Derivatives Foundation puzzles, solved step by step

Puzzles: 100
Traced to a firm: 66
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025A stock trades at 500. The one-year 500-strike call is priced at 40 and the 500-strike put at 45, and interest rates are zero. What dividend is the options market pricing in, and what would you check before trading on it?Option payoffs and no-arbitrageCoreEquity derivativesMarket making
Try it first
Before using any formula: the put costs more than the call at the same at-the-money strike. With zero rates, that tells you
Show the worked solution
A dividend of 5 per share. With zero rates, put-call parity says call minus put equals spot minus the dividend minus the strike. Here 40 minus 45 = 500 minus D minus 500, so D = 5, and the implied forward is 495. Before trading against it, check that the options are European or that early exercise is worth nothing, what it costs to borrow the stock, whether the ex-date falls before expiry, and whether all four prices are live and tradable.
Why does a put costing more than a call point to a dividend?
Suppose you agree today to buy a friend's scooter in a year for its fair price, but the friend will keep using it to deliver parcels and pocket the fees until then. You would pay less than today's price, by the fees they collect. Holding a call and selling a put at the same strike is an agreement to buy the stock at the strike, so call minus put must equal the forward price minus the strike, and the forward is spot minus whatever the holder collects before expiry. With zero rates there is no interest to account for, so a put dearer than the call by 5 says the forward is 5 below spot.
The relationship
$C - P = S - D - K \;\Rightarrow\; 40 - 45 = 500 - D - 500 \;\Rightarrow\; D = 5$
C, P the call and put prices at the same strike and expiry
S the spot price, 500
K the strike, 500
D the cash the stock pays out before expiry, here the implied dividend
What it says in wordsWith zero rates, the call minus the put equals the spot less the dividend less the strike, so the dividend is what balances the equation.
Call minus put is minus 5, so the options imply a forward of 495, and since the stock's forward is 500 minus the dividend, the market is pricing a dividend of 5; if the company pays only 3, the remaining 2 has to be explained by something else, such as the cost of borrowing the stock.
What would you check before trading on it?
Say your own estimate of the dividend is 3. Then a reversal looks attractive: short the stock, buy the call, sell the put, collecting 500 + 45 minus 40 = 505; at expiry the options deliver the stock back at 500 and you owe the 3 dividend, 503 in all, a locked-in 2 per share. That 2 is only yours if you can borrow the stock for less than 0.4% of its price over the year; a hard-to-borrow stock shows up in parity exactly as an extra dividend. Then check the rest: American calls can be exercised just before a dividend, the ex-date must fall before expiry, and four bid-offer spreads can eat a gap of 2 on their own.
What is the general lesson the interviewer wants?
That parity is an accounting identity between three things you can trade, and any gap between what it implies and what you believe is a claim about something you have not yet priced. The options market does not quote a dividend; it quotes a forward, and the dividend, the borrow cost and the interest rate are the pieces you split it into. A good answer gives the 5, names the implied forward of 495, and then lists what could make the 5 something other than a dividend. The limitation to say plainly: with non-zero rates, the strike is discounted and the arithmetic shifts, so state the zero-rate assumption before quoting the number.
Where candidates lose it
The common answer is that the put is dearer because the market expects the stock to fall. At one strike and one expiry, direction cannot make the put dearer than parity allows; anyone could sell the put, buy the call and short the stock against it. The gap is a forward, not a view.
The second loss is stopping at 5. The question asks what you would check, and the borrow cost is the one interviewers wait for: an implied dividend above the announced one is often a stock that is expensive to short.
What the interviewer asks next
Interest rates are now 6% a year. Redo the implied dividend, and say which way it moves.
The company announces a dividend of 8. Which trade would you put on with these four prices, and what is the risk?
