Hedge Funds puzzles, solved step by step

Puzzles: 100
Traced to a firm: 38
Topics: 14
Hard: 30

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083A binary contract pays Rs 100 if the index closes higher tomorrow and nothing otherwise. The market is 55 bid, 60 offered, and your model says the chance of an up close is 50%. What do you do, and what is your edge?Market making and trading gamesCoreProp and quant trading firmsVolatility and relative value funds
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What is the trade?
Show the worked solution
Sell at the 55 bid; the edge is Rs 5 a contract, five points of probability. A contract paying Rs 100 on an up close is worth Rs 100 times the probability, so your model values it at Rs 50. Selling at 55 collects Rs 55 for a liability worth Rs 50 on average. Buying at the 60 offer would give away Rs 10 of expected value. The edge is only as good as the model behind it.
Why is a binary price just a probability?
Think of a bet with a friend: you pay a fixed sum now and get Rs 100 back if it rains tomorrow. If you think rain is a 30% chance, the most you would pay is Rs 30. A contract paying Rs 100 or nothing is worth 100 times the probability of the payout, so a price of 55 is the market saying 55%. Read as probabilities, the quote says between 55% and 60%, and your model says 50%.
With your model value at 50, a bid of 55 and an offer of 60, selling at the bid earns 5 over fair value and buying at the offer loses 10; selling 100 contracts at 55 receives Rs 5,500 against an expected payout of Rs 5,000, an expected profit of Rs 500.
Which side of the quote can you trade?
You buy at the offer and sell at the bid, never the other way round. Buying costs 60 for something your model values at 50, a loss of 10 in expectation, while selling receives 55 for the same thing, a gain of 5. So the trade is to sell at 55. Sell 100 contracts and you receive Rs 5,500; half the time you pay out Rs 10,000 and half the time nothing, so the expected payout is Rs 5,000 and the expected profit Rs 500.
The relationship
$\text{edge} = \text{bid} - 100 \times p = 55 - 100 \times 0.50 = \text{Rs } 5$
bid the price at which you can sell, 55
p your model's probability of an up close, 0.50
100 the payout if the index closes higher
What it says in wordsYour edge on each contract sold is what the market pays you minus what the contract is worth on your numbers.
When would you do nothing?
If your model said 57%, the fair value of 57 would sit inside the quote: selling at 55 loses 2 and buying at 60 loses 3. An edge exists only when your value falls outside the bid and offer, not whenever you disagree with the middle of the market. Then add the honest caveat. A market pricing an up close near 57.5% may know something your model does not, and each contract swings between plus 55 and minus 45, so a 5-point edge needs many independent trades, and a model you trust, before it shows up in the P&L.
Where candidates lose it
The fast wrong answer is to buy because the market is above 50 and so seems to think up is likely. That reads the quote as a forecast to follow rather than a price to trade against, and the candidate who buys at 60 pays 10 more than the contract is worth on their own numbers.
The second loss is trading the wrong side of the quote, selling at 60 or buying at 55. Say which side you would hit before you give the edge.
What the interviewer asks next
Your model says 57%. What do you do?
You are asked to make a two-sided market around your 50%. Where do you quote, and why not 49 to 51?
The contract instead pays Rs 100 for every point the index rises. How does the pricing change?
084You own a stock bought at Rs 500 and sell a call struck at Rs 550 for a premium of Rs 12. At expiry, what is your maximum profit and where is your breakeven?Options and payoffsWarm upVolatility and relative value fundsProp and quant trading firms
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What is the most you can make?
Show the worked solution
Maximum profit is Rs 62 and the breakeven is Rs 488. Above the Rs 550 strike the stock is called away, so you keep the Rs 50 rise from 500 to 550 plus the Rs 12 premium. Below 550 the call expires worthless and you keep the premium, which cushions the first Rs 12 of any fall. You lose money only below 500 minus 12, which is Rs 488.
What have you actually sold?
Think of renting out a flat you own with an agreement that the tenant may buy it at a fixed price within the year. You collect rent now, but if flat prices soar, the tenant buys at the agreed price and the extra gain is theirs. A covered call swaps the upside above the strike for cash today: the premium is the rent, and the strike is the agreed sale price.
