Portfolio Management puzzles, solved step by step

The single six in four rolls, which wins 51.8% of the time against 49.1% for the double six in 24. Work each through its complement. No six in four rolls has chance (5/6) to the 4th, 48.2%. No double six in 24 rolls has chance (35/36) to the 24th, 50.9%. The naive count, trials times chance, gives two thirds for both and is wrong for both.

Why does multiplying the trials by six not keep the odds the same?

A double six is six times rarer than a six, so 24 rolls look like a fair swap for four. The intuition treats the chance of success as growing in a straight line with the number of tries. It does not. What compounds is the chance of failing every time, and a rare event's failure chance, raised to a high power, falls more slowly than the straight line suggests. Think of looking for a friend in a crowd: glancing four times at a small crowd and twenty-four times at a crowd six times the size are not the same search, because each extra glance adds less as the misses pile up.

How many rolls would the double six bet need to be favourable?

Solve for the number of rolls where the chance of no double six drops below a half. (35/36) to the 24th is 0.5086, still above a half; (35/36) to the 25th is 0.4945, just below. So 25 rolls, not 24, is where the double six bet turns favourable, a gap of a single roll between a losing and a winning bet. That is the second way to show you understand the question: the naive rule is off by a small amount, and in a repeated game a small edge is everything.

On a desk the lesson is the difference between a rough count and an exact one: a strategy that looks equivalent on a back-of-envelope scaling can sit on the wrong side of break-even once you do the compounding properly.

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.

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.

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.

The average trade loses 0.2%, and break-even needs a 75% hit rate. Expected value is 0.7 x 1% minus 0.3 x 3%, which is 0.7% minus 0.9%, or minus 0.2%. To break even the winners must pay for the losers: p x 1 = (1 minus p) x 3, so p is 3 over 4, or 75%. The trader is right more often than wrong and still loses money.

Why does a 70% hit rate not settle the question?

A shopkeeper who makes a small profit on seven sales out of ten but sells the other three at a big loss can still close the month in the red. Counting the happy sales tells you nothing until you know how big each one was. Expected return is each outcome's probability times its size, added up, so a hit rate only means something next to the payoff ratio. Here the losers are three times the size of the winners, and that ratio is doing all the damage.

How do you find the break-even hit rate in one line?

Set expected value to zero and solve. The break-even hit rate is the loss size divided by the sum of the win and loss sizes: 3 over 1 plus 3, which is 75%. Every point of hit rate is worth 0.04% a trade here, because moving one trade in a hundred from loser to winner swings 4 points in total. So the trader is 5 points of hit rate, or 0.2% a trade, short of break-even.

Then say what you would change. The trader can cut losers sooner, let winners run further, or be more selective; lifting the average win to 1.5% with the same losses moves break-even to 3 over 4.5, about 67%. Averages also hide transaction costs, which push break-even higher still.

Stake 10% of capital each time; staking 20% drives long-run growth to about zero. For an even-money bet the growth-maximising, or Kelly, stake is p minus q, 0.55 minus 0.45, which is 10%. That earns about 0.50% a bet in log terms. At 20%, 0.55 x log 1.2 plus 0.45 x log 0.8 is about zero, so twice the stake gives no growth at all despite double the expected profit per bet.

Why does betting more of a winning edge ever make you poorer?

A shopkeeper who wins more often than she loses still goes under if one bad month can wipe out half the shop, because rebuilding from half takes a doubling. Wealth compounds, so what matters over many bets is the average log return, and a loss hurts the log more than an equal-sized gain helps it. With a small stake, the edge dominates. With a large stake, the losses' extra damage grows faster than the edge, and past a point it wins.

How do you find the 10% and check the 20%?

Write growth per bet as g(f) = p log(1 + f) + q log(1 minus f), set its slope to zero, and the answer for even money is f = p minus q. The Kelly fraction for an even-money bet is simply the edge: 55% minus 45% is 10%. At 10%, g is 0.55 x 0.0953 minus 0.45 x 0.1054, about 0.50%. At 20% it is 0.55 x 0.1823 minus 0.45 x 0.2231, about -0.014%: essentially zero, and in fact a touch below it. Over 100 bets at the Kelly stake, capital grows by a factor of about 1.65 on the typical path; at 20% it goes nowhere.

