Derivatives Foundation puzzles, solved step by step

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040Five rational pirates, A the most senior down to E, split 100 gold coins. The most senior proposes a split; it passes if at least half of the pirates, his own vote included, agree, otherwise he is thrown overboard and the next most senior proposes. Each pirate wants to survive, then to get the most coins, then to see others thrown overboard. How should A split the coins to survive and keep the most?Games and logicHardOld Mission CapitalNew York · 2014

Try it first

How many coins can A keep?

Show the worked solution

A proposes 98 for himself, 0 for B, 1 for C, 0 for D and 1 for E. Work backwards. With two pirates, D keeps 100, since his own vote is half. With three, C keeps 99 and gives E 1, because E gets nothing if C dies. With four, B keeps 99 and gives D 1. With five, A needs two other votes and buys the cheapest: C and E, who get nothing under B's plan, accept one coin each. A's vote plus two makes three of five.

Why does the answer come from the end of the game, not the start?

When you negotiate a price you think about what the other side does if you walk away; what they do next determines what they will accept now. Each pirate compares the offer in front of him with what he would get under the next proposal, so to know what A must offer you first need to know what B would do, which needs C, which needs D with two pirates left. With two pirates, D proposes 100 for himself; his own vote is one of two, which is half, so it passes and E gets nothing. That is the anchor. Every earlier step is a question of who gets zero in the next round, because those are the pirates whose votes are cheapest to buy.

Working up from two pirates, D keeps 100, C keeps 99 by paying E one coin, B keeps 99 by paying D one coin, and A keeps 98 by paying C and E one coin each, the two pirates who would receive nothing under B's proposal.

How does each proposer decide whose vote to buy?

He buys exactly the number of extra votes he needs, from the pirates who are cheapest, and a pirate is cheap if the next proposal gives him nothing. With three pirates, C needs one more vote; under D's plan E gets zero, so C offers E one coin and keeps 99, and E accepts because one is better than zero. With four, B needs one more; under C's plan D gets zero, so B pays D one coin. With five, A needs two more votes, and under B's plan the pirates with nothing are C and E, so A pays them one coin each and keeps 98; offering anything to B or D is wasted, because B would get 99 and D one coin without A, and neither can be bought for less. The alternation is the pattern to say out loud: each round, the pirates who were paid last time are the ones left out this time.

The relationship

\text{2: }(100, 0) \;\to\; \text{3: }(99, 0, 1) \;\to\; \text{4: }(99, 0, 1, 0) \;\to\; \text{5: }(98, 0, 1, 0, 1)

each tuple	coins to the proposer first, then to the others in order of seniority
the 1s	votes bought for one coin from pirates who would get zero in the next round
votes needed	half of the pirates, rounded up, including the proposer: 1 of 2, 2 of 3, 2 of 4, 3 of 5

What it says in wordsEach proposer keeps everything except one coin for each extra vote he needs, bought from whoever the next round would leave empty-handed.

What assumptions is the answer resting on, and what happens when they change?

Three assumptions, and the interviewer will test at least one. Pirates are perfectly rational and know the others are. A pirate who is indifferent between two outcomes prefers the one where a rival is thrown overboard, which is why a bought pirate must be paid one coin and not zero. And the voting rule is at least half, with the proposer voting. Change the rule to a strict majority of all votes and the two-pirate case flips, D cannot pass anything and dies, so E would reject everything in the three-pirate round, and the whole ladder shifts. With a strict majority of the other pirates' votes the ladder shifts again, and a candidate who re-derives it from the two-pirate anchor is the one who gets hired. The limitation is that real negotiators are not this rational; the puzzle is a model of backward induction, not of pirates.

Where candidates lose it

The common loss is reasoning forward from fairness, offering 20 each or 50 for A, and being unable to say why those numbers rather than others. Fairness is not in the rules; survival and coins are.

The second is paying the wrong pirates. B and D would do well without A and cannot be bought cheaply; C and E would get nothing. Candidates who pay B and D have not worked out the four-pirate round.

What the interviewer asks next

The proposal now needs a strict majority of all votes. What does A propose?
The proposal needs a majority of the other pirates' votes, excluding the proposer. Does A survive?
There are 6 pirates. How does the pattern continue?
What if a pirate who is indifferent prefers to keep the proposer alive?

Asked at Old Mission Capital, Prop Trading, New York, 2014 (Wall Street Oasis): There are 5 pirates and they are trying to split 100 gold coins in a rational way.