The covered call earns Rs 12 more than the plain stock at every price up to Rs 550, is capped at a profit of Rs 62 from Rs 550 upwards, breaks even at Rs 488, and falls behind the plain stock above Rs 562.
How do you find the two numbers quickly?
Take the two regions separately. At or above Rs 550 the position is worth 550 plus the 12 kept, against 500 paid, so Rs 62 whatever the stock does. Below 550 the call is worthless and the position is just the stock plus 12, so it loses money only once the stock falls more than 12 below the purchase price, at Rs 488. At a stock price of 520 you make 32: 20 on the stock and 12 of premium. At 450 you lose 38 instead of 50.
The relationship
$\text{profit} = \min(S_T, K) - S_0 + c \qquad \max = 550 - 500 + 12 = 62 \qquad \text{breakeven} = 500 - 12 = 488$
S_T the stock price at expiry
K the strike of the call sold, Rs 550
S_0 the price paid for the stock, Rs 500
c the premium received, Rs 12
What it says in wordsYou keep the stock's value up to the strike, plus the premium, minus what you paid for the stock.
What is the trade-off in plain terms?
The premium improves every outcome below Rs 562 and worsens every outcome above it. Above Rs 562 the plain stock position beats the covered call, because the gain you gave away exceeds the premium you took in. On the downside the cushion is thin: the stock can fall all the way from 500 and the premium covers only 12 of it. That is the limitation to state: a covered call is income with a cap, not protection.
Where candidates lose it
The common slip is quoting the premium, Rs 12, as the maximum profit, which forgets that you still own the stock and keep its rise up to the strike. The opposite slip is saying unlimited, which forgets the call you sold.
For the breakeven, candidates often add the premium to the purchase price and say Rs 512. The premium is money received, so it lowers the breakeven, to Rs 488.
What the interviewer asks next
At what stock price are the covered call and the plain stock worth the same at expiry?
Why might the same call fetch more than Rs 12 just before a results announcement?
Which position gives the same payoff at expiry as a covered call without owning the stock?
086A stock ticks up or down by Rs 1 each minute with equal probability, starting at Rs 50. On average, how many minutes pass before it first touches Rs 45 or Rs 55?Random walks and Markov chainsCoreQuant and systematic fundsProp and quant trading firms
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Pick your answer before setting up any equation.
Show the worked solution
25 minutes. Let E(k) be the expected minutes to exit from price k. Each minute costs one and moves the price up or down with equal chance, so E(k) = 1 + half E(k + 1) + half E(k - 1), with E(45) = E(55) = 0. The solution is E(k) = (k - 45)(55 - k), the product of the distances to the two barriers. From Rs 50 that is 5 x 5 = 25.
Why is the answer not 5 minutes?
Think of someone pacing a corridor, taking one step forward or back on each coin toss. After 25 steps they are not 25 steps away; typically they are about 5 away, because the steps keep undoing each other. A fair random walk covers distance like the square root of time, so reaching a barrier 5 away takes on the order of 5 squared, 25 steps, not 5.
Three sample paths from Rs 50 leave the Rs 45 to Rs 55 channel after 11, 23 and 45 minutes; averaged over all paths the exit takes (50 - 45) x (55 - 50) = 25 minutes, and from Rs 48 or Rs 46 it takes 21 or 9.
How do you get exactly 25?
Set up the one-step equation. From any price k strictly between the barriers, you spend one minute and then stand at k + 1 or k - 1 with equal chance. E(k) = 1 + half E(k + 1) + half E(k - 1) says the second difference of E is always minus 2, so E is a downward parabola that is zero at both barriers. The only such parabola is (k - 45)(55 - k). Check a point: from Rs 46 it gives 1 x 9 = 9 minutes, and it passes the one-step test, since 1 plus half of E(47), which is 16, plus half of E(45), which is 0, is 9.
The relationship
$E(k) = (k - a)(b - k) \qquad E(50) = (50 - 45)(55 - 50) = 25$
a, b the lower and upper barriers, Rs 45 and Rs 55
k the starting price
E(k) the expected number of one-minute steps before either barrier is touched
What it says in wordsFor a fair walk, the expected time to leave a channel is the distance to the floor times the distance to the ceiling.
What does the shape tell a trader?
Starting in the middle is the slowest place to be, and an off-centre start is much faster: from Rs 48 the answer is 3 x 7 = 21 minutes, and from Rs 46 only 9. Doubling both distances quadruples the expected time: barriers at Rs 40 and Rs 60 give 100 minutes. That is the arithmetic behind why widening a stop and a profit target together makes a trade live much longer. It holds only for a fair walk: if the stock ticks up 60% of the time, the exit comes sooner, about 19.2 minutes, and mostly at the top.
