Derivatives Foundation puzzles, solved step by step

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

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051You have the same 60% coin as before and even money on every flip, but now you bet half your capital every time. After 100 bets, what does your typical outcome look like, even though every single bet has positive expected value?Betting and sizingHardRisk managementHedge funds
Try it first
Each bet has a 10% edge on the money at risk. Where does betting half your stack 100 times leave the typical player?
Show the worked solution
The typical player ends with about 3% of the starting capital, while the average outcome is about 13,781 times the start. With 60 wins and 40 losses the stack is multiplied by 1.5 to the 60 and 0.5 to the 40, which is 0.033. Betting half the stack is 2.5 times the Kelly fraction of 20%, and anything beyond twice Kelly turns a positive-edge game into a losing one for the person actually playing it.
Why can every bet be favourable and the typical result still be a loss?
Think of a shopkeeper who doubles the stock every time a line sells and halves it every time it does not. A good line sells six times in ten, yet after a hundred cycles the shelf is nearly bare, because growth is a chain of multiplications and the order of operations does not rescue you. Expected value is an average across many imaginary players, but your capital follows one path, and a path is a product of factors, not a sum. Betting half the stack makes each win a factor of 1.5 and each loss a factor of 0.5. Sixty wins and forty losses, the most likely split, multiply to 0.033. The arithmetic mean of 13,781 is carried by the rare paths with 80 or 90 wins, which almost nobody lives on.
On a shared log scale, the average path compounds at 1.1 per bet to about 13,781 times the start after 100 bets, while the typical path of 60 wins and 40 losses shrinks by 3.4% per bet to about 3% of the start, because capital multiplies and the half-stack bet makes the loss factor too harsh.
What is the right quantity to look at instead of the mean?
Look at the expected growth rate of the logarithm of capital, because the log turns products into sums and the law of large numbers then works. Each bet changes log capital by 0.6 x log(1.5) + 0.4 x log(0.5), which is -0.034, a negative number, so the typical player loses about 3.4% of capital per bet on average. Over 100 bets that is a factor of exp(-3.4), the 0.033 you saw above. The chance of finishing below the start is 76%, so the mean is a number you will almost never meet.
The relationship
$g(f) = p\,\ln(1+f) + (1-p)\,\ln(1-f) \qquad g(0.5) = 0.6\ln 1.5 + 0.4\ln 0.5 \approx -0.034 \qquad f^* = 2p-1 = 0.2$
g(f) the expected growth of log capital per bet when you stake fraction f
p the chance of winning each even-money bet, here 0.6
f* the Kelly fraction that maximises g, here 20% of capital
What it says in wordsStaking half the stack makes the expected log growth negative, while the Kelly stake of a fifth makes it positive.
Where does the sizing go wrong, and what would fix it?
The Kelly fraction for an even-money bet is 2p minus 1, here 20% of capital, and it maximises the growth rate at +0.0201 per bet, which turns 100 bets into a typical multiple of about 7.5. Growth is zero at exactly twice Kelly, 40% here, and negative beyond it, so a half-stack bettor is on the wrong side of the hill even though the coin is in their favour. The limitation is that the Kelly rule assumes you know p exactly and will play many times; with an uncertain edge, desks size below Kelly on purpose, often at half of it.
Where candidates lose it
Most candidates say the player ends up rich, because they multiply the 10% edge through 100 bets and quote the mean. The interviewer wants to hear the word median, or typical, and the observation that capital compounds multiplicatively, so wins and losses of equal size do not cancel.
The second loss is stopping at the picture without the growth rate. Say the log growth per bet, show it is negative at a half stake and positive at a fifth, and name twice Kelly as the point where the game turns against you.
What the interviewer asks next
What fraction maximises the typical growth, and what is the growth rate there?
At what stake does the typical player break even despite the edge?
If you only get to play 5 times rather than 100, does the answer change?
Your estimate of the 60% is itself noisy. Does that push the stake up or down?
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?
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?
059On a stock that pays no dividend, the three-month 100-strike call trades at 6.50 and the six-month 100-strike call trades at 6.10. Is there an arbitrage, and how would you capture it?Option payoffs and no-arbitrageCoreMarket makingVolatility trading
Try it first
