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Portfolio Management puzzles, solved step by step

Puzzles
100
Traced to a firm
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13
Hard
30
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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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Showing 11–20 of 100
  1. 011A fund compounds at 15% a year for ten years. What share of the total gain over the decade arrives in the last three years?Compounding and fee dragCoreAsset managementWealth management

    Try it first

    Guess before you calculate.

    Show the worked solution

    About 45%. One rupee at 15% grows to 2.66 after seven years and 4.05 after ten. The total gain is 3.05, and the last three years add 4.05 minus 2.66, which is 1.39. That is 45.5% of the decade's gain in 30% of the time, because each year's 15% is earned on a larger base than the year before.

    Why is the gain not spread evenly across the years?

    Think of a snowball rolled down a long slope. In the first few metres it picks up a little snow; near the bottom each turn picks up much more, because the ball itself is bigger. Compounding earns the same rate on a growing base, so every year's rupee gain is 15% larger than the year before. Year 1 adds 0.15 on each rupee. Year 10 adds 0.53, three and a half times as much. The first three years together add only 0.52; the last three add 1.39.

    Each year's gain on Rs 1 at 15%: the last three bars carry almost half the total1.00start+0.15Y1+0.17Y2+0.20Y3+0.23Y4+0.26Y5+0.30Y6+0.35Y7+0.40Y8+0.46Y9+0.53Y104.05endYears 1 to 3 add 0.52Years 8 to 10 add 1.391.39 of 3.05 total gain = 45.5%
    At an assumed 15% a year, each rupee gains 0.15 in year 1 but 0.53 in year 10, so the last three years add 1.39 of the 3.05 total gain, 45.5% of it.
    The relationship
    (1.15)10−(1.15)7(1.15)10−1=4.046−2.6603.046≈0.455\frac{(1.15)^{10}-(1.15)^{7}}{(1.15)^{10}-1}=\frac{4.046-2.660}{3.046}\approx 0.455
    (1.15)^{10}the value of one rupee after ten years
    (1.15)^{7}the value after seven years
    (1.15)^{10}-1the total gain over the decade
    What it says in wordsThe last three years' share is the growth from year seven to year ten, divided by the growth over all ten.

    What does this mean for an investor who leaves early?

    Someone who exits after seven years has sat through 70% of the time but collected only 55% of the decade's gain. Because compounding back-loads the reward, leaving a long plan early costs far more than the fraction of time given up. The same arithmetic runs against the investor with fees: a charge taken every year compounds too, and its cost is also concentrated at the end. Say the limitation plainly: 15% is an assumed rate for the arithmetic, not a forecast, and real returns arrive unevenly, so the actual last three years could be the worst three.

    Where candidates lose it

    The trap is answering 30%, proportional to time, because the question sounds like a fraction of a decade. The interviewer is checking whether you picture compounding as a curve.

    The other slip is dividing the last three years' gain by the final value, {P11_END:.2f}, rather than by the total gain, {P11_TOT:.2f}. Read the question again: it asks for a share of the gain, not of the ending pot.

    What the interviewer asks next

    • At what rate would the last three years carry exactly half the gain?
    • A 1.5% annual fee is taken throughout. What share of the lost wealth falls in the last three years?
    • Why do long-horizon savers care more about the final years' return than the first years'?
  2. 012Estimate how many tonnes of gold Indian households buy in a year.Market sizing and estimationCoreRothschild & CoParis · 2026

    Try it first

    Which split makes this estimable in two minutes?

    Show the worked solution

    About 800 tonnes a year, on stated assumptions. Weddings: about 1 crore a year at 35 grams each is 350 tonnes. Festivals and gifting: 30 crore households, a quarter buying 4 grams, is 300 tonnes. Investment coins and bars: 5% of households buying 10 grams is 150 tonnes. The total is 800 tonnes. Check it against published industry demand figures before using it anywhere.

    How do you start when you have no idea of the answer?

    Ask when a family actually walks into a jeweller. It is not random: a wedding, Dhanteras or Akshaya Tritiya, a birth, or a decision to save in coins. Splitting a total into the occasions that drive it turns one number nobody knows into several that anyone can picture. State the base first: about 140 crore people at four to five per household is roughly 30 crore households. That figure and every weight below are assumptions for the estimate, said out loud so the interviewer can challenge them.