Where candidates lose it
The quick wrong answer is 5 minutes, which treats the walk as if it moved steadily towards one barrier. A fair walk wanders, and the interviewer is checking whether you know that distance grows with the square root of time.
The second loss is reaching 25 from the square-root intuition without being able to show it. Write the one-step equation and the parabola; that is what turns a good guess into an answer.
What the interviewer asks next
What is the probability the stock touches Rs 55 before Rs 45?
From Rs 50, the barriers move to Rs 40 and Rs 55. What is the expected time now?
The stock ticks up with probability 0.6. Why does the expected time fall?
088A stock's true model is: stock return = 0.5 x market return + 1.0 x sector return + noise. Regressing the sector's return on the market gives a slope of 0.4. If you regress the stock on the market alone, what slope do you get?Statistics and estimationCoreQuant and systematic funds
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What does the market-only regression report?
Show the worked solution
About 0.9. The market reaches the stock by two paths: directly, with a coefficient of 0.5, and through the sector, which moves 0.4 for each unit of market and passes all of it on with a coefficient of 1.0. A regression on the market alone cannot separate the two and reports the total, 0.5 + 1.0 x 0.4 = 0.9. The extra 0.4 is omitted variable bias.
Why does leaving the sector out change the market slope?
Suppose you measure how much ice cream sales rise on hot days, but hot days also tend to be holidays, and holidays sell ice cream too. Leave holidays out and the heat gets the credit for both. A regression gives a left-out variable's effect to whichever included variable moves with it, in proportion to how strongly the two move together. Here the sector moves with the market, so the market's slope absorbs part of the sector's effect.
The market reaches the stock directly with a coefficient of 0.5 and through the sector with 0.4 x 1.0 = 0.4, so a regression of the stock on the market alone reports 0.9, of which 0.4 is the sector's effect credited to the market.
How do you compute the bias?
Write the sector as 0.4 x market plus a part unrelated to the market, then substitute. Stock = 0.5 x market + 1.0 x (0.4 x market + other) + noise = 0.9 x market + (1.0 x other + noise). The bracket is unrelated to the market, so a regression on the market alone recovers 0.9. The bias is the omitted coefficient times the slope of the omitted variable on the included one, 1.0 x 0.4. A simulation of 20,000 days with these coefficients gives a slope of 0.897, matching the algebra.
The relationship
$\hat\beta_{\text{short}} = \beta_M + \beta_S\,\delta = 0.5 + 1.0 \times 0.4 = 0.9$
beta_M the stock's true direct loading on the market, 0.5
beta_S the stock's loading on the sector that was left out, 1.0
delta the slope of the sector's return on the market's, 0.4
What it says in wordsThe short regression's slope is the true slope plus the left-out variable's effect times how much that variable moves with the one you kept.
Is 0.9 wrong, or answering a different question?
It depends on what you use it for. If you want to hedge the stock with the market alone, 0.9 is the right hedge ratio, because it captures everything the market drags along with it. If you want the stock's exposure holding the sector fixed, say to build a sector-neutral book, 0.9 overstates it and 0.5 is the number you need. The bias can also run the other way: if the sector moved against the market, or the stock loaded negatively on the sector, the short slope would sit below 0.5. Naming both uses is what the interviewer is listening for.
Where candidates lose it
The fast wrong answer is 0.5: candidates assume a regression recovers the true coefficient whatever else is left out. It does so only when the omitted variable is unrelated to the included one.
The second loss is getting 0.9 and calling it simply wrong. It is the correct total effect of the market and the right number for a market-only hedge; it is wrong only as an estimate of the direct effect.
What the interviewer asks next
What slope do you get if the sector's slope on the market is minus 0.4?
You add the sector to the regression. What happens to the standard error of the market coefficient if the two are highly correlated?
How does this bias show up when you estimate a stock's factor exposures with too few factors?