Two calls, same strike, the longer one cheaper. What do you do?
Show the worked solution
Yes. Sell the three-month call at 6.50, buy the six-month call at 6.10, and lock in 0.40 today with no risk. On a non-dividend stock, a call with more time to expiry is worth at least as much as the same strike with less, because at three months the six-month call is still worth at least its intrinsic value, which is exactly what the expiring call pays out. The position never owes money, so the 0.40 is a pure profit.
Why must the longer call be worth at least the shorter one?
A ticket that lets you buy a train seat any time in the next six months is worth at least as much as one that expires in three; you can always do with the long ticket exactly what you would have done with the short one, and then keep it. At the three-month expiry the short call pays max(S - 100, 0), and the six-month call at that moment is worth at least max(S - 100, 0) too, because a call on a non-dividend stock is never worth less than its intrinsic value and is usually worth more. So holding the long and being short the short can never leave you with a negative balance at three months, and the 0.40 you took in today is yours.
At-the-money call value rises with time to expiry, so the six-month call must be worth at least the 6.50 the three-month call trades at; the quoted 6.10 sits below that floor, and selling the three-month while buying the six-month locks in 0.40 with a spread that can only pay more.
What happens at three months in each case?
Walk through both branches. If the stock is above 100, the short call is exercised against you, you deliver stock for 100, and you still hold a six-month call worth more than S - 100, so you exercise or sell it and finish ahead; if the stock is at or below 100, the short call expires worthless and you are left holding a live six-month call for free. Either way you keep the 0.40 from today and own something worth zero or more. The limitation is the non-dividend assumption: with a large dividend before six months, early exercise of an American three-month call could matter, and the comparison needs the dividend's present value subtracted.
The relationship
$C(K, T_2) \ge C(K, T_1) \quad \text{for } T_2 > T_1 \text{ on a non-dividend stock} \qquad 6.10 < 6.50 \Rightarrow \text{sell } T_1,\ \text{buy } T_2,\ \text{keep } 0.40$
C(K, T) the price of a call with strike K and time to expiry T
T_1, T_2 three and six months
0.40 the cash locked in today, the smallest profit the trade can make
What it says in wordsA longer-dated call at the same strike can never be worth less than a shorter one, so a longer call quoted cheaper is sold against the shorter one for a riskless profit.
Why would a market maker see this quote and what does it say about the vol surface?
The three-month price of 6.50 corresponds to an implied volatility of about 33% with zero rates, and at that volatility a six-month call would be worth around 9.18. A quote of 6.10 for the six-month implies a term structure of volatility so inverted that no volatility at all could justify it, which is why this is an arbitrage and not merely a view on calendar spreads. Real quotes rarely break the bound outright; what you see instead is a long-dated call a few ticks above the floor on an illiquid name, and the question is then whether the bid-ask spread swallows the edge before you can trade both legs.
Where candidates lose it
Candidates treat it as a volatility question and start talking about the term structure. Implied volatility cannot push a longer call below a shorter one at the same strike, so the answer is a bound, not a view.
The second loss is getting the direction wrong under pressure. Sell what is dear, the three-month at 6.50; buy what is cheap, the six-month at 6.10. Then walk the two branches at the first expiry out loud.
What the interviewer asks next
Does the same bound hold for puts on a non-dividend stock?
The stock pays a large dividend in month four. Can the three-month call now be worth more than the six-month?
What if the two calls had different strikes, say 100 and 105?
Both quotes are mid prices and the bid-ask is 0.30 on each. Is the arbitrage still there?
061You and I each show heads or tails at the same time. You win Rs 3 if we both show heads, Rs 1 if we both show tails, and you lose Rs 2 if we show different faces. What mix should you play, and is the game worth playing?Games and logicCoreQuant tradingProp trading firms
Try it first
Two wins and two losses in the table. Is this game good for you?
Show the worked solution
Show heads 3/8 of the time, and do not play unless you are paid at least Rs 0.125 a round. If you show heads with probability p, your expected payoff is 5p - 2 when I show heads and 1 - 3p when I show tails. I will pick whichever is lower, so you choose p to make the lower line as high as possible, which is where they cross: p = 3/8, value - 1/8. Any other p lets me push you below that.