    Split the unknowable total into occasions you can pictureHousehold goldbought a yearabout 800 tWeddings1 crore weddings x 35 g350 tFestivals and gifting30 crore households x 25% buyingx 4 g each300 tCoins and bars30 crore households x 5% buyingx 10 g each150 ttonnes1 crore grams = 10 tonnes. Every input is an assumption for the estimate, not a statistic.
    Splitting household gold buying into weddings at 350 tonnes, festivals and gifting at 300 tonnes and investment coins and bars at 150 tonnes gives an estimate of about 800 tonnes a year, with every input stated as an assumption.
    OccasionCountGrams eachTonnes
    Weddings1 crore a year35350
    Festivals and gifting30 crore households x 25%4300
    Coins and bars30 crore households x 5%10150
    Total800
    The three occasions add to about 800 tonnes, remembering that 1 crore grams is 10 tonnes.

    How do you sanity check it, and where is it weakest?

    Divide back: 800 tonnes over 30 crore households is about 2.7 grams per household per year, a small coin's worth on average, which feels plausible given that most households buy nothing in a typical year and a few buy a lot at a wedding. The weakest input is grams per wedding, because the spread is huge and the average is dragged up by a few large weddings. Moving it by 10 grams moves the total by 100 tonnes. Say that, and say you would check the result against published demand data from the industry before quoting it: an estimate is a structure plus assumptions, not a statistic.

    On a multi-asset or wealth desk, the follow-up is usually why it matters: household gold is a large part of Indian savings, so its demand affects imports, the rupee and how much money is left for financial assets.

    Where candidates lose it

    Candidates start with a number they half remember and then try to justify it. The interviewer cannot tell whether you reasoned or recalled, and a wrong remembered number sinks the answer. Build it from drivers instead.

    The other loss is unit confusion: grams, kilograms and tonnes across crore and lakh. Say the conversion once, 1 crore grams is 10 tonnes, and use it every time.

    What the interviewer asks next

    • How would a sharp rise in the gold price change your estimate, and through which driver?
    • How much of this might be old gold exchanged rather than new buying?
    • Estimate the rupee value of the same purchases, stating the price you assume.

    Asked at Rothschild & Co, Asset Management, Paris, 2026 (Wall Street Oasis): interview were with 2 seperate analysts, first part was more about market sizing and logic reasoning

  3. 013A foreign equity index has 16% volatility in its local currency, and that currency has 8% volatility against the rupee. What is the volatility of an unhedged position if the two correlate at plus 0.3, and if they correlate at minus 0.3?Currency and global returnsHardGlobal investingMulti-asset

    Try it first

    With a correlation of minus 0.3, is the unhedged position riskier than the hedged one?

    Show the worked solution

    About 19.9% at plus 0.3 and 15.6% at minus 0.3. The unhedged return is roughly the local return plus the currency return, so the variances add with a correlation term: 16 squared plus 8 squared, plus or minus 2 x 0.3 x 16 x 8. That is 396.8 or 243.2, with square roots of 19.9% and 15.6%. With negative correlation the unhedged position is less volatile than the hedged one at 16%.

    How can adding a second risk reduce the total?

    Think of a shop that sells umbrellas and sunglasses. Each product's sales swing a lot with the weather, but in opposite directions, so the till is steadier than either product alone. When two sources of return tend to move against each other, combining them lowers risk even though each one is volatile on its own. For an Indian investor holding foreign shares, the currency is the second source. If the foreign currency tends to strengthen against the rupee when that market falls, it cushions the loss, and the unhedged position is steadier: 15.6% instead of 16%.

    Currency adds to or offsets local risk, depending on the sign of the correlationCorrelation +0.3local market 16currency 8unhedged: 19.916 x 16 + 8 x 8 + 2 x 0.3 x 16 x 8 = 396.8square root: 19.9% against 16.0% hedgedCorrelation -0.3local market 16currency 8unhedged: 15.616 x 16 + 8 x 8 - 2 x 0.3 x 16 x 8 = 243.2square root: 15.6% against 16.0% hedged
    Drawn as vectors, local risk of 16 and currency risk of 8 combine to 19.9 when they correlate at plus 0.3, but to only 15.6 at minus 0.3, which is below the 16 of a fully hedged position.
    The relationship
    σunhedged=σL2+σX2+2ρ σLσX=256+64±76.8\sigma_{unhedged}=\sqrt{\sigma_L^2+\sigma_X^2+2\rho\,\sigma_L\sigma_X}=\sqrt{256+64\pm 76.8}
    \sigma_Lthe index's volatility in local currency, 16%
    \sigma_Xthe currency's volatility against the rupee, 8%
    \rhothe correlation between the two, plus or minus 0.3
    What it says in wordsThe unhedged variance is the two variances plus twice the covariance, and the covariance changes sign with the correlation.

