Quant puzzles, solved step by step

Puzzles
100
Traced to a firm
71
Topics
12
Hard
30
Topic
All topicsLogic and algorithmic reasoning10Conditional probability and Bayes7Counting and combinatorics8Continuous and geometric probability9Correlation, regression and linear algebra9Market making, betting and sizing9Expected value and optimal stopping9Statistics and estimation9Pricing, options and index maths7Games and strategic reasoning8Markov chains and random walks7Mental maths and number sense8
Level
AnyWarm upCoreHard
Source
AnyReported at a firmStandard
Showing 1–2 of 2 · filtered from 100Clear filters
  1. 021A contract pays max(X - 3, 0) rupees, where X is one roll of a fair die. What is its fair value? What is a contract paying max(4 - X, 0) worth?Pricing, options and index mathsWarm upBelvedere TradingChicago · 2021

    Try it first

    What is the call paying max(X - 3, 0) worth?

    Show the worked solution

    Both are worth Rs 1. The call pays 0, 0, 0, 1, 2 and 3 on faces 1 to 6, which sum to 6, so its average payoff is 6/6 = Rs 1. The put pays 3, 2, 1, 0, 0 and 0, also summing to 6, so it is worth Rs 1 too. They match because the die is symmetric: face x and face 7 - x are equally likely, which turns one payoff into the other.

    Why is the price just an average of payoffs?

    If a friend offers you a game where you win the number of rupees shown on a die above 3, playing a hundred times earns about Rs 100: some rolls pay nothing, some pay 1, 2 or 3. With no interest and no risk premium in a dice game, the fair value of any payoff is its average over the equally likely outcomes. That is exactly how an option is priced in the simplest world: list the states, write the payoff in each, weight by the probabilities and add.

    Price an option on a die by averaging its payoff over six facesCall struck at 3: pays max(X - 3, 0)000123Sum 6 over 6 faces = worth 1Put struck at 4: pays max(4 - X, 0)321000Sum 6 over 6 faces = worth 1Parity at strike 3: call 1 - put 0.5 = 0.5 = average face 3.5 - strike 3
    A call struck at 3 on one die roll pays 0, 0, 0, 1, 2, 3 and a put struck at 4 pays 3, 2, 1, 0, 0, 0; both payoffs add to 6 over six faces, so each is worth Rs 1, and at a common strike of 3 the call minus the put equals the average face less the strike.

    Why is 0.50 the tempting wrong answer?

    Because it plugs the average face, 3.5, into the payoff: 3.5 - 3 = 0.5. An option's value is the average of the payoff, not the payoff at the average, and because the payoff is floored at zero, the average payoff is always at least the payoff at the average. The floor throws away the losing faces, which is the whole point of owning an option. That gap, here Rs 0.50, is what traders pay for, and it grows with how spread out the outcome is: this is convexityA payoff that curves upward, so averaging over uncertain outcomes gives more than the payoff at the average outcome. in a single roll.

    The relationship
    C3=16(0+0+0+1+2+3)=1P4=16(3+2+1+0+0+0)=1C3−P3=1−12=E[X]−3C_3 = \tfrac{1}{6}(0+0+0+1+2+3) = 1 \qquad P_4 = \tfrac{1}{6}(3+2+1+0+0+0) = 1 \qquad C_3 - P_3 = 1 - \tfrac12 = E[X] - 3
    C_3the call struck at 3
    P_4, P_3puts struck at 4 and at 3
    E[X]the average face, 3.5
    What it says in wordsEach option is the average of its payoff, and a call minus a put at the same strike is the forward, the average face less the strike.

    Where is put-call parity in this?

    Take a call and a put with the same strike, 3. Owning the call and selling the put pays max(X - 3, 0) - max(3 - X, 0) = X - 3 on every face, so its value must be the average of X - 3, which is 0.5. The put struck at 3 pays 2, 1, 0, 0, 0, 0, worth 1/2, and 1 - 0.5 = 0.5 as parity requires. Saying this unprompted shows the interviewer you see a dice game as a model of a real options book, which is why the question is asked.

    Where candidates lose it

    The trap is pricing the option at the average outcome, 3.5 - 3 = 0.5. It treats an option as a linear contract and ignores the floor, and it undervalues the call by half.