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?

045Two players each ante Rs 1 and receive one card from an ace, a king and a queen. Player one may bet Rs 1 or check, and a check goes straight to showdown. Facing a bet, player two may call or fold. How often should player one bluff with the queen, and how often should player two call with the king?Games and logicHardOld Mission CapitalNew York · 2022

Try it first

How often should player one bet the queen?

Show the worked solution

Player one bets the ace always, checks the king, and bluffs the queen one time in three; player two calls with the ace, folds the queen, and calls with the king one time in three. The bluff rate makes one bet in four a bluff, which is the break-even for a call of 1 into a pot of 3. The king's call rate makes player two fold one time in three overall, the break-even for a bluff of 1 into a pot of 2. Player one gains 1/18 of a rupee a hand.

Which decisions are obvious, and which one is the real question?

Start by removing the choices nobody would make. Player one always bets the ace, since it wins every showdown, and player two always calls with the ace and folds with the queen, since those cards win or lose every time; the only real decisions are player one's queen and player two's king. Player one's king should check: if it bets, player two calls only with the ace and folds the queen, so the bet loses 2 half the time and wins 1 half the time, an average of -0.5, worse than the 0 a showdown gives. That leaves two numbers to find, the queen's bluff frequency and the king's calling frequency, and each is set to make the other player's choice a matter of indifference.

How do you find the bluffing frequency?

A goalkeeper who always dives left is easy to beat; a penalty taker mixes his side so the keeper gains nothing by guessing. Poker equilibrium works the same way. Player one bluffs just often enough that player two's king gains nothing by calling over folding, and player two calls just often enough that player one's queen gains nothing by bluffing over checking. Facing a bet with the king, folding loses the ante, -1. Calling wins 2 against a bluff and loses 2 against the ace. With the ace always bet and the queen bet a fraction b of the time, a bet is a bluff with probability b / (1 + b). Setting the call's value equal to -1 gives b = 1/3, so one bet in four is a bluff.

The relationship

\text{King indifferent: } \tfrac{b}{1+b}(+2) + \tfrac{1}{1+b}(-2) = -1 \Rightarrow b = \tfrac13, \qquad \text{Queen indifferent: } \tfrac12(-2) + \tfrac12\big(c(-2) + (1-c)(+1)\big) = -1 \Rightarrow c = \tfrac13

b	how often player one bets the queen
c	how often player two calls with the king
-1	the value of the alternative: folding the king, or checking the queen, each loses the ante
1/2	given the queen, player two holds the ace or the king with equal chance

What it says in wordsEach player's frequency is chosen so that the other player's two choices are worth the same.

Player one bets the ace, checks the king and bluffs the queen one time in three, while player two calls with the ace, folds the queen and calls with the king one time in three, because those frequencies leave each player's marginal choice worth exactly the same either way.

How do you sanity check the frequencies, and who wins the game?

Use pot odds as the check. Player two calls 1 to win a pot of 3, two antes and the bet, so a call breaks even when bluffs are 1 / (3 + 1) of bets, a quarter, which is what b = 1/3 delivers. Player one bluffs 1 to win the pot of 2, so a bluff breaks even when player two folds 1 / (2 + 1) of the time; he folds the queen always and the king two times in three, which averages to one in three. Average over the six deals and player one gains 1/18 of a rupee a hand, because only he can bet; neither player can improve on that by changing his own frequency. Against a player two who never calls with the king, bluffing every queen would earn 1/6 of a rupee a hand instead of 1/18. The limitation is that equilibrium is a defence, not the most profitable play against a predictable opponent; a desk uses it as the baseline and then leans towards the other side's mistakes.

Where candidates lose it

The common loss is saying player one should never bluff because the queen cannot win a showdown. Without bluffs player two simply folds his king to every bet, and player one's ace stops getting paid.

The second is giving the bluff rate as a quarter. A quarter is the share of bets that are bluffs; because the ace is always bet, that needs the queen bet one time in three. Keep the two fractions apart and say which one you mean.

What the interviewer asks next

Player two may also bet after player one checks. How does the equilibrium change?
The bet size doubles to Rs 2. What are the new bluffing and calling frequencies?
Player two never calls with the king. What is player one's best response, and what does it earn?
Why does player one never want to bet the king?