    So should a global portfolio hedge its currency?

    The puzzle gives the risk side of the answer, not the whole decision. A hedge removes the currency's volatility, but it also removes the currency's correlation with the market, and when that correlation is negative the hedge adds risk rather than cutting it. The rest of the decision is cost, which depends on the interest rate gap between the two currencies, and the investor's own liabilities in rupees. Say the limitation: correlations are measured on history and tend to shift in a crisis, so the minus 0.3 that makes the unhedged position look safer is the number least likely to hold when it matters. The formula also ignores the small cross term from multiplying the two returns.

    Where candidates lose it

    Most candidates say hedging always lowers risk, because the hedge removes a volatile exposure. That is true only when the currency correlates positively with the local market. With a negative correlation the currency is itself a hedge.

    The arithmetic trap is adding 16 and 8 to get 24, which assumes perfect correlation. Add variances, include the covariance term with its sign, then take the root.

    What the interviewer asks next

    • At what correlation is the unhedged volatility exactly 16%?
    • What does it cost to hedge, and what drives that cost?
    • Why do some investors hedge their foreign bonds but not their foreign equities?
  4. 014You backtest 20 independent trading strategies, none of which has any real edge, and test each one at the 5% significance level. What is the chance at least one looks significant, and what per-test threshold would hold that overall false alarm rate at 5%?Statistics and forecastingCoreQuantitative researchSystematic investing

    Try it first

    Chance that at least one of the 20 worthless strategies passes?

    Show the worked solution

    About 64%, and a per-test threshold of about 0.25%. Each worthless strategy passes by luck 5% of the time, so all 20 fail with chance 0.95 to the 20th, 35.8%, and at least one passes 64.2% of the time. To hold the overall rate at 5%, test each at 5% divided by 20, which is 0.25%; the exact version, 1 minus 0.95 to the power of one twentieth, is 0.256%.

    Why does testing more ideas throw up a false winner?

    Ask a room of 20 people to each flip a coin five times, and there is a fair chance someone gets five heads. Nobody in the room has a lucky hand; there were simply enough tries. A 5% test lets one worthless idea in twenty through by chance, so a researcher who tests twenty ideas should expect about one false winner, not be impressed by it. The expected number of false positives here is 20 x 0.05, exactly 1, and the chance of at least one is 64.2%.

    Test enough worthless strategies and one will look like a winner25%50%75%100%11020304050Number of strategies tested20 tests: 64.2%each tested at 5%each tested at 0.25%: 4.9% at 20Chance at least one looks significant
    Testing each worthless strategy at 5%, the chance that at least one looks significant reaches 64.2% at 20 strategies, while testing each at 0.25% holds it near 4.9%.
    The relationship
    P(≥1 false)=1−(0.95)20≈0.642αeach=0.0520=0.25%P(\ge 1 \text{ false}) = 1-(0.95)^{20}\approx 0.642 \qquad \alpha_{each}=\frac{0.05}{20}=0.25\%
    0.95the chance a worthless strategy fails a 5% test
    20the number of independent strategies tested
    \alpha_{each}the per-test threshold that caps the overall false alarm rate near 5%
    What it says in wordsThe chance of at least one false winner is one minus the chance that every test correctly fails; dividing the level by the number of tests caps it.

    What does a quant desk actually do about it?

    Dividing the threshold by the number of tests is called the Bonferroni correctionA rule that divides the significance level by the number of tests run, so the chance of any false positive across all of them stays near the original level.. The real discipline is counting every test you ran, including the ones you dropped quietly, because the correction is only as honest as that count. A researcher who tried 200 variants and reports the best 20 has a far bigger multiple testing problem than the 20 suggest. Desks also hold out data the research never touched and demand a reason for the edge before the backtest. Say the limitation: the correction assumes independent tests, and for correlated strategies it is too strict, which costs real ideas.

