Portfolio Management puzzles, solved step by step

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100
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All topicsStatistics and forecasting9Portfolio risk maths10Logic brainteasers7Behavioural and decision traps7Probability and expected value8Bond maths10Valuation riddles8Performance measurement8Private and real asset maths8Funds, ETFs and implementation7Compounding and fee drag7Market sizing and estimation6Currency and global returns5
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  1. 030You have two ropes and a lighter. Each rope takes exactly 60 minutes to burn from one end to the other, but it burns unevenly, so half the length does not mean half the time. How do you measure exactly 45 minutes?Logic brainteasersWarm upAsset managementReal assets

    Try it first

    What does lighting a rope at both ends give you, if it burns unevenly?

    Show the worked solution

    Light rope 1 at both ends and rope 2 at one end, together. When rope 1 burns out, light the other end of rope 2; when rope 2 burns out, 45 minutes have passed. Rope 1 lasts 30 minutes because two flames share its 60 minutes of burning. At that moment rope 2 has 30 minutes left, and lighting its other end halves that to 15. 30 plus 15 is 45.

    Why does uneven burning not spoil the halving?

    Two people eating one plate of food from opposite sides finish it in half the time one person would take, whether the food is piled high on one side or spread evenly. They simply meet off-centre. A rope holds 60 minutes of burning in total, and two flames consume it twice as fast, so it is gone in 30 minutes wherever the flames happen to meet. The unevenness decides the place, never the time. That is the only fact the puzzle needs.

    Light both ends and time halves, however unevenly the rope burns0 min15 min30 min45 minRope 1both endsflames meet here, not in the middlegone at 30Rope 2one end,then bothone end: 30 min of burning usedboth ends30 left, halvedlight rope 2's other endRope 2 gone at45 minutesStopwatch
    Rope 1, lit at both ends, is gone after 30 minutes even though its flames meet away from the middle. Rope 2, lit at one end, has 30 minutes of burning left at that moment, and lighting its other end halves that to 15 minutes, ending at 45.

    How do you know rope 2 has exactly 30 minutes left at the half-hour?

    Rope 2 has been burning from one end for 30 minutes, so it has used 30 of its 60 minutes of material, whatever length that turned out to be. You never measure length; you only ever track time used and time left. Lighting the far end of what remains halves the 30 minutes left, and the rope finishes 15 minutes later. Say the timeline in that order and the answer is audible.

    Interviewers often extend it: with the same two ropes you can also time 15, 30, 60 and 90 minutes, each by deciding which ends are burning at which moment. Answering one extension shows you own the principle rather than a memorised trick.

    Why would a real asset desk ask this? It is a clean test of whether you separate the thing you can measure from the thing that is noisy. A rent roll is lumpy month to month; the annual total is what the valuation rests on. Saying that link in one sentence costs nothing.

    Where candidates lose it

    Candidates try to cut or fold the ropes, or to reason about lengths, which the uneven burning makes useless. The whole question is set up to see whether you let go of length and think only in minutes of burning.

    The second slip is lighting rope 2's second end at the start. Rope 2 must be lit at one end at minute 0 so that exactly 30 minutes of it are used up when rope 1 finishes.

    What the interviewer asks next

    • Using the same two ropes, how do you measure 15 minutes?
    • With three ropes, what is the longest time you can measure beyond 60 minutes?
    • Can you measure 20 minutes with two ropes? Why or why not?
  2. 053You have nine bags of coins that look identical. Eight weigh the same and one is lighter. Using a balance scale with no weights, what is the fewest number of weighings that always finds the light bag?Logic brainteasersWarm upAsset management

    Try it first

    What is the fewest weighings that is guaranteed to work?

    Show the worked solution

    Two weighings. Put three bags on each side and leave three aside. Whichever side rises holds the light bag; if the pans balance, it is among the three on the table. Take that group of three and weigh one bag against another: the one that rises is light, and if they balance, it is the third. A balance gives three outcomes per weighing, and three times three covers nine bags.

    Why is halving the pile the wrong instinct?

    Halving is what you would do with a question that answers yes or no, like guessing a number between 1 and 100. A balance tells you more than that. A balance scale has three outcomes, left pan rises, right pan rises, or level, so each weighing should split the suspects into three equal groups, not two. The group left on the table is not wasted: a level balance is information too. With four and four plus one aside you would learn little from a level result except that the odd bag is the spare one, and an unlucky run needs three weighings.

    A balance has three outcomes, so split the bags into three every timeWeighing 1: bags 1-3 vs 4-6bags 7-9 stay on the tableLeft side riseslight bag in 1, 2 or 3Balancedlight bag in 7, 8 or 9Right side riseslight bag in 4, 5 or 6Weighing 2: bag 7 vs bag 87 risesbag 7balancedbag 98 risesbag 8Same second stepunder the othertwo branches1 weighing:3 outcomes2 weighings:3 x 3 = 9
    Weighing three bags against three splits nine suspects into three groups of three, whichever way the pans move, and weighing one bag against one then names the light bag, so two weighings give nine end points, one for each bag.

    How do you prove two is the minimum?

    Count outcomes. One weighing produces at most three different results, and three results cannot point to nine different bags. With k weighings you can separate at most 3 to the power k bags, so nine bags need at least two weighings, and the method above shows two is enough. The same count answers the natural follow-up: three weighings handle up to 27 bags, four handle 81.

    The relationship
    3k≥n⇒32=9≥93^k \ge n \quad\Rightarrow\quad 3^2 = 9 \ge 9
    kthe number of weighings
    nthe number of bags, one of which is light
    3outcomes of one weighing: left light, right light, balanced
    What it says in wordsEach weighing multiplies the number of distinguishable outcomes by three, so you need enough weighings for the outcomes to cover every bag.

    Why would an asset manager ask this? It is a test of whether you use all the information a measurement gives you. A risk report, a performance attribution or a data screen can be read in more ways than pass and fail, and the candidate who spots the third outcome here is the one who reads the level result as a finding.

    Where candidates lose it

    The trap is answering three or four because you split into halves. It is the natural instinct, and the interviewer is watching whether you notice that a balance can come out level, which is a result in its own right.

    The second loss is giving two without the counting proof. Say the 3 to the power k rule in one sentence: it shows you know the answer is a minimum, not just a method that happened to work.

    What the interviewer asks next

    • How many weighings do you need for 27 bags? For 100?
    • Now you do not know whether the odd bag is heavier or lighter, and there are twelve. How many weighings?
    • You may take any number of coins from each bag and weigh once on a scale that shows grams. How do you find the light bag?
  3. 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?