Portfolio Management puzzles, solved step by step

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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. 036Twenty fund managers have no skill at all: each has a 50% chance of beating the benchmark in any year, independently. What is the chance that at least one of them beats the benchmark five years running?Behavioural and decision trapsCoreFund selectionMulti-manager allocation

    Try it first

    Instinct first: how likely is at least one five-year streak among 20 unskilled managers?

    Show the worked solution

    About 47%, close to a coin toss. One manager beats the benchmark five years running with probability one half to the fifth, or 1 in 32. The chance that none of the 20 does it is 31/32 to the 20th power, about 53%. So the chance that at least one does is about 47%. In a crowd of unskilled managers, a five-year streak is almost as likely as not.

    Why is a five-year streak weak evidence of skill?

    Put 20 people in a room and ask each to call five coin tosses. Somebody calling all five correctly would look gifted, but with that many people, somebody usually does. The chance of a streak for one named person is small, but the chance that someone in a crowd produces one is large, and fund selection looks at the crowd. An investor who screens a category for managers with five straight winning years is sampling exactly the lucky tail of this distribution.

    Halve the crowd each year: a five-year streak survives more often than not20Start10Yr 15Yr 22.5Yr 31.25Yr 40.625Yr 5Expected managers still unbeaten25%50%75%100%02050100Number of managers with no skill20 managers: 47%100: 96%Chance at least one has a 5-year streak
    Twenty unskilled managers thin to an expected 0.625 unbeaten after five years, yet the chance that at least one of them completes the streak is 47%. With 100 managers it rises to 96%, so a streak in a large category says very little on its own.

    How do you compute at least one without adding up cases?

    Use the complement. The chance that at least one of many independent trials succeeds is one minus the chance that every trial fails. Each manager fails the streak with probability 31/32. All 20 fail with probability (31/32) to the 20th, which is about 0.53. One minus that is 0.47. The expected number of streaks is 20 over 32, or 0.625, which is a separate and also useful number: less than one streak on average, yet close to even odds of seeing at least one.

    The relationship
    P(at least one)=1−(1−0.55)20=1−(3132)20≈0.47P(\text{at least one}) = 1 - \left(1 - 0.5^{5}\right)^{20} = 1 - \left(\tfrac{31}{32}\right)^{20} \approx 0.47
    0.5^5one manager's chance of five straight wins, 1 in 32
    20the number of independent managers
    What it says in wordsFind the chance every manager misses the streak, then take it away from one.

    The limitation runs in two directions. Real managers are not independent, since many hold similar stocks, which makes streaks cluster and the calculation rougher. And some managers do have skill; the point is only that a streak on its own cannot tell the skilled from the lucky. That takes longer records, consistent process and a view on why the edge should persist.

    Where candidates lose it

    The common wrong answer is one in 32, about 3%, which answers the question for one named manager rather than for any of twenty. The next most common is adding 20 times 1/32 to get 62.5%, which counts the cases where two managers both succeed twice.

    Say the complement method aloud and give 47%. Then draw the lesson for manager selection in one sentence: a screen for long winning streaks mostly selects luck.

    What the interviewer asks next

    • How many unskilled managers would you need for a 90% chance of at least one ten-year streak?
    • If one manager in the group truly beats the benchmark 60% of the time, how likely is that manager to show a five-year streak?
    • How would you design a fund selection test that is not fooled by this effect?
  2. 062A household keeps Rs 5 lakh in a fixed deposit at 7% while carrying Rs 3 lakh of credit card debt at 36% a year. What does keeping both cost them each year, compared with using the deposit to clear the card?Behavioural and decision trapsCoreIndian wealth managementWealth management

    Try it first

    Roughly what does keeping both cost a year?

    Show the worked solution

    About Rs 87,000 a year, and more after tax on the deposit interest. The card costs 36% of Rs 3 lakh, Rs 1,08,000. Using Rs 3 lakh of the deposit to clear it gives up 7% on that money, Rs 21,000. The difference, Rs 87,000, is a certain loss every year the household keeps both, paid for the comfort of a bigger deposit balance.

    Why do people keep both?

    Because money gets labels. The deposit is the emergency fund or the daughter's education money, and touching it feels like failing; the card is day-to-day spending and feels temporary. A rupee is a rupee whatever label it carries, so money earning 7% while debt costs 36% is simply borrowing at 36% to lend at 7%. The habit has a name, mental accountingTreating money differently depending on which mental category it sits in, a term from the economist Richard Thaler., and it is one of the most common and costly patterns a wealth adviser meets.

    Interest a year, Rs: the labelled-safe money is quietly the expensive choiceKeep bothRs 5 lakh FD at 7%, Rs 3 lakh card at 36%+ Rs 35,000 earned- Rs 1,08,000 paidNet: Rs -73,000Pay the card offRs 2 lakh FD left at 7%, no card balance+ Rs 14,000 earnednothing paid on the cardNet: Rs 14,000Cost of keeping both, a year, before tax on the deposit interestRs 87,000
    Keeping both earns Rs 35,000 on the deposit and pays Rs 1,08,000 on the card, a net of minus Rs 73,000, while clearing the card leaves Rs 14,000 of interest and no card cost, so keeping both costs Rs 87,000 a year.