    Where candidates lose it

    The fast wrong answers are 5%, which ignores that there are 20 tests, and 100%, which adds the chances. Say the complement and the answer arrives in one line.

    The second trap is naming the fix without the cost. A tighter threshold throws away some genuine strategies too, and an interviewer on a systematic desk expects you to say that trade-off out loud.

    What the interviewer asks next

    • If the 20 strategies are highly correlated, is the true chance of a false winner higher or lower than 64%?
    • Of 1,000 strategies, how many worthless ones pass at 5%?
    • Why is out-of-sample testing a better defence than a stricter threshold?
  5. 015A Rs 50 crore equity portfolio has a beta of 1.2 to the index. How much index futures notional must you sell to bring the portfolio's beta down to 0.5?Portfolio risk mathsWarm upPortfolio implementationHedge funds

    Try it first

    How much notional do you sell?

    Show the worked solution

    Sell Rs 35 crore of index futures notional. At a beta of 1.2 the portfolio moves like Rs 60 crore of the index. At the target of 0.5 it should move like Rs 25 crore. The difference, (1.2 minus 0.5) x Rs 50 crore, is Rs 35 crore, assuming the futures move one for one with the index. At an assumed Rs 10 lakh a contract, that is about 350 contracts.

    Why is the portfolio's market exposure not simply Rs 50 crore?

    Think of a car that goes 1.2 km for every km a reference car goes. Holding Rs 50 crore of it is like holding Rs 60 crore of the reference. Beta converts a portfolio's value into index-equivalent exposure, so a Rs 50 crore book at beta 1.2 carries Rs 60 crore of market risk. Once you see the exposure in index rupees, the hedge is a subtraction: you want Rs 25 crore left, so you take away Rs 35 crore by selling futures, which carry a beta of one to the index.

    Hedge the change in beta, not the whole portfolioPortfolio value50its value in rupeesMarket exposure now6050 x 1.2Sell index futures-35(1.2 - 0.5) x 50Exposure left2550 x 0.5 = the target betaAt an assumed Rs 10 lakh a contract, Rs 35 crore is about 350 contracts.
    The Rs 50 crore portfolio at beta 1.2 carries Rs 60 crore of market exposure; selling Rs 35 crore of index futures leaves Rs 25 crore, which is a beta of 0.5 on the portfolio.
    The relationship
    N=(βtarget−βnow)×V=(0.5−1.2)×50=−35N=(\beta_{target}-\beta_{now})\times V=(0.5-1.2)\times 50=-35
    Nfutures notional to trade, Rs crore; negative means sell
    \beta_{now}, \beta_{target}the current beta 1.2 and the target 0.5
    Vthe portfolio's value, Rs 50 crore
    What it says in wordsThe futures notional equals the change in beta times the portfolio value, with a minus sign meaning a sale.

    What does the hedge not do?

    It removes market risk, not stock risk. After the hedge the portfolio still carries every stock-specific bet it had; only its sensitivity to the index has been cut. That is often the point: a manager who likes the stocks but not the market can keep the stock picks and trim the market bet. Say the limitations: beta is estimated from history and drifts, so the hedge is right only on average; futures need margin and must be rolled at expiry, and the futures price can move slightly differently from the index, which is called basis risk.

    Where candidates lose it

    The common answers are Rs 50 crore, hedging the whole value, and Rs 60 crore, hedging the whole beta-weighted value. Both take the beta to zero, not to 0.5. Hedge the change in beta.

    Candidates also forget the direction. Lowering beta means selling futures; raising it means buying them. Say the sign with the number.

    What the interviewer asks next

    • How much would you trade to raise the beta to 1.5 instead?
    • If the portfolio falls 10% in value, is the hedge still right?
    • Why might the hedged portfolio still lose money in a market fall?
  6. 016A stock's prices over six days are 7, 1, 5, 3, 6 and 4. What is the most you can make with one buy followed by one later sell? And what is the most with any number of buy and sell trades, if you can hold at most one share and cannot short?Logic brainteasersCoreMan GroupLondon · 2019

    Try it first

    With unlimited trades, what is the maximum profit?

