - \tfrac{125}{216} = \tfrac{91}{216} \approx 0.421$
(5/6)^3 the chance that each of the three dice avoids a six
91/216 the share of the 216 rolls with at least one six
What it says in wordsThe chance of at least one success is one minus the chance of none.
What does a trader add after the number?
A fair value is the centre of a market, not the market itself. A market maker quotes a bid below 42 paise and an offer above it, and the width depends on how confident they are in the number and how much risk one ticket adds to their book. Here the fair value is exact, so a tight market such as 40 bid, 44 offer is defensible. Saying that sentence turns a probability answer into a trading answer, which is what the question format is inviting.
Where candidates lose it
The fast wrong answer is 50 paise, from adding the chance of a six on each die. It is fast, it feels natural, and it ignores that rolls with two or three sixes get counted more than once.
The second loss is time. In an online assessment where each question has seconds, working exactly one, exactly two and exactly three sixes separately is correct and too slow. The complement is the habit being tested.
What the interviewer asks next
What is the fair price if the ticket pays Rs 1 for each six that appears?
How many dice do you need before at least one six is more likely than not?
Quote me a two-sided market on this ticket and tell me what you do if I lift your offer ten times.
Asked at Akuna Capital, Junior Trader Interview, Chicago, 2026 (Wall Street Oasis): if you win you get 1$. how much money would be a fair bet
007You roll a fair six-sided die six times. What is the expected number of different faces that appear?Expected value and optimal stoppingCoreQuant tradingProp trading firms
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Your estimate before any working?
Show the worked solution
About 3.99. Give each face an indicator that equals 1 if that face appears at least once. A given face is missed on all six rolls with probability (5/6) to the 6th, about 0.335, so it appears with probability 0.665. The expected count of distinct faces is the sum of the six indicators' expectations: 6 x 0.665 = 3.99. No case listing is needed.
Why not list the cases?
You could work out the chance of exactly one, two, up to six distinct faces and average them. It works, but it needs Stirling numbersCounts of the ways to split a set of items into a given number of non-empty groups; they appear when counting surjections. or a lot of careful counting, and it is easy to slip. Linearity of expectation lets you ignore how the faces interact: the expected total of several indicators is the sum of their expectations, whether or not they are independent. Here the six indicators are clearly dependent, since seeing many faces leaves fewer rolls for the others, and it does not matter at all.
Each of the six faces appears at least once with probability 66.5%, so the expected number of distinct faces is six times that, 3.99; the exact distribution peaks at four distinct faces and has all six only 1.5% of the time.
How does the indicator trick work step by step?
Think of a teacher counting how many of six friends turn up to a party. Instead of listing every guest list, she asks for each friend separately how likely that friend is to come, then adds. Write the count as I1 + I2 + ... + I6, where I_k is 1 if face k appears; take expectations; each E[I_k] is just the probability face k appears. Face k is missed on one roll with probability 5/6, on all six with (5/6) to the 6th, 0.335. So each indicator averages 0.665 and the total averages 3.99.
The relationship
$E[D] = \sum_{k=1}^{6} P(\text{face } k \text{ appears}) = 6\left(1 - \left(\tfrac{5}{6}\right)^6\right) \approx 6 \times 0.665 = 3.99$
D the number of distinct faces seen in six rolls
(5/6)^6 the chance a given face never appears in six rolls
What it says in wordsThe expected number of distinct faces is six times the chance that any one face appears.
Where does this pattern reappear?
The same shape answers how many distinct birthdays a group of n people has, how many of n hash buckets get used, and how many different stocks a random sample of trades touches. For n faces and n rolls, the expected share of faces seen is 1 - (1 - 1/n) to the n, which tends to 1 - 1/e, about 63.2%, as n grows. Six faces give 66.5%, already close. The exact enumeration of all 46,656 rolls gives a mean of 3.9906, the same number.
Where candidates lose it
The instinctive answer is 6, or something close to it, because six rolls over six faces feels like one of each. In reality repeats are the norm, and all six different faces happen in under 2% of runs.
The costlier trap is starting to enumerate cases under time pressure. Candidates who try to list exactly four distinct faces lose minutes. Say indicator variables and linearity in the first sentence.
What the interviewer asks next
What is the expected number of faces that appear exactly once?
How many rolls do you need, on average, to see all six faces?
What is the variance of the number of distinct faces?
008Z1 and Z2 are independent standard normal random variables. What are the mean and variance of Z1 squared + Z2 squared, and what is the probability that it exceeds 2?Statistics and estimationCoreQuant researchQuant trading
Try it first
What is P(Z1 squared + Z2 squared > 2)?
Show the worked solution
Mean 2, variance 4, and the probability of exceeding 2 is e to the -1, about 36.8%. Each Z squared has mean 1 and variance 2, so the sum has mean 2 and variance 4. The sum is a chi-squared with two degrees of freedom, which happens to be exactly an exponential with mean 2. Its tail beyond t is e to the -t/2, so beyond 2 it is e to the -1.