Why is the answer a mix rather than a single face?
Two children playing odds and evens learn fast that any pattern is punished: show heads every time and the other child shows tails every time. In a game where my best reply depends on what you do, any fixed choice is exploited, so you protect yourself by randomising in a ratio that leaves me with nothing to exploit. That ratio is found by making me indifferent between my two replies. If you show heads a fraction p of the time, my heads earns you 3p - 2(1 - p) = 5p - 2 and my tails earns you - 2p + (1 - p) = 1 - 3p. They are equal at p = 3/8.
Your expected payoff is 5p - 2 if I show heads and 1 - 3p if I show tails, and since I will always pick the lower line, the best you can do is the crossing at p = 3/8, where both lines give minus 1/8, so the game is worth minus Rs 0.125 to you per round.
How do you know minus 1/8 is the most you can guarantee?
Look at the lower of the two lines across all p. To the left of 3/8 the heads line is lower and rising; to the right the tails line is lower and falling, so the lower envelope peaks exactly at the crossing, and that peak is your guaranteed value. I have the same calculation from my side: if I show heads a fraction q of the time, you are indifferent when 3q - 2(1 - q) = - 2q + (1 - q), which again gives q = 3/8, and at that mix I hold you to - 1/8 whatever you do. Both sides landing on the same number is the minimax theorem, attributed to von Neumann, at work in a two by two table.
The relationship
$5p - 2 = 1 - 3p \;\Rightarrow\; p = \tfrac{3}{8}, \qquad V = 5\cdot\tfrac{3}{8} - 2 = -\tfrac{1}{8}$
p your probability of showing heads
5p - 2 your expected payoff when I show heads
1 - 3p your expected payoff when I show tails
V the value of the game to you per round
What it says in wordsEqualising your payoff across my two replies gives a three-eighths mix and a value of minus one eighth of a rupee per round.
What is the desk version of this question?
Quoting against a counterparty who sees your pattern. A market maker who always leans the same way after a fill is the child who always shows heads, and the counterparty who notices earns the difference, so randomised sizing and skew are the trading-floor form of the 3/8 mix. The limitation of the puzzle answer is that it assumes I play optimally; against an opponent who shows heads half the time out of habit, your best reply is pure heads, with an expected 0.5 x 3 - 0.5 x 2 = + Rs 0.50 a round, and the game becomes worth playing. Ask who you are playing before you quote the value.
Where candidates lose it
The common answer is that the game is fair or favourable, from summing the four cells. The sum of a payoff table says nothing when the opponent chooses the column. Set up the two lines and find where they cross.
The second loss is solving for the right p and then saying the game is fine because 3 and 1 are bigger than 2. State the value, minus 1/8, and say you need a fee of at least that to play.
What the interviewer asks next
What is my optimal mix, and what does it earn me?
Change the heads-heads payoff to Rs 4. Does the game become worth playing?
I am known to show heads 60% of the time regardless. What should you do now?
Why do both players end up with the same 3/8 here, and is that a coincidence?
064A bag holds 9 fair coins and 1 coin with heads on both sides. You pull one out at random and flip it 5 times, getting 5 heads. What is the probability it is the two-headed coin, and what is the probability the next flip is heads?Conditional probability and BayesCoreQuant tradingHedge funds
Try it first
Five heads in a row from a coin that is two-headed one time in ten. How likely is it the two-headed one?
Show the worked solution
78.0% that it is the two-headed coin, and 89.0% that the next flip is heads. Prior odds are 1 to 9. The two-headed coin gives a head with certainty and a fair coin with chance 1/2, so each head multiplies the odds by 2; after five heads the odds are 32 to 9, which is 32/41. The next flip is heads with chance 32/41 x 1 + 9/41 x 1/2 = 73/82.
Why work in odds rather than probabilities?
A doctor who sees the same symptom five mornings running does not recompute the whole diagnosis each day; each new observation multiplies the odds of the condition by one fixed factor. In odds form, Bayes' rule is a multiplication: posterior odds equal prior odds times the likelihood ratio of each observation, and here every head has the same ratio of 1 to 1/2, which is 2. So the odds on the two-headed coin go 1:9, 2:9, 4:9, 8:9, 16:9, 32:9. Convert at the end: 32 divided by 32 plus 9 is 32/41. Doing it in probabilities means dividing by a different normaliser five times, which is where people slip.