    Show the worked solution

    5 with one trade and 7 with any number. With one trade, buy at the lowest price that comes before a higher one: buy at 1 on day 2 and sell at 6 on day 5, for 5. With unlimited trades, collect every day-to-day rise and sit out every fall: 1 to 5 earns 4, 3 to 6 earns 3, a total of 7. The general answer is the sum of the positive daily changes.

    How do you find the best single trade without checking every pair?

    Walk through the prices once, like a shopper who notes the cheapest price seen so far and asks each day how much they would make selling today. The best single trade is the largest gap between today's price and the lowest price seen before today, found in one pass. At day 2 the low is 1. Day 3 offers 5 minus 1, which is 4; day 5 offers 6 minus 1, which is 5; no later day beats it. The trap is subtracting the lowest price from the highest overall: the high of 7 comes before the low of 1, so it cannot be sold after buying.

    Same prices, two rules: one trade catches the widest gap, many trades catch every riseOne trade: buy 1, sell 6 = 5048day 1day 2day 3day 4day 5day 67buy 153sell 64Any number: 4 + 3 = 7048day 1day 2day 3day 4day 5day 6715364+4+3Falls (grey) are skipped: with no shorting, a fall is only avoided, never earned.
    On prices of 7, 1, 5, 3, 6 and 4, one trade catches the widest later gap, buying at 1 and selling at 6 for 5, while unlimited trades catch each rise, 4 and then 3, for 7.

    Why is the unlimited answer just the sum of the rises?

    Any profitable trade from a low to a later high can be split into daily steps, and it gains only on the up days inside it while paying for every down day it sits through. With no limit on trades and no shorting, the best strategy holds the stock on every day it rises and nothing on every day it falls, so profit equals the sum of positive daily changes. Here that is 4 plus 3, which is 7. One pass through the prices gives the answer, which is the point of asking a coding-flavoured candidate.

    The relationship
    one trade=max⁡j>i(pj−pi)=6−1=5many=∑tmax⁡(pt+1−pt,0)=4+3=7\text{one trade}=\max_{j>i}(p_j-p_i)=6-1=5 \qquad \text{many}=\sum_t \max(p_{t+1}-p_t,0)=4+3=7
    p_tthe price on day t
    \max(p_{t+1}-p_t, 0)a day's rise, or zero on a falling day
    What it says in wordsOne trade is the widest later gap; unlimited trades collect every daily rise.

    Say the limitation as a portfolio manager would: this is perfect hindsight with no costs. Add a transaction cost per trade and the two answers move toward each other, because catching a small rise is no longer worth paying for.

    Where candidates lose it

    The common mistake is answering 6, the highest price minus the lowest, without checking that the high comes after the low. The interviewer is testing whether you respect the order of time.

    For the second part, candidates sometimes add the falls as well, answering 9 or more, as if they could short. Reread the constraint: no shorting means falls are only avoided, never earned.

    What the interviewer asks next

    • What if you are allowed at most two trades?
    • Each trade now costs 1. What is the best total?
    • Write the one-pass algorithm for the single trade and say its running time.

    Asked at Man Group, Alternative Investments, London, 2019 (Wall Street Oasis): Given a series of prices, find the one buy/sell trade pair which gives the maximum profit

  7. 017A coin-flip bet wins Rs 1.5 lakh or loses Rs 1 lakh. Many people refuse to play it once. Why, and what is the chance of ending with an overall loss if you play it ten times?Behavioural and decision trapsHardWealth managementAsset management

    Try it first

    Over ten independent plays, how often does the total end in a loss?

    Show the worked solution

    People refuse because a loss hurts more than an equal gain pleases, but ten plays lose money only about 17% of the time. Each play is worth plus Rs 25,000 on average. Over ten plays, four wins and six losses break even, so only three or fewer wins lose: 176 of 1,024 sequences, 17.2%. Judging each bet alone makes a good repeated bet feel bad.

    Why does a bet with a positive expected value get refused?

    Ask someone whether they would bet their weekend on a coin flip to win a second weekend, and most say no, because losing the one they have feels worse than gaining a new one feels good. This is loss aversionThe tendency, documented by Kahneman and Tversky in prospect theory, to feel a loss more strongly than a gain of the same size.. If a loss feels twice as bad as an equal gain feels good, this bet's felt value is 0.5 x 1.5 minus 0.5 x 2 x 1, which is minus 0.25 lakh, so refusing it once is consistent with how the person feels, even though its expected value is plus Rs 25,000. The factor of two is an assumption for the arithmetic, not a measured constant.

