Portfolio Management puzzles, solved step by step

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  1. 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?
  2. 094A strategy stakes Rs 1 lakh per bet and wins each bet with probability 55%, winning or losing the stake. Starting with Rs 5 lakh, what is the chance of reaching Rs 10 lakh before losing everything?Probability and expected valueHardHedge fundsRisk management

    Try it first

    Roughly what is the chance of reaching Rs 10 lakh?

    Show the worked solution

    About 73%, against 50% for a fair bet. This is the gambler's ruin problem. With the ratio of loss to win odds r = 0.45 / 0.55, the chance of reaching 10 units from 5 is (1 minus r to the 5) over (1 minus r to the 10), about 73.2%. Smaller bets make it higher still: with Rs 50,000 bets it is about 88%.

    Why is the answer so much higher than 55%?

    A slightly better tennis player might win 55% of points, yet win most matches, because a match needs many points and the small edge keeps adding up. Reaching Rs 10 lakh before zero needs a net five wins, which takes many bets, and every bet tilts the race a little in your favour, so the edge compounds into a much bigger survival advantage.

    A 55% edge bows the curve up: from the middle, 73% instead of 50%01234567891025%50%75%100%55% win rate: 73.2%fair coin: 50%Starting capital, Rs lakh (target 10, bet 1)Start 5, target 10at 55%, by bet sizeRs 5 lakh bets55%Rs 2.5 lakh bets60%Rs 1 lakh bets73%Rs 0.5 lakh bets88%smaller bets, surer edge
    With Rs 1 lakh bets and a Rs 10 lakh target, a fair coin gives a straight line and a 50% chance from Rs 5 lakh, while a 55% win rate bows the curve up to 73.2%; shrinking the bet to Rs 50,000 lifts it to about 88%.
    The relationship
    P(reach N from i)=1−r i1−r N,r=qp=0.450.55P(\text{reach } N \text{ from } i) = \frac{1 - r^{\,i}}{1 - r^{\,N}}, \quad r = \frac{q}{p} = \frac{0.45}{0.55}
    istarting capital in bets, 5
    Ntarget in bets, 10
    p, qchance of winning and losing each bet, 0.55 and 0.45
    What it says in wordsThe chance of hitting the target first depends on the edge through r, and on how many bets away each barrier is.

    Why does bet size matter so much?

    Keep the edge and the rupee distances fixed and change only the stake. With Rs 5 lakh bets the whole outcome rides on one toss, 55%; with Rs 1 lakh bets it is 73%; with Rs 50,000 bets it is 88%, because more bets give the law of large numbers room to work. That is the portfolio lesson: a real edge is only worth something if position sizes are small enough that an unlucky run cannot end the game first.

    Bet sizeBets to ruinBets to targetChance of reaching Rs 10 lakh
    Rs 5 lakh1155.0%
    Rs 2.5 lakh2259.9%
    Rs 1 lakh5573.2%
    Rs 0.5 lakh101088.1%
    Holding the 55% edge and the Rs 5 lakh start fixed, smaller bets raise the chance of reaching Rs 10 lakh from 55% to about 88%, because each barrier is more bets away.

    Where candidates lose it

    The common slip is answering 55%, the per-bet win rate, which treats a long race as a single toss. The interviewer wants to see that you recognise the gambler's ruin structure.

    The second loss is missing the sizing point. The follow-up is almost always what happens with bigger or smaller bets, so volunteer it.

    What the interviewer asks next

    • With a fair coin, what is the chance of reaching Rs 10 lakh from Rs 3 lakh?
    • What if the target were unlimited? What is the chance of never going broke?
    • How does this connect to the Kelly criterion for sizing bets?