Portfolio Management puzzles, solved step by step

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  1. 043You have n cars, each fuelled to drive exactly 1,000 miles, and fuel can be moved from one car to another along the way. Tanks cannot be overfilled. How far can one car get, and how does that distance grow as n becomes very large?Logic brainteasersHardMillennium ManagementLondon · 2024

    Try it first

    With four cars, how far can one car get?

    Show the worked solution

    1,000 x (1 + 1/2 + 1/3 + ... + 1/n) miles, which grows without limit but only like the logarithm of n. Drive all n cars 1,000/n miles; together they have burned one full tank, so one car can refill the rest and be abandoned. Repeat with n minus 1 cars for 1,000/(n minus 1) miles, and so on. Four cars reach 2,083 miles, 100 cars about 5,187, and the distance tracks 1,000 x (ln n + 0.577).

    When should a car drop out of the convoy?

    Think of friends sharing water on a long walk, where every bottle is full at the start and nobody can carry more than one. The moment the group has drunk exactly one bottle's worth, one friend can pour the rest of theirs into everyone else's bottles and head home. A car should drop out the instant the convoy has burned exactly one tank in total, because that is the first moment its remaining fuel exactly fills the others. With n cars that happens after 1,000/n miles, since n cars burn fuel n times as fast as one.

    Each extra car adds a shorter leg: 1000/n, then 1000/(n-1), and so on1000/4 = 2501000/3 = 3331000/2 = 5001000/1 = 1,000Car 4tops up, left emptyCar 3tops up, left emptyCar 2tops up, left emptyCar 12,083 miles1,0003,0005,0001255075100Number of carsMiles reachable10 cars: 2,929100 cars: 5,187
    Four cars travel 250, 333, 500 and 1,000 miles in successive legs as one car at a time tops up the rest and drops out, reaching 2,083 miles. The distance with n cars keeps growing but ever more slowly, from 2,929 miles with 10 cars to 5,187 with 100.

    Why does the distance grow only like a logarithm?

    The total is 1,000 times the harmonic sum 1 + 1/2 + ... + 1/n. Each extra car adds the shortest leg of the journey, 1,000/n miles, so the gains shrink as the convoy grows, and the harmonic sum rises like ln n plus about 0.577. It never stops growing, so any distance is reachable in principle, but slowly: 10 cars reach about 2,929 miles, and getting past 5,000 miles takes 83 cars. Doubling the fleet adds only about 1,000 x ln 2, roughly 693 miles.

    The relationship
    D(n)=1000∑k=1n1k≈1000 (ln⁡n+0.577)D(n) = 1000\sum_{k=1}^{n}\frac{1}{k} \approx 1000\,(\ln n + 0.577)
    nthe number of cars at the start
    1000/kthe leg driven while k cars remain
    0.577the Euler-Mascheroni constant
    What it says in wordsEach leg is one tank shared among the cars still running, and the legs add up to a harmonic series.

    Why a hedge fund asks it: the structure is the same as scaling a strategy. Each extra unit of capital or effort buys a smaller increment, and a candidate who sees the diminishing returns and names the rate of decay is showing the instinct the desk wants. Also be ready to argue optimality in one sentence: any plan that drops a car earlier wastes fuel it cannot hand over, and dropping later wastes the fuel spent carrying a car that is no longer needed.

    Where candidates lose it

    The quick wrong answers are n times 1,000, as if all the fuel could be pooled into one car, or a flat 1,000 because tanks cannot be overfilled. Both skip the key idea that the convoy itself consumes fuel while carrying the reserve.

    Work n equals 2 out loud first: drive 500, pour the rest of car 2 into car 1, and drive 1,000 more, for 1,500. Then generalise. The interviewer wants the harmonic series and the words grows like log n.

    What the interviewer asks next

    • Roughly how many cars do you need to travel 10,000 miles?
    • What changes if cars can come back to a depot and cache fuel along the road?
    • Where do you see diminishing returns of this shape in portfolio construction?

    Asked at Millennium Management, Investments, London, 2024 (Wall Street Oasis): Suppose you have n cars, each fueled so that they can drive for 1000 miles.

  2. 066Five pirates, ranked A to E by seniority, must split 100 gold coins. The most senior pirate proposes a split and everyone votes. If at least half the votes, including his own, are in favour, the split stands; otherwise he is thrown overboard and the next pirate proposes. Pirates are perfectly rational, want to survive first and maximise coins second, and vote no when indifferent. What does A propose?Logic brainteasersHardHedge fundsQuantitative asset management

    Try it first

    How many coins does A keep?

    Show the worked solution

    A proposes 98 for himself, 0 for B, 1 for C, 0 for D and 1 for E. Solve from the end. With two pirates, D keeps all 100 because his own vote is half. With three, C buys E for 1 coin. With four, B buys D for 1. With five, A needs two votes and buys the two pirates who get nothing under B's plan, C and E, for one coin each, keeping 98.

    Where do you start?

    At the end, where there is no choice left. Think of planning a train journey with connections: you start from the time you must arrive and work back to when you must leave. A sequential game is solved backwards, because each pirate's vote depends only on what he would get if the current proposal failed. With two pirates left, D proposes 100 for himself; his own vote is half, which is enough. So E gets nothing if it ever comes to that, and E knows it.

    Solve it backwards: each proposer buys the pirates who would get nothing nextPirate APirate BPirate CPirate DPirate Esenior to junior5 pirates leftneeds 3 votes9801014 pirates leftneeds 2 votesgone990103 pirates leftneeds 2 votesgonegone99012 pirates leftneeds 1 votesgonegonegone1000readupwarda vote bought with 1 coin: this pirate gets 0 in the row belowthe proposer, who votes for himself
    Read the grid from the bottom up: each proposer keeps everything except one coin for each vote he needs, and he buys the pirates who would get nothing in the row below, so A ends with 98 and pays C and E one coin each.

    How does each step follow from the one below?

    With three pirates, C needs two votes, his own and one more. E gets 0 if C dies, so one coin buys E: C proposes [99, 0, 1]. With four, B needs two votes; under C's plan D gets 0, so B buys D for one coin: [99, 0, 1, 0]. A vote is worth exactly one coin more than that pirate's fallback, so a proposer always buys the cheapest voters, the ones left with nothing in the next round. With five, A needs three votes. Under B's plan C and E get nothing, so one coin each buys them, and A proposes [98, 0, 1, 0, 1].

    State the assumptions, because the answer rests on them. If an indifferent pirate voted yes, A could buy votes for zero coins. If the rule needed a strict majority, the counts change. Interviewers often change one rule as a follow-up to see whether you rebuild the chain or reach for a memorised answer. The buy-side lesson is about incentives: what someone will accept depends on their alternative, not on fairness.

    Where candidates lose it

    The trap is reasoning forwards from fairness, proposing an even split or generous bribes to the next in line. Without the backward chain you cannot know who is cheap to buy, and B, the obvious ally, is in fact the most expensive vote because he inherits the power if A dies.

    The second loss is skipping the stated assumptions. Say that indifferent pirates vote no and that exactly half passes; they decide whether the bribe is one coin or zero.

    What the interviewer asks next

    • What if a proposal needs a strict majority rather than half?
    • With the same rules, what happens with 200 pirates and 100 coins?
    • Where do you see the same logic, what someone accepts depends on their outside option, in a debt restructuring?