    How do you get the number right?

    Compare two whole choices. Keep both: the household earns Rs 35,000 and pays Rs 1,08,000, a net of minus Rs 73,000. Clear the card: it earns 7% on the remaining Rs 2 lakh, Rs 14,000, and pays nothing. The cost of a choice is the difference between the two outcomes, not the interest bill on its own. That difference is Rs 87,000, the same as 29 points of rate gap on Rs 3 lakh.

    The relationship
    cost=(36%−7%)×3 lakh=29%×3 lakh=87,000\text{cost} = (36\% - 7\%) \times 3\text{ lakh} = 29\% \times 3\text{ lakh} = 87,000
    36%the card's annual rate, an illustration; confirm the card's actual rate
    7%the deposit rate
    3 lakhthe money that could move from deposit to card
    What it says in wordsThe yearly cost is the gap between the two rates applied to the amount that could be moved.

    Tax widens the gap. Deposit interest is taxed at the household's slab while card interest is paid from income already taxed. At an illustrative 30% slab, confirm the current rates, the cost rises to about Rs 93,300. The honest limit is liquidity: a family with no buffer at all may need some cash on hand. Here the Rs 2 lakh left in the deposit is that buffer, so the case for keeping the debt is gone.

    Where candidates lose it

    The trap is answering Rs 73,000, the household's net interest bill, or Rs 1,08,000, the card interest alone. Neither is the cost of the choice. The cost is what changes between keeping both and clearing the card.

    The quieter miss is accepting the labels, treating the deposit as untouchable. The interviewer is watching whether you see through the label to the rate gap, and whether you can say it to a client without making them feel foolish.

    What the interviewer asks next

    • How much of the deposit would you keep as a buffer, and why?
    • How does the answer change if the deposit has a premature withdrawal penalty of 1%?
    • What other everyday decisions share this pattern of borrowing dear while lending cheap?
  3. 091A trader bets on a fair coin, starting at Rs 1,000 and doubling the stake after every loss until a win, then starting again. The trader has Rs 63,000. What does each cycle win, how often does it wipe out, and what is the expected value?Behavioural and decision trapsCoreHedge fundsRisk management

    Try it first

    What is the expected value of one cycle?

    Show the worked solution

    Each cycle wins Rs 1,000 with probability 63 in 64 and loses Rs 63,000 with probability 1 in 64, so the expected value is exactly zero. The capital covers six stakes, Rs 1,000 up to Rs 32,000, so six straight losses, a 1 in 64 chance, wipe it out. Doubling down turns many small wins into one rare, large loss without creating any edge.

    Why does it feel like a sure thing?

    A driver who never buys insurance saves the premium every month and feels clever for years, until the one accident. Doubling after each loss produces a long run of small wins because the one outcome that loses, six tails in a row, is rare, but when it comes it is sixty three times the usual win. Short track records of such strategies look excellent, which is exactly the danger.

    Each loss doubles the next stake until the capital runs outLoss 1stake 1kLoss 2stake 2kLoss 3stake 4kLoss 4stake 8kLoss 5stake 16kLoss 6stake 32kRs 63,000 gone64k?No moneyfor 64kbars: stake on that tossline: total lost so farOne cycleWin Rs 1,000chance 63 in 64Lose Rs 63,000chance 1 in 64Expected value0Over 50 cycles, chanceof at least one wipe-out54%
    Six straight losses with doubling stakes of Rs 1,000 to Rs 32,000 use up the whole Rs 63,000, and the seventh stake cannot be placed; the cycle wins Rs 1,000 sixty three times in sixty four and loses Rs 63,000 once, for an expected value of zero.
    The relationship
    E=6364(1,000)−164(63,000)=984.4−984.4=0E = \tfrac{63}{64}(1{,}000) - \tfrac{1}{64}(63{,}000) = 984.4 - 984.4 = 0
    63/64chance of at least one head in six tosses
    1/64chance of six tails in a row
    What it says in wordsFrequent small wins and a rare large loss cancel exactly on a fair coin.

    What does this teach about judging a strategy?

    Run it 50 times and the chance of at least one wipe-out is about 54%, yet most individual traders running it for a few weeks will show a smooth profit. A win rate says nothing on its own; what matters is the size of the losses relative to the wins, and whether the worst case can end the game. Selling far out-of-the-money options has the same shape: steady premium income, and occasionally a loss that erases years of it.

    Say the limitation of the fix people suggest. More capital only pushes the wipe-out further out and makes it bigger: with infinite money and no stake limit the strategy would work, and nobody has either.

    Where candidates lose it

    The trap is being impressed by the 98.4% win rate. The interviewer is checking whether you weigh outcomes by size as well as frequency.

    The second loss is saying the strategy is negative in expectation. On a fair coin it is exactly zero; the harm is in the shape, a small chance of ruin, not in the average. Say both halves.

    What the interviewer asks next

    • The coin pays slightly less than even money. What happens to the expected value?
    • Where do you see this payoff shape in real portfolios?
    • How would you size risk so that no single loss can end the strategy?