Why might an American call on this stock be worth more than its European twin, and how would that distort the implied dividend?
031A stock is at 1,000, volatility is 20% and rates are near zero. Estimate the three-month at-the-money call in your head, and the straddle.Option pricing intuitionWarm upMarket makingVolatility trading
Try it first
Say the call price before you reach for a formula.
Show the worked solution
Call about 40, straddle about 80. The rule is 0.4 x S x sigma x sqrt T. Three months is a quarter of a year, so sqrt T is 0.5 and the volatility over the period is 10%; 0.4 x 1,000 x 0.1 = 40. With rates at zero the at-the-money put is worth the same, so the straddle is 80. The full model gives 39.88 for the call, because the exact constant is 1 over sqrt(2 pi), 0.3989, not 0.4.
Where does the 0.4 come from?
A tailor who knows a customer's height is normally distributed around 170 cm with a spread of 10 cm can say how far above 170 the average tall customer stands: about 0.4 of the spread, which is 4 cm, because the mean of the positive half of a normal is sigma over sqrt(2 pi). An at-the-money call pays the positive half of the stock's move, and the average of the positive half of a normal is 0.4 of its standard deviation, so the call is worth 0.4 times the standard deviation of the move over its life. The standard deviation of the move is S x sigma x sqrt T, which is 1,000 x 0.2 x 0.5 = 100 here, and 0.4 of 100 is 40. The exact constant is 1 / sqrt(2 pi) = 0.3989, so the rule gives 40 where the precise version gives 39.89, and the model 39.88.
The relationship
$C_{ATM} \approx 0.4\, S\, \sigma \sqrt{T} = 0.4 \times 1000 \times 0.20 \times 0.5 = 40 \qquad C_{exact} = S\left(2N\!\left(\tfrac{\sigma\sqrt{T}}{2}\right) - 1\right)$
S the stock price, 1,000
sigma sqrt T the volatility scaled to the option's life: 20% x 0.5 = 10%
0.4 the approximation to 1 over root 2 pi, which is 0.3989
N the standard normal distribution function
What it says in wordsAn at-the-money call is about four tenths of one standard deviation of the stock's move over its life.
Across volatilities from 0 to 100% the rule 0.4 x S x sigma x sqrt T sits almost on top of the full model price for a three-month at-the-money call, giving 40 against 39.88 at 20% volatility, and bends below it only at high volatility where the model's price curves.
Why is the straddle just double, and when is it not?
With rates at zero and no dividends, the forward equals the spot, so an at-the-money call and put have the same value by put-call parity, and the straddle is simply two calls, about 80. The straddle costs 8% of the stock for a three-month bet, which is the whole quarter's one-standard-deviation move of 10% times 0.8: that is the number a volatility trader carries in their head. With rates or dividends the forward moves away from spot, at-the-money means at-the-forward, and the call and put split the straddle unevenly, though their sum barely changes. The rule also assumes the volatility is the right one for this strike, which on a real surface with skew it may not be.
How far can you push the rule?
The rule is linear in volatility and time, the model is not. For three months the gap is 0.12 on 40 at 20% volatility, and even at 60% volatility the rule gives 120 against 119.2. Over one year at 20% the rule gives 80 against the model's 79.66. Up to a total move of about 50% the rule is within a couple of percent, which is every interview case and most of the real book; beyond that the model price flattens because a call can never be worth more than the stock. Say the limitation and then use the rule anyway: on a desk the question is never whether 40 is exactly right, it is whether 44 on the screen is rich or cheap.
Where candidates lose it
The common loss is forgetting to scale the volatility to the horizon and quoting 0.4 x 1,000 x 0.2 = 80 for the call, which is the one-year number. Say sqrt T out loud: three months is a half.
The second is giving the call and then stalling on the straddle, or doubling the call without saying why. The put equals the call only because rates are zero and there is no dividend; name parity and the interviewer knows you understand what at the money means.
What the interviewer asks next
Rates are now 8%. Which is worth more at the money, the call or the put, and by roughly how much?
The stock pays a 2% dividend before expiry. What changes?
Quote me the one-month straddle on the same stock, then the one-year.
The market is paying 44 for the call. What volatility is it implying, roughly?