Asked at Old Mission Capital, Quantitative Research, New York, 2022 (Wall Street Oasis): Asking to find the game theory optimal strategy in a simplified poker game

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?
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?
057The correlation of X and Y is 0.2 and the correlation of Y and Z is 0.5. What range of values can the correlation of X and Z take?Volatility and correlationHardTower Research CapitalNew York · 2019
Try it first
Two correlations are given. Before any algebra: is the third one free to be anything between minus 1 and 1?
Show the worked solution
Between about -0.75 and 0.95. The three correlations must form a positive semi-definite matrix, and for three variables that means its determinant 1 - a^2 - b^2 - c^2 + 2abc cannot be negative. Solving for c with a = 0.2 and b = 0.5 gives c = ab plus or minus sqrt((1 - a^2)(1 - b^2)) = 0.1 plus or minus sqrt(0.96 x 0.75) = 0.1 plus or minus 0.849.
Why can the third correlation not be anything it likes?
If Amit is tall when Bela is tall, and Bela is tall when Chitra is tall, then Amit and Chitra cannot be perfectly opposite; the middle person ties them together. Three pairwise correlations describe one joint set of three variables, and a joint set has a variance for every combination of them that cannot be negative, which is the positive semi-definite condition on the correlation matrix. With weak links, 0.2 and 0.5, the tie through Y is loose and the band is wide; with links of 0.9 and 0.9, X and Z would be forced to be strongly positively correlated.
The correlation of X and Z can only sit in the green band from -0.75 to 0.95, centred on 0.2 x 0.5 = 0.1, because outside it the determinant of the 3 x 3 correlation matrix, plotted below the axis, turns negative and no three variables can have those correlations.
How do you get the band without writing out eigenvalues?
Use the determinant. For a 3 x 3 correlation matrix with off-diagonal entries a, b and c, the determinant is 1 - a^2 - b^2 - c^2 + 2abc, and the matrix is valid exactly when that is at least zero. Treat it as a quadratic in the unknown c: c^2 - 2ab c + (a^2 + b^2 - 1) must be at most zero, so c lies between the two roots ab plus or minus sqrt((1 - a^2)(1 - b^2)). With a = 0.2 and b = 0.5 the centre is 0.1 and the half-width is sqrt(0.96 x 0.75) = sqrt(0.72), about 0.849. The geometric reading is the same thing: think of each variable as a unit vector, correlations as cosines of the angles between them, and the third angle is bounded by the sum and difference of the first two.
The relationship
$\rho_{XZ} \in \left[\rho_{XY}\rho_{YZ} - \sqrt{(1-\rho_{XY}^2)(1-\rho_{YZ}^2)},\; \rho_{XY}\rho_{YZ} + \sqrt{(1-\rho_{XY}^2)(1-\rho_{YZ}^2)}\right] = [0.1 - 0.849,\; 0.1 + 0.849]$
rho_XY, rho_YZ the two given correlations, 0.2 and 0.5
rho_XZ the correlation being bounded
the square root term the half-width of the feasible band, which shrinks as the given correlations strengthen
What it says in wordsThe third correlation sits within a band centred on the product of the other two, with a width that vanishes only when the other two are plus or minus one.
Where does this show up on a desk?
In any correlation matrix that was assembled by hand or from mismatched histories. A desk that marks the stock-to-index correlation at 0.9, the index-to-sector correlation at 0.9, and the stock-to-sector correlation at 0.3 has written down a matrix that no world can produce, and a basket option priced from it will give nonsense, often a negative variance for some combination. The fix is to project the matrix back to the nearest valid one before pricing. The limitation of the puzzle is that it covers only three variables; with more, every principal sub-matrix must pass, and the feasible set has no simple closed form.
Where candidates lose it
The common answer is that the third correlation is unconstrained, or that it must equal 0.1. Both miss that three variables share one joint distribution, and that its correlation matrix must be valid.
The second loss is knowing the condition and failing to turn it into a number. Have the determinant formula ready, solve the quadratic in the unknown, and quote the band to two decimals.
What the interviewer asks next
Now the two given correlations are 0.9 and 0.9. What is the band?
Can the X Z correlation be negative if X Y and Y Z are both 0.8?
Explain the geometric picture using angles between unit vectors.
What does a desk do when its estimated correlation matrix fails this test?
Asked at Tower Research Capital, Prop Trading, New York, 2019 (Wall Street Oasis): What if the correlation between X and Y is 0.2 and the correlation between Y and Z is 0.5. What is the range for the correlation of X and Z?
060A bowl holds 100 cooked noodles. You repeatedly pick two free ends at random and tie them together, until no free ends remain. What is the expected number of loops in the bowl?Probability and countingHardD.E. ShawNew York · 2026
Try it first
100 noodles, 100 ties. Roughly how many loops do you expect at the end?
Show the worked solution
About 3.28 loops. Each tie either closes a loop or joins two strands into one longer strand, so after every tie there is one strand fewer. With n strands left there are 2n free ends; pick one, and of the other 2n - 1 ends exactly one belongs to the same strand, so that tie closes a loop with chance 1/(2n - 1). Linearity of expectation adds those chances over n = 100 down to 1: 1/199 + 1/197 + ... + 1/3 + 1.
Why does every tie either close a loop or shorten the list by one strand?
Think of 100 pieces of string on a table. Tie two ends from different pieces and you now have 99 pieces, one of them longer. Tie the two ends of the same piece and you have a ring and 99 pieces left as well. Whatever happens, the number of strands with free ends falls by exactly one per tie, so there are exactly 100 ties, and the only question at each tie is whether it closed a loop. That makes the count of loops a sum of 100 indicator events, which is the signal to use linearity of expectation rather than to enumerate outcomes.
The chance that a tie closes a loop is 1/(2n - 1) with n strands left, so it is 1/199 at the first tie and stays near zero until the last handful, reaching 1/3 and then 1; the running expected total climbs slowly and reaches 3.28 only because the last ten ties alone contribute 2.13.
Where does 1/(2n - 1) come from?