Each head doubles the odds on the two-headed coin, lifting the probability from 10% to 78% after five heads, and the chance the next flip is heads, 89%, mixes a certain head from the two-headed coin with a coin flip from a fair one.
Why is the next flip not simply 78% heads?
Because the fair coin also produces heads. The next flip is heads if the coin is two-headed, with chance 32/41, or if the coin is fair and lands heads, with chance 9/41 times 1/2, and the two routes add to 73/82, about 89%. Candidates who answer 78% have confused the probability of the hypothesis with the probability of the outcome. The gap between the two is the fair coin's half chance of a head, weighted by the 22% chance you are holding a fair coin.
The relationship
$\frac{P(\text{two-headed}\mid 5H)}{P(\text{fair}\mid 5H)} = \frac{1}{9}\times\left(\frac{1}{1/2}\right)^5 = \frac{32}{9} \;\Rightarrow\; P = \frac{32}{41}, \qquad P(\text{next } H) = \frac{32}{41}\cdot 1 + \frac{9}{41}\cdot\frac{1}{2} = \frac{73}{82}$
1/9 the prior odds of drawing the two-headed coin
(1 / (1/2))^5 the likelihood ratio of five heads, two to the fifth
73/82 the chance of a sixth head, mixing both coins
What it says in wordsFive heads multiply the prior odds by thirty-two, giving thirty-two to nine, and the next flip mixes a sure head with a fair flip in those proportions.
How many heads would it take to be nearly sure, and what does a desk take from this?
Each head doubles the odds, so after 10 heads the odds are 1,024 to 9, about 99.1%, and after 5 you are only at 78%. Evidence that is merely consistent with a hypothesis moves you slowly when the alternative also produces it often, which is why five good months from a new trading strategy prove far less than people feel they do. The limitation is the prior: if the bag held 99 fair coins and one two-headed, five heads would leave you at 32 to 99, still under 25%, and no amount of looking at the flips alone tells you the composition of the bag.
Where candidates lose it
The fast wrong answer is 1 minus 1/32, about 97%, which is the chance a fair coin would not have done this. That number ignores that there are nine fair coins for every two-headed one. Start from the prior odds and double.
The second loss is giving 78% for the next flip. The next flip is a mixture: a certain head from the two-headed coin and a half chance from a fair one.
What the interviewer asks next
After how many heads does the probability it is the two-headed coin pass 99%?
The sixth flip is tails. What is the probability it is the two-headed coin now?
The bag has 99 fair coins and one two-headed coin. What are the two answers after five heads?
Why is the probability of the next head always between the fair coin's 1/2 and 1?
066A company is worth a uniformly random amount between Rs 0 and Rs 100 crore to its owner, who knows the exact value. It is worth 1.5 times that amount to you. You make one take-it-or-leave-it offer, and the owner accepts if your offer exceeds the value. What should you offer?Games and logicHardHedge fundsMarket making
Try it first
The company is worth 50% more to you than to the owner. What do you bid?
Show the worked solution
Offer nothing. Every positive offer loses money on average. If you offer b and the owner accepts, you learn the value is below b, so its average is b/2, worth 1.5 x b/2 = 0.75b to you. You pay b, so each accepted deal loses 0.25b, and the deal is accepted a fraction b/100 of the time. Expected profit is minus 0.0025 b squared, negative for every b above zero. The 1.5 multiplier is not enough to overcome what acceptance tells you.
Why is the owner saying yes bad news for you?
A friend sells you their old scooter for any price you name, but only if your price beats what they privately think it is worth. If they take Rs 20,000 instantly, you have just learned the scooter is worth less than that to someone who knows it well. Acceptance is information: it tells you the true value lies below your bid, so the only companies you ever buy are the ones worth less than you paid, and the ones worth more walk away. Averaging over all possible values, as if you bought every company, is the mistake; you must average only over the values at which the owner says yes.
Expected profit from an offer b is minus a quarter of b times the acceptance chance b/100, a parabola that is zero at b = 0 and falls to minus 25 crore at b = 100, so no positive offer earns anything and the best move is not to bid; an offer of 60, for instance, buys a company worth 45 to you on average and loses 9 in expectation.
How do you set up the expected profit cleanly?
Condition on acceptance, then multiply by its probability. Given a bid b that is accepted, the value V is uniform on 0 to b, so E[V | accepted] = b/2, the company is worth 1.5 x b/2 = 0.75b to you, and the profit on an accepted deal is 0.75b - b = - 0.25b. Acceptance happens with probability b/100, so expected profit is - 0.25b x b/100 = - b squared over 400. At b = 50 that is minus 6.25 crore; at b = 100 it is minus 25 crore. The derivative is negative everywhere above zero, so the maximum is at b = 0.