032One-month implied volatility is 20% and three-month implied volatility is 25%. What volatility is implied for the period from month one to month three?Volatility and correlationCoreVolatility tradingEquity derivatives

Try it first

What is the forward volatility for months two and three?

Show the worked solution

About 27.2%. Variance is volatility squared times time, and variances add across periods. The three months carry 0.25 squared x 3 = 0.1875 of variance; the first month carries 0.20 squared x 1 = 0.04; so months two and three carry 0.1475 between them, 0.0737 per month on an annualised basis. The square root is 0.272. The forward volatility must sit above 25% because the quarter's average has to be pulled up from the 20% start.

Why do you add variances and not volatilities?

If you walk a random distance each hour, your spread after three hours is not three times the hourly spread, because some hours cancel others. What grows in a straight line is the variance, the spread squared. Implied volatility is quoted per year, so the variance an option carries is volatility squared times its life, and a longer option's variance is the sum of the variances of the pieces of time inside it. That is the whole mechanism. Twenty per cent for one month is 0.04 of variance; twenty-five per cent for three months is 0.1875; the difference belongs to the two months in between, and dividing by two months and taking the square root turns it back into a volatility.

Drawn as area, the first month's variance of 0.04 plus the two forward months' variance of 0.0737 each must fill the quarter's 0.1875, so the forward box has height 0.0737 and its square root is a forward volatility of 27.2%.

The relationship

\sigma_{1,3}^2 = \frac{\sigma_3^2 T_3 - \sigma_1^2 T_1}{T_3 - T_1} = \frac{0.0625 \times 3 - 0.04 \times 1}{2} = 0.07375 \qquad \sigma_{1,3} = \sqrt{0.07375} \approx 27.2\%

sigma 3, sigma 1	the three-month and one-month implied volatilities, 25% and 20%
T	the option life in months; the units cancel as long as both use the same one
sigma 1,3	the forward volatility for the period between the two expiries

What it says in wordsSubtract the short period's variance from the long period's and spread what is left over the time in between.

What does the number tell a trader?

The 27.2% is the volatility you would lock in for months two and three by selling the one-month option and buying the three-month one in variance-weighted sizes, a calendar spread. If you believe realised volatility in those two months will be well below 27.2%, the three-month option is rich relative to the one-month, and the calendar is the trade that expresses it. Say the rounding: an upward sloping term structure, 20% then 25%, hides a forward that is steeper than either quote, 27.2%, and the trap of reading 25% as the forward is what the question is built to catch. The calculation uses calendar time in months; a desk would use trading days or variance-weighted business days, which shifts the number by a few tenths.

What is the limit of the calculation?

The subtraction has to leave something positive. If the three-month volatility were below 11.5% with the one-month still at 20%, the quarter would carry less variance than its first month alone, which is impossible without arbitrage: you could sell the one-month, buy the three-month and hold a position with negative forward variance. Forward variance can be small but never negative, so a term structure that inverts too sharply is a mispricing, not a forecast. The second limit is that both quotes must refer to the same strike in forward terms; mixing an at-the-money one-month with a three-month that has rolled away from the money adds skew to the comparison and the forward you compute is no longer clean.

Where candidates lose it

The common loss is averaging volatilities: 25% over three months with 20% in the first month gives 27.5% for the rest if you treat volatility as additive. It is close, which is why it survives, but it is the wrong quantity and on a steeper curve the gap is large.

The second is forgetting to weight by time: subtracting 0.04 from 0.0625 and taking the root. Write variance times time for each leg, every time.

What the interviewer asks next

Six-month volatility is also 25%. What is the forward volatility for months four to six?
The one-month is 30% and the three-month is 25%. What is the forward, and what does it say about the market?
How would you actually lock in that forward volatility with listed options, and what would break the hedge?
Why do desks compute this in trading days rather than calendar months?