    The second loss is averaging only over the paying faces: 1, 2 and 3 average to 2. The three faces that pay nothing still happen half the time and must be in the average.

    What the interviewer asks next

    • What is the call worth if you may reroll once after seeing the first roll?
    • Price a call struck at 7 on the sum of two dice.
    • Quote a market in the call struck at 3 and say how you would hedge it.

    Asked at Belvedere Trading, Generalist, Chicago, 2021 (Wall Street Oasis): Pricing an option contract on a game involving rolling a die.

  2. 086A price-weighted index holds three stocks priced 50, 100 and 150, with a divisor of 3. The 150 stock splits 3 for 1. What is the new divisor, and how does a market-cap-weighted index handle the same split?Pricing, options and index mathsWarm upMizuhoHong Kong · 2024

    Try it first

    What must the new divisor be?

    Show the worked solution

    The new divisor is 2. Before the split the prices sum to 300, and 300 / 3 is 100. After a 3 for 1 split the 150 stock trades at 50, the sum is 200, and only a divisor of 2 keeps the index at 100. A cap-weighted index needs no adjustment at all, because a split triples the share count as it cuts the price to a third, leaving market value unchanged.

    Why must the divisor change when nothing about the company changed?

    Cut a pizza into twelve slices instead of four and you have not made more pizza. A stock split does the same to a company: three times the shares, each worth a third. A price-weighted index adds up share prices, so a split drops the sum even though no value was lost, and the divisor must be cut to stop a fake fall in the index. Solve for it by keeping the index level fixed: 200 divided by the new divisor must equal 100, so the divisor is 2.

    A split changes the sum of prices, so the divisor must change to hold the indexBefore: C at 15050stock Aweight 17%100stock Bweight 33%150stock Cweight 50%sum 300 / divisor 3 = index 100After a 3 for 1 split: C at 5050stock Aweight 25%100stock Bweight 50%50stock Cweight 25%sum 200 / divisor 2 = index 100split
    Before the split the prices sum to 300 and the index is 300 / 3 = 100; after C splits 3 for 1 the sum is 200, so the divisor falls to 2 to hold the index at 100, and C's weight falls from 50% to 25%.
    The relationship
    I=∑iPid3003=200d′⇒d′=2I = \frac{\sum_i P_i}{d} \qquad \frac{300}{3} = \frac{200}{d'} \Rightarrow d' = 2
    P_ithe price of stock i
    dthe divisor before the split, 3
    d'the divisor after the split
    What it says in wordsChoose the new divisor so the index is the same the moment after the split as the moment before.

    What else changes in a price-weighted index after the split?

    The weights. In a price-weighted index a stock's weight is its price over the sum of prices, so the expensive stock dominates whatever the size of the company. Before the split C carried 50% of the index; after it, C carries only 25% and B, untouched, jumps to 50%. A 10% rise in C used to add 5 index points; now it adds 2.5. Nothing about C's business changed; the index simply started caring less about it, which is the main criticism of price weighting.

    How do the other common methods treat the split?

    A market-cap-weighted index sums price times shares, and a 3 for 1 split multiplies shares by 3 while dividing price by 3, so the stock's market value, its weight and the index are all unchanged; no divisor adjustment is needed for a split. Cap-weighted divisors still change for events that alter total market value without a price move, such as share issuance, buybacks or a constituent being replaced. An equal-weighted index is also untouched by a split, since weights are reset to equal at each rebalance, but it has to trade at every rebalance to get back to equal, which costs money.

    Where candidates lose it

    The common slip is dividing the old divisor by the split ratio and answering 1. The divisor is fixed by keeping the index level unchanged, and only one stock split, so the adjustment is smaller than the ratio.

    The second loss is saying a cap-weighted index needs the same adjustment. Market value does not change in a split, so a cap-weighted index does nothing; say that, then name the events that do change its divisor.

    What the interviewer asks next

    • Stock B now pays a special dividend of 20. How does each index type handle it?
    • Replace stock A with a new stock priced 200. What is the new divisor?
    • Which stock has the most influence on a price-weighted index, and why is that a flaw?

    Asked at Mizuho, Sales and Trading, Hong Kong, 2024 (Wall Street Oasis): Different index methodology - need to know all of them with examples.