The relationship
$E[\pi(b)] = \frac{b}{100}\left(1.5\cdot\frac{b}{2} - b\right) = -\frac{b^2}{400} < 0 \quad \text{for all } b > 0$
b your offer in Rs crore
b/100 the chance the owner accepts, since the value is uniform on 0 to 100
1.5 x b/2 what the company is worth to you on average once you know the value is below b
What it says in wordsThe expected profit from any offer is a negative multiple of the offer squared, so the best offer is zero.
At what multiplier does a bid start to make sense?
Replace 1.5 with a general m. Given acceptance, the company is worth m x b/2 to you against the b you pay, so the deal breaks even when m/2 = 1, that is m = 2. Unless the company is worth more than twice as much to you as to the owner, acceptance always costs you, and at exactly double every bid is a wash. That is the winner's curse in its purest form: the party with less information loses whenever the informed party decides whether to trade. The limitation is the uniform prior and the single offer; with a floor on the value, or a negotiation that reveals information, positive bids can be profitable.
Where candidates lose it
Candidates bid around 50 or 75 by averaging over the whole range of values, forgetting that they only buy when the owner accepts. The interviewer wants you to say that acceptance is information before you touch any arithmetic.
The second loss is getting zero and not being able to say what would change it. Give the general condition: the multiplier must exceed two for any positive bid to pay.
What the interviewer asks next
At what multiplier does a positive bid first break even, and what is the best bid at a multiplier of 3?
The value is uniform on Rs 50 to Rs 100 crore instead. Does a positive bid make sense now?
How is this the same problem as a market maker being hit only when they are wrong?
You get two offers rather than one, and the owner rejects the first. Does that change anything?
068Your book is delta neutral with gamma of 2,000 shares per rupee on a stock trading at Rs 500. The stock jumps Rs 10. Roughly what is your P&L before you rehedge, and how many shares do you now need to trade?Option pricing intuitionCoreEquity derivativesVolatility trading
Try it first
Delta zero, gamma 2,000 shares per rupee, a Rs 10 jump. P&L?
Show the worked solution
About Rs 1,00,000 profit, and you need to sell about 20,000 shares. P&L from gamma is one half of gamma times the move squared: 0.5 x 2,000 x 10 squared = Rs 1,00,000. The delta picked up during the move is gamma times the move, 2,000 x 10 = 20,000 shares long, which you sell to get back to neutral. A Rs 10 fall would earn the same amount and leave you 20,000 shares short to buy back.
Why does a delta-neutral book make money on a move?
A cyclist at the bottom of a valley is on flat ground, but every metre up either slope gets steeper. Delta neutral means the P&L is flat at the current price only; gamma is how fast the slope changes, so as the stock moves the book acquires delta in the direction of the move and earns on it the whole way. With gamma of 2,000 shares per rupee, after the first rupee you are 2,000 shares long, after the fifth 10,000, after the tenth 20,000. The P&L is the area under that rising delta, a triangle with base 10 and height 20,000, which is 1,00,000.
A delta-neutral book with gamma of 2,000 shares per rupee earns one half of gamma times the move squared, Rs 1,00,000 on a Rs 10 move in either direction, and at the new price its slope is gamma times the move, 20,000 shares long, which is what must be sold to be flat again.
What is the arithmetic, and where does the one half come from?
Expand the book's value as a Taylor series in the stock price. The first-order term is delta times the move, zero here; the second-order term is one half of gamma times the move squared, 0.5 x 2,000 x 100 = Rs 1,00,000; and the new delta is the derivative of that, gamma times the move, 20,000 shares. The one half is the same one half as in the area of a triangle: delta started at zero and finished at 20,000, so on average it was 10,000 shares over the Rs 10 move. The limitation is that a jump also changes implied volatility and burns a day of theta, both ignored here.
The relationship
$\Delta P\&L \approx \Delta\cdot\delta S + \tfrac{1}{2}\Gamma\,(\delta S)^2 = 0 + \tfrac{1}{2}\times 2000\times 10^2 = 1{,}00{,}000, \qquad \Delta_{\text{new}} = \Gamma\,\delta S = 20{,}000$
delta the book's share-equivalent exposure, zero before the move
Gamma the change in delta per rupee of stock move, 2,000 shares
delta S the stock move, Rs 10
What it says in wordsThe profit is half of gamma times the move squared, and the delta to be hedged afterwards is gamma times the move.
What happens if you rehedge and the stock comes back?