035There are 100 coins on the table. Players take turns removing 1 to 10 coins, and whoever takes the last coin wins. Do you want to go first, and what is your first move?Games and logicWarm upQuant trading
Try it first
Go first or second, and what is the opening?
Show the worked solution
Go first and take 1, leaving 99. Work backwards: whoever faces 11 coins loses, because any take of 1 to 10 leaves 1 to 10 for the other player to finish. The same holds for 22, 33 and every multiple of 11. From 100, taking 1 leaves 99, a multiple of 11; after that, whatever the opponent takes, you take 11 minus it, stepping down 88, 77, 66 and so on to 0, where you take the last coin.
Why work backwards from the last coin?
If you are climbing stairs with a friend and the rule is that the person who steps onto the top stair wins, you do not plan from the bottom; you ask which stair you must leave your friend on so that they cannot reach the top in one go. Games with a fixed last move are solved from the end: find the positions where the player to move loses, then find the positions from which you can push your opponent onto one of them. With 1 to 10 coins allowed, facing 1 to 10 coins is a win, you take them all. Facing 11 is a loss, because every move leaves between 1 and 10. Facing 12 to 21 is a win, since you can reduce to 11. Facing 22 is a loss again. The losing positions repeat every 11.
Every multiple of 11 from 0 to 99 is a losing position for the player who must move, so the first player takes 1 to leave 99 and then answers every take of t with 11 minus t, stepping down through 88, 77 and 66 until the last coin.
How do you find the period without listing every position?
The period is the largest take plus one, 11, because that is the one total a pair of moves can always be made to add up to: whatever your opponent takes between 1 and 10, you can take the balance of 11. The losing positions are the multiples of the largest take plus one, and the winning opening move is the remainder when the pile is divided by that number. 100 divided by 11 is 9 remainder 1, so take 1. If the rule allowed 1 to 7 coins, the period would be 8 and the opening would be 100 mod 8, which is 4. If the pile had been 99 to start with, you would want to go second, because the first player cannot leave a multiple of 11.
The relationship
$\text{losing positions} = \{0, 11, 22, \ldots, 99\}, \qquad \text{opening} = 100 \bmod 11 = 1$
11 the largest allowed take plus one, the amount you can always complete in a pair of moves
100 mod 11 the remainder when 100 is divided by 11; take exactly this many
What it says in wordsTake the remainder on your first move, then keep each pair of moves summing to 11.
What changes if the last coin loses instead of wins?
Then you want to hand your opponent the last coin, so the position you avoid facing is 1 coin, and the losing positions shift up by one: 1, 12, 23 and so on up to 100. Facing 100 in that version you are already lost, so you would want to go second, which shows the interviewer that you re-derive the pattern rather than remember it. The method is the same in every variant: name the terminal position, step back one move at a time to find the first losing position, then find the period. The limitation of the trick is that it needs a game with perfect information and no chance; add a die that sets each turn's maximum and the clean period disappears.
Where candidates lose it
The common loss is taking 10, because a bigger move feels like a stronger start. It leaves 90, which is not a multiple of 11, and a prepared opponent takes 2 to leave 88 and wins from there.
The second is knowing the answer and not the reason. Say why 11 is the period: any take of 1 to 10 can be completed to 11. Without that sentence the interviewer will change the numbers and watch you stall.
What the interviewer asks next
Players may take 1 to 7 coins instead. Do you go first, and what is the opening?
The player who takes the last coin loses. Do you go first?
There are two piles, 100 and 60, and you may take from either pile. Who wins?
Each turn a die sets the maximum take. Is there still a strategy, and what is it?
038Box A holds 3 red and 1 black ball, box B holds 1 red and 3 black. You pick a box at random and draw a red ball. You then draw again from the same box without replacement. What is the probability the second ball is red?Conditional probability and BayesCoreProp trading firmsMarket making
Try it first
Pick the probability of a second red before you compute.
Show the worked solution
Exactly 1/2. The first red is evidence: box A produces a red with probability 3/4, box B with 1/4, so after seeing red the odds are 3 to 1 for A, a posterior of 3/4. Then update the contents. Box A has 2 red of 3 left, so a second red has probability 2/3; box B has 0 red of 3, probability 0. Average over the updated boxes: 3/4 x 2/3 + 1/4 x 0 = 1/2. Forgetting the update gives 1/3, which is wrong.
Why does the first red change which box you think you hold?
If a friend's cooking is spicy three times out of four and yours one time in four, and a dish picked at random from one of you turns out spicy, you would guess it was your friend's, and you would be right three times in four. The first red ball is the same clue: it comes from box A three times as easily as from box B, so the box you are holding is A with probability 3/4, not 1/2. The arithmetic is Bayes. The weight for A is 1/2 x 3/4 = 3/8, for B it is 1/2 x 1/4 = 1/8, and A's share of the total 1/2 is 3/4. The question is built so that candidates who skip this update get a clean wrong answer.
After a red first draw the box is A with posterior 3/4 and B with 1/4, box A has 2 red of 3 left and box B none, so the chance of a second red is 3/4 x 2/3 plus 1/4 x 0, which is exactly one half.
What has to be updated besides the belief about the box?
The box itself. One red ball has left it. Box A now holds 2 red and 1 black, so a red is 2/3; box B now holds 0 red and 3 black, so a red is impossible. Both updates matter and they pull in opposite directions: believing more in A pushes the answer up, removing a red from A pushes it down, and here they land on exactly 1/2. Say that out loud, then check it by counting. Imagine 8 rounds, 4 with each box. Box A yields a first red in 3 of its 4 rounds and box B in 1 of its 4: four red first draws in all. In the three from A the second draw is red 2/3 of the time, 2 rounds; in the one from B, never. Two of four is 1/2.
The relationship
$P(R_2 \mid R_1) = P(A \mid R_1)\,P(R_2 \mid A, R_1) + P(B \mid R_1)\,P(R_2 \mid B, R_1) = \tfrac34 \cdot \tfrac23 + \tfrac14 \cdot 0 = \tfrac12$
R1, R2 the first and second draws are red
P(A | R1) the posterior that the box is A after one red, 3/4
P(R2 | A, R1) a second red from box A with one red already gone, 2/3
What it says in wordsWeight each box's chance of a second red by how likely that box is after the first red.
How do the variants separate candidates who understand from those who memorised?
Put the first ball back and the box contents do not change but the belief still does: 3/4 x 3/4 + 1/4 x 1/4 = 5/8, which is above 1/2 because the posterior leans to A and A stays rich in red. Forget the belief update and keep the contents update, and you get 1/2 x 2/3 + 1/2 x 0 = 1/3. The gap between 1/3, 1/2 and 5/8 is the whole lesson: the first draw tells you about the box and changes the box, and you need both. If a second red does appear, the posterior for A becomes 1, since B has no reds left, and a third red is then 1/2. The limitation is the usual one for Bayes: the 1/2 prior for each box is given here, and in a trading setting the prior is the thing you are least sure of.
Where candidates lose it
The common loss is 1/3: keeping the 1/2 prior on each box and only updating the contents. The first red is evidence about the box, and leaving it out throws away half the information in the question.
The second is 2/3: updating to box A and then forgetting that B is still possible with probability 1/4, or forgetting that A has lost a red. Carry both boxes through to the end, then average.
What the interviewer asks next
The first ball is put back before the second draw. What is the probability now?
The second ball is also red. What is the probability the box is A, and that a third draw is red?
Box B held 0 red and 4 black instead. What changes, and what is the answer?
A trader sees one winning trade from a new signal. What is the analogue of the box update?