How do you get the mean and variance without the distribution?
E[Z squared] is the variance of Z, which is 1. For the variance of Z squared you need the fourth moment: E[Z to the 4] is 3 for a standard normal. So Var(Z squared) = 3 - 1 = 2, and because Z1 and Z2 are independent the variances add, giving a mean of 2 and a variance of 4. Say the fourth moment of 3 aloud; it is the number interviewers check you know, and it is why normal kurtosis is quoted as 3.
The sum of two squared standard normals has density one half times e to the minus x over 2, an exponential with mean 2 and variance 4, and the area beyond 2 is exactly e to the minus 1, about 36.8%.
Why is this particular sum exponential?
Think of a dart thrown at a board where both the horizontal and vertical errors are independent standard normals. The joint density depends only on the distance from the centre, so the dart's direction is uniform and all the information is in the radiusThe distance of the point (Z1, Z2) from the origin, the square root of Z1 squared plus Z2 squared.. Switching to polar coordinates, the chance that the squared distance exceeds t is e to the -t/2, which is the tail of an exponential with mean 2. At t = 2 the answer is e to the -1.
The relationship
$P(Z_1^2+Z_2^2 > t) = \int_0^{2\pi}\!\!\int_{\sqrt t}^{\infty} \frac{1}{2\pi} e^{-r^2/2}\, r\,dr\,d\theta = e^{-t/2} \qquad P(\,\cdot > 2) = e^{-1}$
r the distance of (Z1, Z2) from the origin
\frac{1}{2\pi} e^{-r^2/2} the joint density of two independent standard normals
e^{-t/2} the tail of an exponential distribution with mean 2
What it says in wordsIn polar coordinates the angle integrates out and the radius gives an exponential tail.
Why is 50% the tempting wrong answer?
Because 2 is the mean and people read the mean as the middle. For a right-skewed distribution the mean sits above the median, so less than half the mass lies beyond it; here the median is 2 ln 2, about 1.39. A simulation of 200,000 pairs gives a mean of 1.997, a variance of 3.96 and a tail share of 0.368, matching the exact results. The same polar trick is what powers the Box-Muller method for generating normal random numbers.
Where candidates lose it
The common wrong answer is 50%, from treating the mean as the median. Chi-squared variables are skewed to the right, and the skew is largest with few degrees of freedom.
The second trap is the variance. Candidates who say the variance of Z squared is 1 have confused it with the variance of Z. The fourth moment of 3 is the step, and missing it gives a variance of 2 for the sum instead of 4.
What the interviewer asks next
What is the distribution of the square root of Z1 squared + Z2 squared?
How would you use this to generate normal random numbers from uniforms?
What are the mean and variance of a chi-squared with k degrees of freedom?
009A stock trades at 100 and in one period will be either 120 or 80. Interest rates are zero. Price a call option struck at 100 by building a portfolio of shares and borrowing that copies it, and explain why the real-world probability of the up move does not appear in the price.Pricing, options and index mathsCoreOptions market makingQuant trading
Try it first
If you believe the stock goes up with probability 90%, what is the call worth?
Show the worked solution
The call is worth 10. It pays 20 if the stock goes to 120 and 0 at 80. Half a share pays 60 or 40, so half a share with a loan of 40 pays 20 or 0, exactly the call. That portfolio costs 50 - 40 = 10 today. If the call traded at any other price, you could buy the cheap one and sell the dear one for a riskless profit, so no probability is needed.
How do you build the copy?
Match the swing first. The call's payoff moves by 20 between the two states while the stock moves by 40, so the copy needs 20/40 = 0.5 of a share: that ratio is the option's deltaHow much an option's value changes for a one-unit change in the underlying price; here, the number of shares that copies the option.. Half a share is worth 60 or 40 at the end, which is 40 more than the call in both states. Borrow 40 today, repay 40 at the end with zero interest, and the copy pays exactly 20 or 0.
A call struck at 100 on a stock that moves to 120 or 80 is copied by half a share and a loan of 40, which pays 20 or 0 exactly as the call does and costs 10 today, so the call is worth 10 with no probability used.
Why does the chance of the up move not matter?
Think of a shop selling a bundle of two items that you can also buy separately. The bundle's price is pinned by the parts, whatever you think about how useful the items are. The call is a bundle of half a share and a loan; the share price already reflects everyone's views about the up move, so the option inherits them and adds none of its own. A 90% view is a reason to hold the stock itself, not a reason to pay more for the call than its parts cost.
The relationship
$\Delta = \frac{C_u - C_d}{S_u - S_d} = \frac{20-0}{120-80} = 0.5 \qquad C_0 = \Delta S_0 - B = 50 - 40 = 10 = q\,C_u + (1-q)\,C_d,\; q = \frac{S_0 - S_d}{S_u - S_d} = 0.5$
\Delta shares held in the copy
B the amount borrowed, 0.5 x 80 - 0 = 40
q the risk-neutral weight on the up state, fixed by the prices, not by beliefs
What it says in wordsThe copy's cost gives the price, and the same price is an average of the payoffs using weights set by today's stock price.
What would you do if the call traded at 12?