You sell 20,000 shares at Rs 510. If the stock then falls back to Rs 500, the options give back their Rs 1,00,000 but the short stock earns 20,000 x Rs 10 = Rs 2,00,000, so you net Rs 1,00,000 from the round trip. That is what long gamma means in practice: each rehedge locks in half of gamma times the move squared, and a stock that moves a lot and comes back pays you twice. The cost is theta, the daily decay you pay for holding the options, and the trade only works if realised movement is larger than the implied volatility you paid for.
Where candidates lose it
Candidates say zero because the book is delta neutral, or they give gamma times the move squared without the one half and double the answer. Say the triangle: delta climbs from zero to 20,000, average 10,000, times Rs 10.
The second loss is confusing the two numbers. The P&L is in rupees and uses the move squared; the delta to trade is in shares and uses the move once.
What the interviewer asks next
The stock falls Rs 10 instead. What is the P&L and what do you trade?
You rehedge at Rs 510 and the stock returns to Rs 500. What have you made on the round trip?
What daily theta would make this book break even on a Rs 10 move per day?
The book is short gamma instead. Describe the same Rs 10 move.
069An index of ten equally weighted stocks has 15% implied volatility, and each of the ten stocks has 30% implied volatility. What average correlation is the market implying?Volatility and correlationHardVolatility tradingExotics trading
Try it first
Index vol is half the single-stock vol. Rough instinct for the implied correlation?
Show the worked solution
About 0.17, exactly one sixth. For n equally weighted stocks with the same volatility and the same pairwise correlation rho, index variance is stock variance times (1/n + (1 - 1/n) rho). Index variance over stock variance is 0.15 squared over 0.30 squared = 0.25, so 0.25 = 0.1 + 0.9 rho and rho = 0.15/0.9 = 1/6. The large-n shortcut of 0.25 overstates it because it ignores the 1/n term.
Why does the index have less volatility than its members?
Ten shopkeepers in one market each have noisy daily takings, but the market's total takings are steadier, because one shop's bad day is often another's good day. Only the part of the noise they share, the weather or a festival, survives the averaging. Index variance has two parts: the stocks' own noise, which averages away as 1/n, and the common movement, which survives in proportion to the correlation. With ten stocks at 30% and no correlation at all the index would still have 9.5% volatility, and that floor is why the implied correlation is below the naive 0.25.
For ten equally weighted stocks at 30% volatility, index volatility rises from 9.5% at zero correlation to 30% at a correlation of one, and a 15% index volatility is reached at an average correlation of one sixth, about 0.17.
How do you derive the formula on the spot?
Write the index as the average of n returns and expand its variance. There are n variance terms, each sigma squared over n squared, and n(n - 1) covariance terms, each rho sigma squared over n squared, so index variance is sigma squared times (1/n + (n - 1) rho / n). Set that equal to 0.15 squared with sigma = 0.30 and n = 10: 0.0225 = 0.09 x (0.1 + 0.9 rho), so 0.25 = 0.1 + 0.9 rho and rho = 1/6. The derivation takes four lines, and the interviewer wants to hear the count of covariance terms, n(n - 1), said out loud.
The relationship
$\sigma_I^2 = \sigma^2\left(\frac{1}{n} + \frac{n-1}{n}\rho\right) \;\Rightarrow\; \rho = \frac{\sigma_I^2/\sigma^2 - 1/n}{1 - 1/n} = \frac{0.25 - 0.1}{0.9} = \frac{1}{6}$
sigma_I index volatility, 15%
sigma single-stock volatility, 30% for every stock
n the number of equally weighted stocks, ten
rho the average pairwise correlation implied by the two volatilities
What it says in wordsIndex variance is stock variance times one over n plus the correlation times the rest, which solves to a correlation of one sixth.
What does a desk do with implied correlation?
It compares it with realised correlation and trades the gap. Selling index options and buying single-stock options is a short correlation position, called a dispersion trade, and it pays when the stocks move more independently than the 1/6 the market has priced in. The limitation of the puzzle is its symmetry: real indices have unequal weights and unequal volatilities, so the implied correlation is a weighted average that can differ from the simple formula, and the implied figure also moves with the skew of the index options, which the single number cannot show.
Where candidates lose it
The common wrong answers are 0.5 and 0.25: the first divides volatilities and the second divides variances but forgets the 1/n floor from the stocks' own noise. With only ten stocks the floor is a tenth of the variance, which is a big correction.