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

Try it first

Which portfolio of plain options replicates the chooser?

Show the worked solution

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

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

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

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

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

The relationship

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

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

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

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

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

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

Where candidates lose it

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

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

What the interviewer asks next

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

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

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

Try it first

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

Show the worked solution

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

What order should you test patterns in?

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

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

How does each of the three give way?

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

The relationship

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

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

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

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

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

Where candidates lose it

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

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

What the interviewer asks next

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

047Your fair value on a contract is 50 and you quote 49 at 51. You have been lifted until you are short 20 lots against a risk limit of 25. Where do you quote now, and why not simply widen?Market makingCoreMarket makingProp trading firms
Try it first
Short 20 of a 25-lot limit, with fair value still 50. What do you do with the quote?
Show the worked solution
Skew the whole quote up, to about 50 at 52, keeping the width of 2. The short is a risk you want to shed, so make your bid attractive to sellers and your offer less attractive to buyers. A simple rule moves the mid 0.05 per lot of inventory, so 20 lots short moves it up 1. Widening to 48 at 52 also stops the buying, but it pushes sellers away too, so the short stays on your book and you earn less while you wait.
What is the position telling you to do?
A fruit seller who has run short of mangoes by mid-morning does not shut the stall; she raises the price she pays suppliers and nudges up the price to customers, so more mangoes arrive and fewer leave. Inventory changes what you want to trade next, not what the contract is worth: with fair value still 50, a short of 20 lots means your next trade should be a buy, so the quote should lean towards buying. You have used 80% of the risk limit, and five more lifts would put you at it. The question is testing whether you separate the two numbers a market maker carries: the fair value, which has not moved, and the price at which you want to trade, which has.
Moving both sides up 0.05 per lot of short keeps the width at 2 while the quote climbs from 49 at 51 to 50 at 52 at 20 lots short, so sellers find your bid at fair value and buyers find a dearer offer, whereas widening to 48 at 52 pushes both sides away.
How far should you skew, and what does it cost?
A common rule moves the mid in proportion to the position: here 0.05 per lot, so 20 lots short lifts the mid by 1 and the quote becomes 50 at 52. The bid now sits at fair value, so you buy back with no edge, and the offer is 2 above fair, so a buyer who still lifts it pays you well for adding to the risk. Skewing trades some edge for risk reduction, and the closer the position is to the limit, the more edge you should be willing to give up. In a toy model where a quote d away from fair trades with probability 0.6 x e to the minus d each period, the unchanged quote earns 0.44 a period and never reduces the short; the full skew earns 0.16 but buys back a net 0.52 lots a period, about 39 periods to flatten; the half skew sits between at 0.38 and 0.23. The numbers are illustrative; the direction is not.
Quote Bid fill Offer fill Net lots bought Edge a period
49 at 51, unchanged 0.22 0.22 +0.00 0.44
49.5 at 51.5, half skew 0.36 0.13 +0.23 0.38
50 at 52, full skew 0.60 0.08 +0.52 0.16
48 at 52, widened 0.08 0.08 +0.00 0.32
In the toy fill model only a skewed quote buys back the short; the widened quote earns more per fill but leaves the position where it is.
Why not simply widen?
Widening to 48 at 52 makes the offer as unattractive as the skew does, but it also moves the bid two points below fair, so sellers go elsewhere. In the toy model both sides fill 0.08 of the time and the expected change in the position is zero: a wider quote stops new risk arriving but does nothing about the risk you already hold, and the 20-lot short keeps moving with the market while you wait. Widening has its place, when you think fair value is uncertain or that the people lifting you know something, because then both sides are dangerous. Say that distinction, and then say the last resort: if skewing does not bring sellers fast enough, hedge the short in a related market, or cross the spread and buy, rather than drift up to the limit.