090Each of your analysts calls the direction of a stock correctly 70% of the time, independently of the others, and your prior is 50/50. Two analysts disagree. What is your probability now that the stock goes up? What if a third analyst then sides with the one who said up?Conditional probability and BayesHardQuant and systematic fundsProp and quant trading firms
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Where are you after the disagreement, and after the third call?
Show the worked solution
50% after the disagreement, and 70% once the third analyst sides with up. Work in odds. Each analyst's call multiplies the odds by 0.7/0.3 = 7/3 in the direction called. One up and one down multiply by 7/3 and 3/7, which cancel, leaving the prior of 1:1. The third call multiplies by 7/3 again: odds of 7:3, a probability of 70%. A two-to-one split is worth one analyst, not two thirds.
Why switch from probabilities to odds?
Think of two friends who read the weather equally well and disagree about rain: you are back where you started, however good they are. With independent signals, each one multiplies your odds by its likelihood ratioHow much more likely a piece of evidence is if the claim is true than if it is false., so in odds form Bayes' rule is just multiplication. An analyst who is right 70% of the time says up with chance 0.7 if the stock will rise and 0.3 if it will fall, so an up call multiplies the odds by 7/3 and a down call by 3/7.
Starting from 50%, analyst A's up call moves the chance of up to 70%, analyst B's down call returns it to 50%, and analyst C's up call moves it to 70% again, not to the 67% a vote count suggests; three up calls and none down would give 92.7%.
How do the three calls combine?
Start at 1:1. The first analyst says up: 7:3, or 70%. The second says down: 7:3 times 3:7 is 1:1, back to 50%. The third says up: 7:3 again, 70%. Only the net count of calls matters, not the total. Three for up and none against would give 343:27, about 92.7%, which shows how much the one dissenter costs.
The relationship
$\text{odds}_{\text{post}} = 1 \times \tfrac{7}{3} \times \tfrac{3}{7} \times \tfrac{7}{3} = \tfrac{7}{3} \quad\Rightarrow\quad p = \frac{7}{7+3} = 70\%$
1 the prior odds of up against down, 50/50
7/3 the likelihood ratio of an up call from a 70% accurate analyst
3/7 the likelihood ratio of a down call
What it says in wordsMultiply the prior odds by one likelihood ratio per independent call, then turn the odds back into a probability.
What assumption carries all of this?
Independence. If the analysts read the same research and speak to the same management teams, their errors are correlated, and a second agreeing call adds much less than a factor of 7/3. In the extreme where the second analyst simply copies the first, it adds nothing at all. The rule also assumes each analyst is right 70% of the time whichever way the stock moves; an analyst who calls up too often tells you more when calling down. State independence as the assumption, then say how you would test it: by checking how often the analysts' past calls agreed with each other.
Where candidates lose it
The vote-counting answer, two out of three so 67%, is the usual loss. It treats a majority as a probability, when a two-to-one split carries exactly one net call of evidence, 70%.
The second loss comes after the disagreement: candidates reach for something like 58%, feeling that two good analysts must add something. With equal accuracy and opposite calls, the evidence cancels exactly.
What the interviewer asks next
One analyst is right 80% of the time and the other 60%. The better one says up, the other down. Where are you?
Five analysts split three to two. What is your probability?
How would you estimate how correlated your analysts' calls are, and how would you adjust for it?
091Cards are turned over one at a time from a well-shuffled 52-card deck. On average, at what position does the first ace appear?Counting and combinatoricsCoreQuant and systematic fundsProp and quant trading firms
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What is the expected position of the first ace?
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Position 10.6. The four aces split the 48 non-aces into five gaps: before the first ace, between each pair of aces, and after the last. By symmetry, a given non-ace is equally likely to fall in any of the five gaps, so each gap holds 48/5 = 9.6 cards on average. The first ace comes straight after the first gap, at 9.6 + 1 = 10.6.
Why do the gaps have the same size on average?
Picture four red beads threaded at random among 48 white ones on a string. Pick any one white bead: it is as likely to sit before all four reds as between the first and second, or after the last, because a shuffle favours no order. Each non-ace lands in each of the five gaps with probability 1/5, so the expected size of every gap is 48/5 = 9.6 cards. Any single shuffle has uneven gaps; it is only the average that is equal.