With n strands there are 2n free ends. Pick the first end; it belongs to some strand. Of the remaining 2n - 1 ends, exactly one is the other end of that same strand, and all are equally likely, so the tie closes a loop with chance 1/(2n - 1). It does not matter how long the strands have become or how many loops already sit in the bowl, because closed loops have no free ends and are out of the picture. The expectation is therefore the sum of 1/(2n - 1) for n from 100 down to 1, which is the odd-denominator half of the harmonic series.
The relationship
$E[\text{loops}] = \sum_{n=1}^{100} \frac{1}{2n-1} = 1 + \tfrac{1}{3} + \tfrac{1}{5} + \cdots + \tfrac{1}{199} \approx 3.284 \approx \tfrac{1}{2}\ln(4n) + \tfrac{\gamma}{2}$
n the number of strands with free ends before a tie
1/(2n - 1) the chance that tie closes a loop
gamma the Euler constant, about 0.577, in the logarithmic approximation
What it says in wordsThe expected number of loops is the sum of the odd reciprocals up to 1/199, which grows only like half the logarithm of the number of noodles.
What is the approximation, and why is the answer so small?
The sum of odd reciprocals up to 1/(2n - 1) is close to half of ln(4n) plus half the Euler constant, which for n = 100 gives 3.28 against the exact 3.284. Doubling the number of noodles adds only about 0.35 to the expected number of loops, so a bowl of a thousand noodles still gives only about four loops. The intuition is that early ties almost never close a loop: they build a few very long strands, and the loops appear at the end when there are only two or three strands left. The limitation is that this is an expectation only; the distribution is skewed, and ending with a single loop is the most likely outcome.
Where candidates lose it
Most candidates try to track the configuration of strands, which explodes. The interviewer wants the one observation that each tie closes a loop with chance 1/(2n - 1) regardless of history, and then linearity of expectation.
The second loss is guessing a large number. Say out loud that the early ties almost never close a loop, so the answer is a slowly growing sum, and that the harmonic-style sum of 100 terms is around 3, not 50.
What the interviewer asks next
What is the probability that all 100 noodles end up in a single loop?
Approximately how many noodles would you need for the expected number of loops to reach 5?
What is the variance of the number of loops?
Now you tie ends only across different strands when you can. How many loops then?
Asked at D.E. Shaw, Research, New York, 2026 (Wall Street Oasis): What is the expected number of loops from tying 100 noodles' ends together randomly
062I am going to roll a die six times. Make me a market on the number of different faces that show up.Market makingHardOptiverSan Francisco · 2026
Try it first
Before quoting: where is the fair value of the number of distinct faces?
Show the worked solution
Fair value is about 3.99, so quote something like 3.9 bid, 4.1 offer. A given face is absent from all six rolls with probability (5/6) to the sixth, about 0.335, so it appears at least once with probability 0.665. The number of distinct faces is the sum of six such indicators, and linearity of expectation gives 6 x 0.665 = 3.99, without listing a single case.
Why does linearity let you skip the cases?
Six friends each toss a letter into one of six boxes at random, and you want the expected number of boxes that end up non-empty. Counting the ways the boxes can fill is a mess; asking each box whether it got anything is easy. The number of distinct faces is one plus one plus one over the six faces, each one counting if that face appeared, and the expectation of a sum is the sum of the expectations even though the six events overlap. Each face is missed on every roll with chance (5/6) to the sixth, so the expected count is 6 times one minus that, about 3.99.
Each of the six faces appears at least once with probability 0.665, so the expected number of distinct faces is 6 x 0.665 = 3.99, and the exact distribution peaks at four faces with standard deviation 0.78, which is what a market of 3.9 at 4.1 is priced around.
How wide should the market be, and how do you defend it?
Width comes from how uncertain the outcome is and how much the other side may know. The exact distribution puts about 50% of the mass on four faces, 23% on three and 23% on five, with a standard deviation of 0.78, so a market 0.2 wide around 3.99 is tight relative to the noise but still symmetric around fair. If the interviewer lifts your offer at 4.1, you are short at a price above fair and should keep the quote where it is; if they lift it twice, ask yourself whether they know something, such as that the die is loaded, and move the market up rather than argue with the flow.
The relationship
$E[D] = \sum_{i=1}^{6} P(\text{face } i \text{ appears}) = 6\left(1 - \left(\tfrac{5}{6}\right)^6\right) \approx 6 \times 0.6651 = 3.991$
D the number of distinct faces in six rolls
(5/6)^6 the chance a particular face is missed on all six rolls
6 the number of faces, each contributing one indicator
What it says in wordsThe expected number of distinct faces is six times the chance that any one face appears at least once.
What is the check, and what changes with more rolls?
Check the ends. With one roll there is exactly one distinct face and the formula gives 6(1 - 5/6) = 1. With many rolls the missed chance collapses and the expectation approaches 6. At six rolls you are at 3.99, meaning two faces are typically missing, which surprises people who expect six rolls to nearly cover six faces. The limitation of the quote is that it prices only the mean; a counterparty who wants to bet on exactly six distinct faces needs a different market, and that chance is only 6!/6^6, about 1.5%.
Where candidates lose it
Candidates start enumerating outcomes, or quote 6 because there are six rolls. The interviewer is looking for the indicator trick: one event per face, sum the probabilities, no cases.
The second loss is a market with no reasoning behind its width. Say the standard deviation, say the market is symmetric around fair, and say what you would do when the other side trades with you twice in the same direction.
What the interviewer asks next
What is the probability that all six faces appear in six rolls?
Make a market on the number of distinct faces in twelve rolls.
The interviewer hits your bid three times in a row. What do you do with the quote?
What is the variance of the number of distinct faces, and does it matter for the quote?
Asked at Optiver, Quant Research Interview, San Francisco, 2026 (Wall Street Oasis): They do ask one round of market making game-like question
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?
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?