Where candidates lose it
The common loss is widening, because it feels cautious. It protects against new risk, but the 20 lots you already hold are the problem, and a wide quote does not bring sellers to buy them back.
The second is moving fair value. Nothing about the contract has changed; only your position has. Keep 50 as fair, move the quote, and say why the two are now different numbers.
What the interviewer asks next
You suspect the buyers lifting you know something. Does that change skew into widen?
You reach the 25-lot limit. What do you do with the offer?
How would you choose the skew per lot of inventory?
A related contract is liquid and moves with yours. How does that change your quote?
050You can buy the 90 call, sell the 110 call, buy the 110 put and sell the 90 put, all European and expiring in one year, for a net 19.20. What does the position pay at expiry, what interest rate does it imply, and when is that attractive?Option payoffs and no-arbitrageHardMarket makingRates derivatives
Try it first
What does the four-leg position pay at expiry?
Show the worked solution
It pays exactly 20 at any stock price, so 19.20 today implies a one-year rate of 20/19.20 - 1, about 4.17%. The 90 call and short 90 put make a long forward at 90; the short 110 call and long 110 put make a short forward at 110. The stock cancels and 110 - 90 = 20 is left. Buying the box is lending at 4.17% and selling it is borrowing at that rate, so it suits a lender whose other return is lower, or a borrower whose funding costs more.
Why does the payoff not depend on the stock?
Agree to buy a scooter from one friend for 90 and to sell it to another for 110, both next year, and you will make 20 whatever scooters cost by then. The box is the same pair of agreements: long the 90 call and short the 90 put is a promise to buy at 90, and short the 110 call and long the 110 put is a promise to sell at 110, so the stock comes in and goes out and only the gap between the strikes, 20, is left. Seen as spreads, the call spread pays from 0 below 90 up to 20 above 110, and the put spread pays the mirror image, 20 below 90 down to 0 above 110. Wherever the stock ends, the two add to 20.
The long call spread rises from 0 to 20 between 90 and 110 while the long put spread falls from 20 to 0 over the same range, so their sum is a flat 20 at every stock price, and paying 19.20 for it is lending at 4.17%.
How do you turn the price into a rate?
You pay 19.20 today and receive 20.00 in a year with no market risk, so it is a deposit. The simple rate is 20 / 19.20 - 1 = 4.17%, and the continuously compounded rate is ln(20 / 19.20) = 4.08%. A box is a zero-coupon bond built from options, and its price is the strike gap times the discount factor, so any box that trades away from the market's interest rate is mispriced. Prices consistent with this rate, for an illustrative stock at 100 with 20% volatility, are a 90 call at 16.11, a 110 call at 5.69, a 110 put at 11.29 and a 90 put at 2.51: 16.11 - 5.69 + 11.29 - 2.51 = 19.20. Volatility does not enter the box price at all, because every volatility effect in the calls is cancelled by the puts.
The relationship
$\underbrace{(S - 90)^+ - (90 - S)^+}_{S - 90} \; - \; \underbrace{\big[(S - 110)^+ - (110 - S)^+\big]}_{S - 110} = 20, \qquad r = \frac{20}{19.20} - 1 = 4.17\%$
(S - K)+ a call payoff struck at K
(K - S)+ a put payoff struck at K
call minus put at the same strike a forward to buy at that strike, paying S - K
r the simple one-year rate implied by paying 19.20 for 20
What it says in wordsA long forward at 90 and a short forward at 110 leave a fixed 20, and its price gives the interest rate.
When is it attractive, and what can go wrong?
Compare 4.17% with your own rates. If cash would otherwise earn 3.5%, buying the box lends at a better rate: 19.20 at 3.5% grows to only 19.87, against 20 from the box. If your funding costs 5%, selling the box borrows more cheaply: you receive 19.20 today and owe 20, while 20 owed at 5% would have raised only 19.05. Whether a box is cheap or dear is never a property of the box alone; it depends on the rate you can otherwise lend or borrow at, which is why boxes are a funding trade. The risks to name are practical. American options can be exercised early, which breaks the box, so use European index options. Four bid-offer spreads and fees can eat the 0.80 of interest. And the counterparty, or the clearing house, must still be there in a year.
Where candidates lose it
The common loss is analysing the four legs as a view on the stock, describing a bull call spread and a bear put spread and forgetting to add them. Add them: the stock cancels and the position is a loan.
The second is calling a box at 19.20 an arbitrage on its own. It is only cheap or dear against a rate, so say which rate you are comparing with, your deposit rate if you buy and your funding rate if you sell.
What the interviewer asks next
The same box trades at 19.80. What rate does that imply, and who would sell it?
Why does an American-style box carry early-exercise risk, and which leg is the danger?
Build a box with strikes 95 and 105. What should it cost at the same rate?
How is a box related to put-call parity at each strike?