One shuffle leaves uneven gaps of 4, 5, 14, 11, 14 non-aces around its four aces, but averaged over all shuffles every one of the five gaps holds 9.6 cards, so the aces sit at 10.6, 21.2, 31.8, 42.4.
How do you turn gap sizes into a position?
The first ace sits immediately after the first gap. Expected position = expected size of the first gap + 1 = 9.6 + 1 = 10.6. The same logic puts the second ace at 21.2, the third at 31.8 and the last at 42.4, evenly spaced 10.6 apart. A quick check: the last ace at 42.4 plus a final gap of 9.6 reaches exactly 52.
The relationship
$E[\text{first ace}] = \frac{n - k}{k + 1} + 1 = \frac{n + 1}{k + 1} = \frac{53}{5} = 10.6$
n cards in the deck, 52
k special cards, here the 4 aces
(n - k)/(k + 1) the expected number of ordinary cards in each of the k + 1 gaps
What it says in wordsWith n cards and k special ones, the first special card is expected at position (n + 1) divided by (k + 1).
Why is 13 the wrong instinct?
Thirteen is the average wait when every draw has a fresh 1-in-13 chance, as if each card were put back and the deck reshuffled. Drawing without replacement makes the aces easier to reach, because every non-ace you turn over raises the share of aces in what is left. The general rule, (n + 1)/(k + 1), is worth keeping: the first spade is expected at 53/14, about card 3.79, and the same shape answers how many trades you expect to check before finding the first of a handful of booking errors in a batch.
Where candidates lose it
The fast wrong answer is 13, borrowed from a wait with replacement. The interviewer wants to see you notice that the deck is finite, so the first ace must appear by card 49 at the latest and is pulled earlier than a with-replacement wait would suggest.
The second loss is starting the direct sum, position times the chance the first ace is there, which is correct but slow and easy to get wrong at a whiteboard. The gap argument gets there in one line.
What the interviewer asks next
What is the expected position of the last ace?
What is the expected number of cards between the first and second ace?
How many cards do you expect to turn before the first spade?
092A casino offers the St Petersburg game: a fair coin is tossed until the first tail, and if that takes n tosses you are paid Rs 2 to the power n. The casino can pay out at most Rs 1 crore. What is a fair price to play?Expected value and dice gamesHardQuant and systematic fundsProp and quant trading firms
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Roughly what is the capped game worth?
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About Rs 24.19. Round n pays Rs 2^n with probability 1/2^n, so each round adds exactly Rs 1 of expected value. That holds up to n = 23, since 2^23 is about Rs 84 lakh and 2^24 is above the Rs 1 crore cap. From round 24 on, the payout is stuck at Rs 1 crore, with total probability 1/2^23, which adds about Rs 1.19. So 23 + 1.19, about Rs 24.
Why is the uncapped game worth an infinite amount?
Each round is a doubling bet: the payout doubles while the chance of reaching it halves. Think of a raffle where a ticket twice as valuable is half as likely to win: every prize tier is worth the same to you. Round n pays 2^n with probability 1/2^n, so every round contributes exactly Rs 1 to the expected value, and there are infinitely many rounds. That is the famous paradox: few people would pay even Rs 100, yet the expected value has no bound. The resolution that matters on a desk is not psychology; it is that nobody can pay out an unlimited amount.
Rounds 1 to 23 each add exactly Rs 1 to the expected value, and once the Rs 1 crore cap binds the later rounds add 0.60, 0.30, 0.15 and so on, Rs 1.19 in all, so the capped game is worth about Rs 24.19.
How does the cap change the sum?
Find the round where the cap starts to bind. 2^23 is Rs 83,88,608, under Rs 1 crore; 2^24 is Rs 1,67,77,216, over it. So rounds 1 to 23 each add Rs 1, and every round from 24 on pays the capped Rs 1 crore, which together happen with probability 1/2^23 and add 1,00,00,000 / 83,88,608, about Rs 1.19. The fair price is about Rs 24.19. Nearly all of the textbook infinity lives in outcomes the casino cannot pay.
The relationship
$E = \sum_{n=1}^{23} \frac{2^n}{2^n} + \frac{10^7}{2^{23}} = 23 + 1.19 = 24.19$
2^n the payout if the first tail arrives on toss n
1/2^n the chance the first tail arrives on toss n
10^7 / 2^23 the capped Rs 1 crore times the chance of reaching round 24 or later
What it says in wordsEvery uncapped round is worth one rupee; the capped tail is worth the cap times the chance of getting that far.