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

rho_XY, rho_YZ	the two given correlations, 0.2 and 0.5
rho_XZ	the correlation being bounded
the square root term	the half-width of the feasible band, which shrinks as the given correlations strengthen

n	the number of strands with free ends before a tie
1/(2n - 1)	the chance that tie closes a loop
gamma	the Euler constant, about 0.577, in the logarithmic approximation

D	the number of distinct faces in six rolls
(5/6)^6	the chance a particular face is missed on all six rolls
6	the number of faces, each contributing one indicator

Try it first

Why does the answer come from the end of the game, not the start?

How does each proposer decide whose vote to buy?

What assumptions is the answer resting on, and what happens when they change?

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

Which decisions are obvious, and which one is the real question?

How do you find the bluffing frequency?

How do you sanity check the frequencies, and who wins the game?

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

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 can the third correlation not be anything it likes?

How do you get the band without writing out eigenvalues?

Where does this show up on a desk?

Where candidates lose it

What the interviewer asks next

Try it first

Why does every tie either close a loop or shorten the list by one strand?

Where does 1/(2n - 1) come from?

What is the approximation, and why is the answer so small?

Where candidates lose it

What the interviewer asks next

Try it first

Why does linearity let you skip the cases?

How wide should the market be, and how do you defend it?

What is the check, and what changes with more rolls?

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 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