    How do you get the chance of a loss over ten plays?

    Count wins. With k wins and 10 minus k losses the total is 1.5k minus (10 minus k), which is 2.5k minus 10 lakh. That is negative only when k is three or less. Because each win outweighs each loss, you can lose six flips out of ten and still break even, which is why pooling the bets makes a loss so much rarer than a single flip suggests. The number of ways to get 0, 1, 2 or 3 wins is 1 plus 10 plus 45 plus 120, which is 176, out of 1,024 equally likely sequences: 17.2%.

    Ten plays of a +1.5 / -1 lakh coin flip: how often the total ends in a loss0.1%-10.01.0%-7.54.4%-5.011.7%-2.520.5%024.6%+2.520.5%+5.011.7%+7.54.4%+10.01.0%+12.50.1%+15.0Total after ten plays, Rs lakhLoss, 3 wins or fewer: 17.2%Break even, 4 wins: 20.5%Ahead, 5 or more wins: 62.3%Average: +2.5 lakh10 plays x Rs 25,000
    Over ten plays the total ends in a loss only with three or fewer wins, a chance of 17.2%, breaks even at four wins with 20.5%, and ends ahead 62.3% of the time, with an average gain of Rs 2.5 lakh.
    The relationship
    P(loss)=∑k=03(10k)(12)10=1+10+45+1201024≈17.2%P(\text{loss})=\sum_{k=0}^{3}\binom{10}{k}\left(\tfrac12\right)^{10}=\frac{1+10+45+120}{1024}\approx 17.2\%
    kthe number of winning flips out of ten
    \binom{10}{k}the number of sequences with exactly k wins
    What it says in wordsAdd up the sequences with three or fewer wins and divide by all 1,024 possible sequences.

    What does this mean for how clients see a portfolio?

    A client who checks a portfolio every day sees each day as a separate bet and feels every loss. Looking at the same holdings less often pools the bets, and the pooled result loses far less often than any single period does. The economist Paul Samuelson told the story of a colleague who refused one such bet but would take a hundred; the lesson for a wealth desk is to frame decisions at the horizon the money actually has. State the limitation: pooling helps only when the bets are independent and the investor can survive the bad runs, and ten plays still lose 17% of the time.

    Where candidates lose it

    Candidates call refusing the single bet irrational and move on. The interviewer wants the mechanism, loss aversion, and then the arithmetic that shows why the same person might accept the bet repeated.

    On the numbers, the slip is setting the loss threshold at fewer than five wins, as if wins and losses were the same size. Write the total as 2.5k minus 10 before you count anything.

    What the interviewer asks next

    • How many plays until the chance of an overall loss falls below 5%?
    • What if the loss were Rs 1.4 lakh instead of Rs 1 lakh?
    • Why might a client rationally refuse even the ten-play version?
  8. 018You must interview 10 fund managers one at a time, in random order, and hire or reject each on the spot, with no going back. What rule gives you the best chance of hiring the single best manager, and what is that chance?Probability and expected valueHardMulti-manager allocationFund selection

    Try it first

    How many managers should you see and pass on before you are willing to hire?

    Show the worked solution

    Interview and pass on the first 3, then hire the first manager better than all of them. You get the best one about 40% of the time. The first three set the bar at no cost but the chance the best is among them. Passing on 4 gives almost the same, 39.8%. As the number of candidates grows, the rule becomes: pass on about 37%, one over e, and the chance of success tends to 37% too.

    Why pass on anyone at all?

    Think of house hunting in a fast market where every flat is gone the moment you walk away. If you sign the first one, you have no idea whether it was good. If you look at every flat before deciding, the best has already been taken. A short look-only phase buys you a benchmark; after that, the first candidate who beats the benchmark is likely to be the best overall. The cost is that the best might be inside the look-only phase, which happens with chance r in 10 if you pass on r. Hiring the first manager blindly succeeds only 10% of the time.

    Chance of hiring the best of 10, by how many you pass on first10.0%028.3%136.6%239.9%339.8%437.3%532.7%626.5%718.9%810.0%9Managers interviewed and passed on before you are willing to hirePass on 3, then hire the firstwho beats all of them
    With 10 managers, the chance of hiring the best rises from 10% if you pass on none to a peak of 39.9% if you pass on 3, then falls back to 10% if you pass on 9, because waiting too long is as costly as not waiting.