What does a bigger casino buy you?
Very little. The value grows only with the logarithm of the cap: each doubling of the casino's bankroll adds about Rs 1. A cap of Rs 1,000 crore, a thousand times larger, lifts the fair price only to about Rs 34. That is the lesson a risk manager takes away: a payoff whose expected value rests on rare, enormous outcomes is worth what the other side can actually pay, and any estimate built on the tail should be checked against who stands behind it.
Where candidates lose it
Answering infinity is the trap for anyone who knows the textbook game. The interviewer added the cap to see whether you can find where it binds and redo the sum, not recite the paradox.
The other loss is dropping the tail beyond the cap and saying Rs 23, or valuing it crudely at the full Rs 1 crore times a guessed chance. The tail is a clean sum: probability 1/2^23 of receiving Rs 1 crore.
What the interviewer asks next
The cap rises to Rs 1,000 crore. What is the fair price now?
With logarithmic utility and wealth of Rs 1 lakh, roughly what would you pay for the uncapped game?
How is a book short deep out-of-the-money options like the casino in this game?
093Four people must cross a narrow bridge at night with one torch. They take 1, 2, 5 and 10 minutes to cross, at most two can cross at a time, a pair moves at the slower person's pace, and the torch must be carried on every crossing. What is the fastest time for all four to get across?Logic and brainteasersCoreProp and quant trading firmsLong-short equity funds
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What is the fastest crossing?
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17 minutes. 1 and 2 cross (2 minutes), 1 returns (1), 5 and 10 cross together (10), 2 returns (2), and 1 and 2 cross again (2). The total is 17. The obvious plan, with the fastest person escorting everyone, takes 19, because the 5 and the 10 each cost a separate crossing. Pairing the two slowest hides the 5 inside the 10.
Why is the obvious plan not the fastest?
The natural plan uses the quickest person as a shuttle: 1 walks each person over and comes back. That costs 2 + 1 + 5 + 1 + 10 = 19. Every slow person who crosses on a separate trip pays their own time in full, so the 5 and the 10 together cost 15 minutes of crossing. Think of sending two slow parcels in one courier van instead of two: the second rides along for free.
The shuttle plan spends 5 and 10 minutes on separate crossings and takes 19 minutes, while sending 5 and 10 together costs 10 minutes for both and brings the total to 17, even after the 2-minute person makes one extra return trip.
How does the 17-minute plan pay for pairing the slow two?
Pairing 5 and 10 costs 10 instead of 15, a saving of 5. But someone fast must already be waiting on the far side to bring the torch back afterwards. That setup costs one extra return by the 2-minute person and an extra crossing of the pair 1 and 2, so the net saving is 2 minutes: 19 down to 17. The general rule: pair the two slowest when twice the second-fastest time is less than the fastest plus the second-slowest, here 4 against 6.
The relationship
$\underbrace{2 + 1 + 10 + 2 + 2}_{\text{pair the slow two}} = 17 \qquad \underbrace{2 + 1 + 5 + 1 + 10}_{\text{shuttle}} = 19$
10 the one crossing that carries both the 5-minute and the 10-minute person
2 + 2 the extra cost of having the 2-minute person bring the torch back and cross again
What it says in wordsPairing the two slowest saves 5 minutes of crossing at a cost of 3 extra minutes of torch returns and re-crossing.
How do you show 17 is the minimum?
Five crossings are unavoidable: three over and two back. If the 5 and the 10 cross separately, the forward trips cost at least 10 + 5 + 2 and the two returns at least 1 each, which is 19; if they cross together, the person who returns next must already be across, which forces the returns to be the 1 and the 2, and the best total is 17. Talking through that bound is what separates a remembered answer from a reasoned one, and it is what lets you handle the follow-up with different times.
Where candidates lose it
19 minutes is the answer most people reach, and it comes so quickly that they stop there. The interviewer is waiting to see whether you ask what the 5-minute person costs and whether that cost can be hidden.
The second loss is finding 17 by trial and error without the reason. Name the saving, 5 minutes from pairing, and its price, the extra return and re-crossing; that is what makes the answer hold when the interviewer changes the times.
What the interviewer asks next
The times are 1, 5, 6 and 10. Does pairing the two slowest still help?
