Portfolio Management puzzles, solved step by step
- Puzzles
- 100
- Traced to a firm
- 31
- Topics
- 13
- Hard
- 30
005Which is the better bet: at least one six in four rolls of a single die, or at least one double six in twenty-four rolls of a pair of dice?Quantitative asset managementHedge funds
Try it first
Which bet has the better odds?
Show the worked solution
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.
The chance of no six in four rolls is 48.2%, so that bet wins 51.8%, while the chance of no double six in 24 rolls is 50.9%, so that bet wins only 49.1%, even though the naive count gives two thirds for both. 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.
The relationship5/6 the chance a single roll is not a six 35/36 the chance a roll of two dice is not a double six What it says in wordsEach bet wins with one minus the chance of missing on every roll.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.
Where candidates lose it
The trap is the naive count. Four sixths and twenty-four thirty-sixths are both two thirds, and a candidate who answers "the same" has fallen for the exact error the puzzle was built to catch.
The other way to lose it is getting the complements right but rounding both to about 50% and calling it a tie. The whole answer lives in the gap between 51.8% and 49.1%, so say both to one decimal.
What the interviewer asks next
- How many rolls of one die give better than even odds of at least one six?
- What is the expected number of sixes in four rolls, and why is it not the same as the chance of at least one?
- If you win Rs 100 on the double six bet and lose Rs 100 otherwise, what is your expected value over 24 rolls?
040A bet pays even money and you win it 55% of the time. You can bet any fraction of your capital, as often as you like. What fraction maximises long-run growth, and what happens to growth if you stake double that?Hedge fundsQuantitative asset management
Try it first
What happens to long-run growth if you stake twice the growth-maximising fraction?
Show the worked solution
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.
Long-run growth peaks at 0.50% a bet when 10% is staked and falls back to about zero at a 19.9% stake, so doubling the growth-maximising bet throws away the whole benefit of the edge. Half the Kelly stake keeps about three quarters of the peak growth. 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.
The relationshipf the fraction of capital staked on each bet p, q the chances of winning and losing, 55% and 45% g(f) the long-run growth rate per bet f^{*} the growth-maximising stake What it says in wordsLong-run growth is the probability-weighted log of each outcome, and it peaks when the stake equals the edge.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.
Where candidates lose it
Candidates reason from expected profit, which rises in a straight line with the stake, and conclude that more is always better with a positive edge. Others say bet everything, or bet 55%, confusing the win probability with the stake.
Frame it as compounding from the first sentence, give f equals p minus q, and then show the 20% case going to zero. That second number is usually why the question is asked.
What the interviewer asks next
- What is the Kelly stake if the bet pays 2 to 1 and wins 40% of the time?
- Why do most practitioners bet half Kelly or less?
- How does uncertainty about the true win probability change the stake you choose?
056You roll a fair die and are paid its face value in thousands of rupees. After seeing the first roll you may reroll once, but then you must take the second result. On which faces do you reroll, and what is the game worth?Quantitative asset managementHedge funds
Try it first
What is the most you should pay to play?
Show the worked solution
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.
Faces 4, 5 and 6 beat the 3.5 a fresh roll is worth and are kept, while 1, 2 and 3 are rerolled, which lifts the value of the game from 3.5 to 4.25 thousand rupees. The relationshipV the value of the game, thousand rupees (4+5+6)/6 the kept faces, each weighted by its one-in-six chance 3/6 x 3.5 half the time you reroll and get a fresh roll's average What it says in wordsAdd the value of each face you keep, weighted by its chance, to the chance of rerolling times what a reroll is worth.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.
Where candidates lose it
The trap is anchoring on the average face and saying the game is worth 3.5, as if the reroll were decoration. The option to reroll is worth three quarters of a thousand rupees, and missing it is missing the point of the question.
The second loss is the wrong threshold. Candidates reroll only a 1, or reroll up to 4, because they compare the face with the middle of the die rather than with the value of the alternative. Say the rule first, then the faces.
What the interviewer asks next
- What is the game worth with two rerolls, and on which faces do you stop?
- What if you must pay Rs 500 for each reroll?
- Now you may keep the higher of your two rolls. What is that worth?
069A stock trades at Rs 100. In a year it will be worth Rs 150 with probability 40% or Rs 70 with probability 60%, and it pays no dividend. Your required return for this risk is 10%. Is it worth buying at Rs 100?Fundamental asset managementAsset management
Try it first
The expected price is above today's price. Does that make it a buy?
Show the worked solution
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.
A 40% chance of Rs 150 and a 60% chance of Rs 70 give an expected price of Rs 102, which discounted at the required 10% is worth Rs 92.7 today, below the Rs 100 price. The relationshipV_0 the value today, Rs 0.4, 0.6 the chances of the up and down cases 1.10 one plus the required return What it says in wordsWeight each outcome by its chance, then discount the expected price at the return you require.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.
Where candidates lose it
The trap is stopping at the expected price. Rs 102 is above Rs 100, candidates say buy, and they have ignored the required return the question hands them.
The second trap is comparing the Rs 50 upside with the Rs 30 downside without weighting them. A bigger upside that happens less often is not automatically better.
What the interviewer asks next
- What required return would make Rs 100 a fair price?
- If the down case became Rs 80, would it be worth buying?
- How would you size a position in a stock with this payoff if you did like it?
081On average, how many tosses of a fair coin does it take to see two heads in a row? And why is the answer for a head followed by a tail smaller?Quantitative asset managementHedge funds
Try it first
Which pair of averages is right?
Show the worked solution
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.
For HH, a tail from state H sends you back to Start, which makes the expected wait 6 tosses; for HT, a head from state H keeps you at H, so no progress is ever lost and the expected wait is only 4 tosses. 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.
The relationshipE_0 expected tosses still needed from the start E_1 expected tosses still needed after one head What it says in wordsEach state's wait is one toss plus the average of the waits in the states it can lead to.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.
Where candidates lose it
The fast wrong answer is 4 for both, from reasoning that each pattern has a one-in-four chance per pair of tosses. That treats tosses as separate non-overlapping pairs, which they are not.
The other loss is getting 6 by recalling it without the state equations. The follow-up is always a new pattern, so show the method, not the memory.
What the interviewer asks next
- How many tosses on average to see three heads in a row?
- In a race between HH and HT, which appears first more often?
- Where does the idea of losing progress on a miss show up in trading or risk?