Practitioners usually bet a fraction of Kelly. Half Kelly, 5%, keeps about 75% of the peak growth with half the volatility, and it protects against the real problem: in markets you never know p exactly. If the true edge were 52.5% rather than 55%, the 10% stake would already sit near zero growth.

Reroll on 1, 2 or 3; keep 4, 5 or 6. The game is worth Rs 4,250. A reroll is worth 3.5 on average, so you keep any face above 3.5 and swap any face below it. The kept faces contribute 4, 5 and 6 each with a one-in-six chance, 2.5 in all; the other half of the time you get 3.5. The total is 2.5 plus 1.75, which is 4.25 thousand.

How do you decide whether to reroll?

Think of a job offer in hand against the option to keep interviewing. You take the offer if it beats what you expect to get by walking away. The rule is to keep a result only when it is worth more than the option you give up, and here the option is a fresh roll worth 3.5. A 3 is below 3.5, so reroll it. A 4 is above, so keep it. Nothing about how you feel about a 4 enters into it; the comparison is one number against another.

Why is this called working backwards?

Because you value the last decision first. The second roll has no choice left in it, so it is worth 3.5. Only once you know that can you decide the first roll. Every multi-stage game is solved from the final stage back to the first, because each decision depends on what the later stages are worth. Allow two rerolls and the same logic stacks: the last reroll is worth 3.5, the middle stage is worth 4.25, so on the first roll you keep only a 5 or 6, and the game is worth 4.67.

Rerolling only on 1 and 2, a common cautious answer, gives 4.17: you keep the 3 when a reroll would have averaged half a point more. The portfolio version is the same idea. A position is kept only while its expected return beats what the capital would earn redeployed, which is an opportunity cost, not a sunk cost.

No. It is worth about Rs 92.7 to you, against a price of Rs 100. The expected price in a year is 0.4 times 150 plus 0.6 times 70, which is Rs 102, an expected return of only 2%. You need 10% for this risk, so you should pay no more than Rs 102 divided by 1.10. To justify Rs 100, the chance of the up case would have to be 50%.

Why is a positive expected gain not enough?

Imagine lending a friend Rs 100 for a year and expecting Rs 102 back, when a bank deposit would pay you more with no worry at all. You would not do it. An investment has to beat the return you require for its risk, not merely beat zero. Here the expected price of Rs 102 is a 2% expected return, and you have said this risk needs 10%. The shortfall is the whole answer.

What would change your mind?

Turn the question round and ask what probability makes Rs 100 fair. At a 10% required return the expected price must be Rs 110, and p times 150 plus one minus p times 70 equals 110 when p is 0.50. Stating the break-even probability turns a yes or no into a view you can argue about: do you believe the up case is at least a coin flip? That is exactly the conversation a portfolio manager has about a stock pitch, where the scenarios and their weights matter more than the single target.

Say the limits of the model too. Two outcomes are a sketch; real outcomes spread across a range. The required return itself is a judgement, and a lower one, 2% or less, would make the same stock look fair. The method survives those caveats: expected value first, then compare it with the return the risk demands.

Two heads in a row takes 6 tosses on average; head then tail takes 4. Each pattern has a one-in-four chance at any pair of tosses, but after a head, a tail wrecks HH and sends you back to the start, while for HT an extra head keeps you exactly where you were. Losing progress on a miss is what costs HH the extra two tosses.

Why do two equally likely patterns take different times to appear?

Picture two ladders. On one, slipping off the second rung drops you to the ground; on the other, slipping leaves you on the second rung. Both ladders need the same two good steps, but you will climb the second one faster. For HH, a tail after a head wipes out your progress; for HT, a head after a head leaves your progress intact, so HT arrives sooner.

How do you solve it in the room?

Name the states, write one equation per state, solve. Let the expected tosses still needed be E0 from the start and E1 after a head. Every equation says: one toss, plus the expected wait from wherever that toss sends you. For HH, E1 = 1 + half of 0 + half of E0, and E0 = 1 + half of E1 + half of E0. Substituting gives E0 = 6 and E1 = 4.

For HT the second equation changes: after a head, a tail finishes and a head leaves you in the same state, so E1 = 1 + half of E1, which is 2, and E0 is 4. The same state method handles any short pattern, which is the point an interviewer at a quantitative fund wants to hear.

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.

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.