S	the stock price, 1,000
sigma sqrt T	the volatility scaled to the option's life: 20% x 0.5 = 10%
0.4	the approximation to 1 over root 2 pi, which is 0.3989
N	the standard normal distribution function

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

Try it first

Why does a put costing more than a call point to a dividend?

What would you check before trading on it?

What is the general lesson the interviewer wants?

Where candidates lose it

What the interviewer asks next

Try it first

Where does the 0.4 come from?

Why is the straddle just double, and when is it not?

How far can you push the rule?

Where candidates lose it

What the interviewer asks next

Try it first

Why do you add variances and not volatilities?

What does the number tell a trader?

What is the limit of the calculation?

Where candidates lose it

What the interviewer asks next

Try it first

Why work backwards from the last coin?

How do you find the period without listing every position?

What changes if the last coin loses instead of wins?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the first red change which box you think you hold?

What has to be updated besides the belief about the box?

How do the variants separate candidates who understand from those who memorised?

Where candidates lose it

What the interviewer asks next

Try it first

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

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

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

Where candidates lose it

What the interviewer asks next

Try it first

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

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

Where does the no-arbitrage check come in?

Where candidates lose it

What the interviewer asks next

Try it first

What order should you test patterns in?

How does each of the three give way?

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

Where candidates lose it

What the interviewer asks next

Try it first

What is the position telling you to do?

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

Why not simply widen?

Where candidates lose it

What the interviewer asks next

Try it first

Why does the payoff not depend on the stock?

How do you turn the price into a rate?

When is it attractive, and what can go wrong?

Where candidates lose it

What the interviewer asks next