The second loss is knowing the formula but failing to say what trade it supports. Say dispersion, and say which side is short correlation.
What the interviewer asks next
Same numbers with 50 stocks instead of 10. What is the implied correlation?
What is the lowest possible index volatility for ten stocks at 30%, and at what correlation?
Implied correlation is 1/6 and realised correlation turns out to be 0.4. Which side of a dispersion trade made money?
The stocks have different weights and volatilities. How does the formula change?
072A stock at 100 will be at 80, 100 or 130 at expiry, each equally likely in your view. What is the expected payoff of a 100-strike call and of a 100-strike put, and why is the expected payoff not what a market maker would charge?Option payoffs and no-arbitrageWarm upSell-side sales and trading
Try it first
Three equally likely outcomes. Expected call payoff and expected put payoff?
Show the worked solution
Expected payoffs are 10 for the call and 6.67 for the put, and neither is a price. The call pays 0, 0 and 30 across the three outcomes, averaging 10; the put pays 20, 0 and 0, averaging 6.67. A market maker charges the cost of hedging, and with zero rates call minus put must equal stock minus strike, which is 0, so prices of 10 and 6.67 would be an arbitrage against the maker. Hedge-implied odds that keep the stock worth 100 give both options the same price.
Why is the expected payoff under your odds not the price?
A shopkeeper who believes the monsoon will be good does not price umbrellas off that belief; the price is set by what the umbrellas cost to stock and what the shop next door charges. An option maker does not hold the option to expiry hoping for the payoff; they hedge it with the stock, so the price is the cost of that hedge, and the cost of the hedge depends on the stock's current price of 100, not on anyone's forecast of where it goes. Your equal odds imply the stock is expected to be worth 103.33, above today's 100; that optimism is yours to trade, not something the maker will pay you for inside an option price.
Under equal odds on 80, 100 and 130 the call's expected payoff is 10 and the put's is 6.67, but with zero rates call minus put must equal stock minus strike, which is zero, so those two numbers cannot both be prices; hedge-implied odds that keep the stock worth 100, such as 0.40, 0.33 and 0.27, price both options at 8.
What is the one-line test that catches the mistake?
Put-call parity. Buying the call and selling the put with the same strike gives you the stock minus 100 in every outcome, which with zero rates is worth 100 - 100 = 0 today, so the call and the put must have the same price. Expected payoffs of 10 and 6.67 fail that test by 3.33, and a maker who quoted them would be lifted on the put and hit on the call until the prices met. The test needs no probabilities at all, which is the point: parity is enforced by hedging, not by views.
The relationship
$C - P = S - K\,e^{-rT} = 100 - 100 = 0 \qquad \text{but } E[C] - E[P] = 10 - 6.67 = 3.33 = E[S] - K$
C, P the prices of the 100-strike call and put
S - K stock minus strike, the value of a long call and short put in every state
E[S] - K the drift your odds put on the stock, 103.33 - 100
What it says in wordsPrices must satisfy parity, and the gap between the two expected payoffs is exactly the drift your personal odds assign to the stock.
What odds would a maker use, and are they unique here?
Odds that make the stock worth its forward, 100 with zero rates. Any set of probabilities on 80, 100 and 130 with an average of 100 prices the two options consistently; one such set is 0.40, 0.33 and 0.27, which gives the call 0.267 x 30 = 8 and the put 0.40 x 20 = 8, equal as parity demands. The limitation is that with three outcomes and only the stock to hedge with, the set is not unique: the common level of the call and put price is pinned down by parity only up to a range, and in practice it is the market's volatility quote that chooses the point inside it.
Where candidates lose it
The arithmetic is easy and candidates get 10 and 6.67 quickly; the loss comes in the second half, where they say a market maker would add a spread to the expected payoff. The real answer is that the probabilities themselves are wrong for pricing, because the price is the cost of a hedge.
Say parity: call minus put equals stock minus strike, zero here, so the two prices must be equal. That one sentence shows you know why risk-neutral pricing exists.
What the interviewer asks next
Find a set of probabilities on 80, 100 and 130 under which the stock is worth 100, and price both options.
Interest is 5% for the period. What does parity say the call minus the put is worth now?
Why is the set of hedge-implied probabilities not unique with three outcomes and one stock?
The stock will be at 80 or 130 only. Price the call by replication.