Add a fifth person who takes 20 minutes. What is the fastest crossing now?
State the general rule for when to pair the two slowest walkers.
094You buy an at-the-money option at 20% implied volatility and delta-hedge it. Over its life the stock realises 30% volatility. Where does your P&L come from, and does the direction of the moves matter?Options and payoffsHardVolatility and relative value fundsProp and quant trading firms
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What drives the hedged position's P&L?
Show the worked solution
From gamma: the hedged option earns on the size of moves, not their direction, and pays theta for the privilege. Each day the P&L is about half gamma times the squared move, minus theta. Theta is priced for 20% volatility, so a daily move of about 1.26% breaks even. At 30%, typical moves are 1.89%, and the squared gain is 2.25 times the decay. Up or down makes no difference; the path still does.
What is left after the delta hedge?
Think of a street vendor who sells umbrellas and sunglasses from the same cart: she no longer cares whether it rains or shines, only whether the weather changes enough to bring people out. A delta hedge sells enough stock to cancel the option's first-order bet on direction, so what remains is the curvature: the option gains a little more on the way up than the hedge loses, and loses a little less on the way down than the hedge gains. That curvature is gamma, and it pays on moves either way.
The option's value curves above the straight delta hedge line on both sides of Rs 1,000, so a move either way earns the shaded gap; a day at 30% realised volatility earns about Rs 0.71 of it per option against Rs 0.32 of time decay priced at 20%.
How big is the daily P&L in numbers?
Take a stock at Rs 1,000 and a three-month at-the-money call priced at 20% volatility, with zero interest rates. Its gamma is about 0.0040 per rupee and its time decay about Rs 0.32 a day over 252 trading days. A day's move of 1.89%, typical of 30% volatility, is Rs 18.9, and half of 0.0040 times 18.9 squared is a gain of Rs 0.71, against Rs 0.32 of decay: about Rs 0.40 a day per option. The decay is exactly what a move of 1.26%, typical of 20% volatility, would pay back.
The relationship
$\text{P\&L}_{\text{day}} \approx \tfrac{1}{2}\,\Gamma\,(\Delta S)^2 - \Theta = \tfrac{1}{2}\,\Gamma S^2\left(\sigma_r^2 - \sigma_i^2\right)\Delta t$
Gamma how fast the option's delta changes with the stock price
Delta S the day's move in the stock price
Theta the option's daily time decay
sigma_r, sigma_i realised and implied volatility, 30% and 20%
What it says in wordsEach day a hedged long option earns half its gamma times the squared move and pays its time decay; on average that is the gap between realised and implied variance.
Why does direction not matter, but the path does?
The move enters squared, so plus Rs 18.9 and minus Rs 18.9 earn the same. The P&L depends on realised volatility against implied, not on where the stock ends up. But gamma is largest near the strike and shrinks as the stock drifts away, so 30% realised in a trend that carries the stock far from the strike early earns less than 30% realised while the stock chops around the strike. Hedging frequency matters too: hedge rarely and the P&L becomes noisy, even if its average barely changes.
Where candidates lose it
The common loss is saying the option made money because the stock went up. With a delta hedge in place, direction has been sold away; a candidate who talks about direction has not understood what the hedge does.
The second loss is saying a hedged position has no P&L. It has exactly one exposure left, realised against implied volatility, and naming it is the whole answer.
What the interviewer asks next
Realised volatility comes in at 15% instead. What happens to the P&L, and why?
Why is the same realised volatility worth more while the stock stays near the strike?
You think implied volatility is too low but have no view on direction. What position expresses that?
096You roll a fair die repeatedly and keep a running total. What is the probability that the total is ever exactly 10? What does the answer approach for large targets?Random walks and Markov chainsHardQuant and systematic fundsProp and quant trading firms
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Roughly what is the chance the running total ever hits exactly 10?
Show the worked solution
About 0.289, and it settles at 2/7, about 0.286, for large targets. Let p(n) be the chance the total ever equals n. To hit n, the total must first land on one of n - 1 to n - 6 and then roll exactly the gap, each with chance 1/6, so p(n) is the average of the six values before it, with p(0) = 1. Working up gives p(10) = 0.2893. Totals advance 3.5 a roll on average, so they land on 1 number in 3.5.