Sell the dear thing and buy the cheap one. Sell the call for 12, buy half a share for 50 and borrow 40, a net cash inflow of 2 today; at the end the portfolio pays exactly what you owe on the call in either state. The 2 is kept whatever happens. The weight q = 0.5 that reproduces the price is called the risk-neutral probability, but it is a pricing weight backed out of the stock price, not a forecast. Say that distinction; interviewers listen for it.
Where candidates lose it
The trap is pricing the call as an expected payoff under your own view: 90% of 20 is 18. That price can be arbitraged against the stock, so nobody could trade it for long, and the interviewer wants to hear that the copying portfolio pins the price.
The second loss is getting 10 by assuming a 50% chance. The number is right by coincidence of the symmetric tree; ask yourself what happens with an up move to 130, and the risk-neutral weight changes to 1/2.5 = 0.4.
What the interviewer asks next
Price the put struck at 100 and check put-call parity.
What changes if interest rates are 5% for the period?
The stock can go to 130 or 80 instead. Price the call again.
010A company's value to its current owner is equally likely to be anything from Rs 0 to Rs 100 crore, and only the owner knows the figure. In your hands the company would be worth 1.5 times that value. You may make one take-it-or-leave-it offer, which the owner accepts only if it is at least the company's value to them. What should you bid?Games and strategic reasoningHardQuant tradingQuant research
Try it first
Which bid maximises your expected profit?
Show the worked solution
Bid nothing. If a bid of b is accepted, the owner has told you the company is worth less than b to them, so its value is uniform on 0 to b and averages b/2. In your hands that is 1.5 x b/2 = 0.75b, a quarter less than you paid. Expected profit is (b/100) x (0.75b - b) = -b squared/400, negative for every positive bid. This is the winner's curse in its purest form.
Why does 75 look right and fail?
Picture buying a used car from someone who knows its history while you do not. If they agree to your price at once, that is itself news: sellers of good cars refuse low offers. Acceptance is not random; it happens exactly in the states where the company is worth less than you offered, so the average value you actually receive is the average below your bid, not the average overall. The naive 75 uses the unconditional average of 50 and forgets that you only trade when the owner is happy to sell.
The naive line values the company at its overall average and shows profit for any bid under 75, but conditioning on the owner accepting gives expected profit of minus b squared over 400, which is below zero for every positive bid, minus 6.25 crore at a bid of 50.
How do you set up the expected profit?
Split it into the chance of a deal and the profit given a deal. A bid of b is accepted with probability b/100; given acceptance the owner's value is uniform on 0 to b, averaging b/2, so your value averages 0.75b and your profit averages minus 0.25b. Multiply: minus 0.25b x b/100, which is minus b squared over 400. At a bid of 50 that is minus 6.25 crore: you win half the time and lose 12.5 crore on average when you do.
The relationship
$E[\pi(b)] = \underbrace{\frac{b}{100}}_{\text{accepted}}\;\Big(1.5\cdot\frac{b}{2} - b\Big) = -\frac{b^2}{400} < 0 \text{ for all } b > 0$
b your bid in Rs crore
b/100 the chance the owner's value is below b
b/2 the owner's average value, given that they accepted
What it says in wordsThe chance of winning times the loss when you win is negative for every positive bid.
When would bidding make sense, and where does this show up on a desk?
The multiplier is the lever. With a multiplier m, the profit given a deal is (m/2 - 1)b, so bidding pays only if you add more than double the owner's value; at exactly 2 you break even, and above 2 you should bid the full 100. On a trading desk the same logic is called adverse selectionThe tendency for the trades you actually get to come from counterparties with better information than you, so they are worse on average than a random trade.: the orders that fill against you are disproportionately the ones from people who know more. A quote that looks profitable against the average counterparty loses against the ones who choose to trade.
Where candidates lose it
Most candidates bid somewhere between 50 and 75, reasoning from the unconditional average value. That ignores the information in the owner's acceptance, which is the entire point of the question.
The second loss is a partial fix: realising acceptance is informative but then bidding a little lower, such as 60, to leave a margin. Any positive bid loses here. Write the expected profit as a function of b and let the algebra say zero.
What the interviewer asks next
What multiplier would make you willing to bid, and how much would you then bid?
What if the owner's value is uniform on 50 to 100 instead?
How does this relate to a market maker who gets filled on their quotes?

a, b	the two fastest times, here 1 and 2
c, d	the two slowest times, here 5 and 10

R, D	rain and dry
YYY	all three friends say yes
(2/3)^3 and (1/3)^3	the chance of three yeses when it rains, and when it is dry

39/51	cards of a new suit among those left after one draw
26/50	cards of a third suit after two draws
\binom{52}{3}	the number of possible three-card hands

N	the number of draws needed
P(N > n)	the chance that n draws were not enough
U_i	the uniform draws

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