g(f)	the expected growth of log capital per bet when you stake fraction f
p	the chance of winning each even-money bet, here 0.6
f*	the Kelly fraction that maximises g, here 20% of capital

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

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

C(K, T)	the price of a call with strike K and time to expiry T
T_1, T_2	three and six months
0.40	the cash locked in today, the smallest profit the trade can make

Try it first

Why can every bet be favourable and the typical result still be a loss?

What is the right quantity to look at instead of the mean?

Where does the sizing go wrong, and what would fix 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 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

Try it first

Why must the longer call be worth at least the shorter one?

What happens at three months in each case?

Why would a market maker see this quote and what does it say about the vol surface?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the answer a mix rather than a single face?

How do you know minus 1/8 is the most you can guarantee?

What is the desk version of this question?

Where candidates lose it

What the interviewer asks next

Try it first

Why work in odds rather than probabilities?

Why is the next flip not simply 78% heads?

How many heads would it take to be nearly sure, and what does a desk take from this?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the owner saying yes bad news for you?

How do you set up the expected profit cleanly?

At what multiplier does a bid start to make sense?

Where candidates lose it

What the interviewer asks next

Try it first

Why does a delta-neutral book make money on a move?

What is the arithmetic, and where does the one half come from?

What happens if you rehedge and the stock comes back?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the index have less volatility than its members?

How do you derive the formula on the spot?

What does a desk do with implied correlation?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the expected payoff under your odds not the price?

What is the one-line test that catches the mistake?

What odds would a maker use, and are they unique here?

Where candidates lose it

What the interviewer asks next