How do you set up the recursion?
Think about the last roll before the total reaches n. The total can land exactly on n only by first landing on one of n - 1 down to n - 6 and then rolling exactly the gap, each with chance 1/6, so p(n) = (1/6)[p(n - 1) + ... + p(n - 6)]. Start with p(0) = 1, because you begin at zero, and p of any negative number = 0. Then p(1) = 1/6, p(2) = 7/36, p(3) = 0.227, and so on up to p(10) = 0.2893.
The chance of ever landing on a total climbs from 1/6 at 1 to a peak of 0.360 at 6, then wobbles and settles onto 2/7, about 0.286; the target of 10 is hit with probability 0.289.
Why does the answer settle at 2/7?
Picture stepping stones across a river, where each stride covers 1 to 6 stones with equal chance. Over a long walk you touch about one stone in every 3.5, because that is your average stride. The running total advances 3.5 per roll on average, so in the long run it lands on a fraction 1/3.5 = 2/7 of all numbers, and each far-off target is hit with probability close to 2/7. The early values wobble: p(6) is the highest, 0.360, because 6 is the last total a single roll from zero can reach directly, and the wobbles die out by about 20.
The relationship
$p(n) = \frac{1}{6}\sum_{k=1}^{6} p(n-k), \quad p(0) = 1 \qquad \lim_{n\to\infty} p(n) = \frac{1}{E[\text{roll}]} = \frac{1}{3.5} = \frac{2}{7}$
p(n) the chance the running total ever equals n
p(n - k) the chance of standing k below the target, one roll away
E[roll] the average roll of a fair die, 3.5
What it says in wordsEach total's chance is the average of the six before it, and in the long run the totals land on one number in every 3.5.
Why would a quant interviewer ask for a table or code here?
Because the recursion is dynamic programmingSolving a problem by building up answers to smaller versions of it and reusing them, instead of recomputing from scratch.: each value reuses the six before it, so a table of ten numbers is faster and safer than listing every sequence of rolls that sums to 10. Say the recursion, compute a few terms out loud, and give the limit with its reason; that is the complete answer. Do not try to enumerate paths: there are 492 ordered ways to reach 10 with rolls of 1 to 6, each with its own probability.
Where candidates lose it
The quick answers are 1/6, reasoning that some roll must land on 10 with one chance in six, and 2/7 stated as exact. The first ignores that most runs skip straight over 10; the second is close but is the long-run limit, and 10 is not yet far enough out for it to be exact.
The second loss is trying to count paths. Set up the recursion in one line instead and let it do the counting.
What the interviewer asks next
What is the probability that the running total ever equals exactly 6?
With a coin that adds 1 or 2 instead of a die, what does the hit probability approach?
How would you write this as a dynamic programme in a few lines of code?

bid	the price at which you can sell, 55
p	your model's probability of an up close, 0.50
100	the payout if the index closes higher

S_T	the stock price at expiry
K	the strike of the call sold, Rs 550
S_0	the price paid for the stock, Rs 500
c	the premium received, Rs 12

a, b	the lower and upper barriers, Rs 45 and Rs 55
k	the starting price
E(k)	the expected number of one-minute steps before either barrier is touched

beta_M	the stock's true direct loading on the market, 0.5
beta_S	the stock's loading on the sector that was left out, 1.0
delta	the slope of the sector's return on the market's, 0.4

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Why is a binary price just a probability?

Which side of the quote can you trade?

When would you do nothing?

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What have you actually sold?

How do you find the two numbers quickly?

What is the trade-off in plain terms?

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Why is the answer not 5 minutes?

How do you get exactly 25?

What does the shape tell a trader?

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Why does leaving the sector out change the market slope?

How do you compute the bias?

Is 0.9 wrong, or answering a different question?

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Why switch from probabilities to odds?

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How do you turn gap sizes into a position?

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What does a bigger casino buy you?

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How do you set up the recursion?

Why does the answer settle at 2/7?

Why would a quant interviewer ask for a table or code here?

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