    How do you compute the chance for a given rule?

    Suppose you pass on r and the best manager sits at position i, after r. You hire them only if nobody between r and i beat the first r, which happens when the best of the first i minus 1 candidates is among the first r: a chance of r over i minus 1. Averaging over where the best one sits gives r over 10 times the sum of 1 over (i minus 1), for i from r plus 1 to 10. For r equals 3 that is 0.3 times (1/3 + 1/4 + ... + 1/9), which is 0.3987. The curve is flat near the top: 3 and 4 differ by less than half a point.

    The relationship
    P(r)=rn∑i=r+1n1i−1P(3)=0.3(13+14+⋯+19)≈0.399P(r)=\frac{r}{n}\sum_{i=r+1}^{n}\frac{1}{i-1} \qquad P(3)=0.3\left(\tfrac13+\tfrac14+\dots+\tfrac19\right)\approx 0.399
    nthe number of candidates, 10
    rhow many you interview and pass on first
    ithe position of the best candidate
    What it says in wordsThe chance of success is the share you skip times the sum, over later positions, of the chance that no one between beats your benchmark.

    Say the limitation, because allocators hear this puzzle and then ask about real selection. The rule maximises the chance of the single best and scores a second-best hire as a total failure. Real allocators care about hiring someone good, can often revisit a manager, and rarely see candidates in random order, and each of those changes the rule.

    Where candidates lose it

    Candidates either hire early because a manager looks strong, or propose looking at half the field before deciding. Both lose most of the value. The interviewer wants the look-then-leap structure and a number.

    The other lost point is stopping at 3 without the general rule. Offer the one-over-e result for large fields: pass on about 37% and win about 37% of the time.

    What the interviewer asks next

    • What changes if you only need a manager in the top three?
    • With 100 candidates, how many should you pass on?
    • How would you adapt the rule if you could call back a rejected manager with some probability?
  9. 019The three-year government yield is 7.0% and the two-year yield is 6.6%. You buy a three-year bond with a 7% coupon at par. If the yield curve does not move over the next year, roughly what does the bond return?Bond mathsCoreFixed income

    Try it first

    Pick the closest one-year return.

    Show the worked solution

    About 7.7%. You collect the 7 coupon, and a year later the bond is a two-year bond. If the curve has not moved, it is priced at the two-year yield of 6.6%, which makes it worth 100.73. Coupon 7.00 plus a price gain of 0.73 is 7.73%. A shortcut gives the same: 7% plus the two-year duration of about 1.82 times the 0.4 point fall in yield.

    Why does an unchanged curve still move the bond's price?

    Think of walking down a gentle slope while standing still relative to the hillside: the ground stays put, but you end up lower because you moved along it. A bond ages along the curve; if the curve slopes up and stays still, the bond's yield falls as its maturity shortens, and its price rises. Today it is a three-year bond at 7.0%. In a year it will be a two-year bond, and two-year bonds yield 6.6%. That 0.4 point fall in yield is the roll downThe price gain a bond earns as it ages into a shorter maturity with a lower yield on an upward sloping curve..

    An upward sloping curve pays twice: the coupon and the roll down6.0%6.5%7.0%1 yr2 yr3 yrYears to maturitytoday: 3 yr at 7.0%in a year: 2 yr at 6.6%rolls downcoupon7.00price gain +0.737.73%curve unchanged7.00%if curve flat
    The three-year bond bought at 7.0% becomes a two-year bond priced at 6.6% a year later, so an unchanged curve returns the 7.00 coupon plus a 0.73 price gain, 7.73% against 7.00 on a flat curve.

    How do you check it without a calculator?

    Use duration. A two-year bond with a 7% coupon has a modified duration of about 1.82, so a 0.4 point fall in yield lifts its price by about 1.82 x 0.4, which is 0.73. Add the 7 coupon and the one-year return is about 7.73%, within a hair of the exact 7.73%. Say the limitation: the answer depends entirely on the curve staying put. If two-year yields rise to 7.0% by next year, the roll down disappears and the return is the coupon alone.

    The relationship
    P1=71.066+1071.0662≈100.73r=7+(100.73−100)100≈7.73%P_1=\frac{7}{1.066}+\frac{107}{1.066^2}\approx 100.73 \qquad r=\frac{7+(100.73-100)}{100}\approx 7.73\%
    P_1the bond's price in a year, as a two-year bond at a 6.6% yield
    7the annual coupon
    100the price paid today, at par
    What it says in wordsThe one-year return is the coupon plus the price gain from repricing at the lower two-year yield, over the price paid.

    Where candidates lose it

    The common answer is 7%, the yield at purchase, which is right only if the curve is flat. The interviewer asked about an unchanged curve precisely to see whether you notice that the bond moves along it.

    The overshoot is adding the whole slope of the curve or forgetting that the price gain depends on duration. The gain is duration times the yield change, not the yield change alone.

    What the interviewer asks next

    • What would the return be if the curve were inverted, with the two-year at 7.4%?
    • Which point on this curve gives the most roll down per unit of duration?
    • How much must the two-year yield rise over the year to wipe out the roll down?
  10. 020A stock trades at 60 times earnings and pays no dividend, and its investors require 12% a year. If it should trade at 20 times earnings in ten years, what annual earnings growth does today's price require?Valuation riddlesHardFundamental asset managementAsset management

    Try it first

    Which growth rate does the 60x multiple imply?

    Show the worked solution

    About 25% a year for ten years. With no dividend, all of the 12% return must come from price, so the price in ten years must be 60 x 1.12 to the tenth, about 186 times today's earnings. If the stock then trades at 20 times, earnings must be 186 divided by 20, or 9.32 times today's. That is growth of 25.0% a year. At 15% growth the investor would earn only about 3% a year.

    How do you turn a multiple into a growth forecast?

    Think of paying 60 years of a shop's current profit for the shop. That only makes sense if the profit is going to be far bigger soon. A high multiple is a forecast you can read: work forward from the return investors want, and back from the multiple the stock should end at, and the growth in between is what the price assumes. Here the investor wants 12% a year with no dividend, so the price must grow 3.11 times in ten years, to about 186 times today's earnings. At a mature 20 times, earnings must reach 9.32 times today's level.

    What a 60x multiple needs: earnings growth of 25% a year for a decade2468100246810YearsEarnings, today = 1needed: 9.32at 15%: 4.05gapPrice in 10 years60 x 1.12^10 = 186.4At 20x, earningsmust be 9.32Growth needed25.0% a yearAt 15%: returnonly 3.0% a year
    To justify 60 times earnings today and 20 times in ten years at a 12% return, earnings must grow 25% a year to 9.32 times today's level, while a 15% path reaches only 4.05 and would leave the investor with about 3% a year.
    The relationship
    g=(PE0(1+r)10PE10)1/10−1=(60×3.10620)1/10−1≈25%g=\left(\frac{PE_0(1+r)^{10}}{PE_{10}}\right)^{1/10}-1=\left(\frac{60\times 3.106}{20}\right)^{1/10}-1\approx 25\%
    PE_0, PE_{10}the multiple today, 60, and in ten years, 20
    rthe required return, 12%, all from price since there is no dividend
    gthe annual earnings growth the price requires
    What it says in wordsThe earnings growth needed is the required price growth, adjusted for the multiple shrinking from 60 to 20.

    What do you do with the number once you have it?

    You ask how often a company sustains that growth for a decade, which is rarely, and you say so. The question the interviewer wants answered is not whether the company is good but whether the price has already paid for more than the company is likely to deliver. At a still strong 15% a year, earnings reach 4.05 times today's, the price at 20 times is about 81, and the investor earns about 3.0% a year rather than 12%. Say the limitations: the exit multiple of 20 is an assumption, and buybacks, dividends or a higher exit multiple would lower the growth needed.

    Where candidates lose it

    The common error is saying the stock needs to grow earnings at 12%, the required return. That ignores the multiple falling from 60 to 20, which on its own costs about 10% a year of price.

    The other trap is stopping at the arithmetic. On a fundamental desk the interviewer wants the judgement: 25% for a decade is a demanding assumption, and a reverse calculation like this is how you show a price is stretched without claiming to know the future.

    What the interviewer asks next

    • What growth is needed if the stock still trades at 40 times in ten years?
    • How does paying a 2% dividend change the answer?
    • What required return does today's price imply if earnings grow at 15%?
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