Mutual Fund Mastery puzzles, solved step by step

Puzzles: 100
Traced to a firm: 29
Topics: 13
Hard: 30

Showing 1–8 of 8 · filtered from 100Clear filters

009A thousand fund managers have no skill at all: each has a 50% chance of beating the index in any year, independently. How many beat it five years in a row by luck, and what is the chance that at least one beats it ten years running?Probability and expected valueCoreFund research and ratingsIndian AMCs
Try it first
Guess the chance that at least one of the thousand posts a ten-year streak.
Show the worked solution
About 31 managers beat the index five years running by luck alone, and there is about a 62% chance that at least one beats it ten years running. Each year halves the survivors: 1,000, 500, 250, 125, 62.5, 31.25. For ten years, one manager's chance is 1 in 1,024, so the chance none of the thousand does it is (1023/1024) to the 1,000th, about 38%.
Why do perfect records appear even when nobody is skilled?
Ask a stadium of a thousand people to toss a coin and sit down on tails. After five rounds about 31 are still standing, and each of them has a perfect record. A streak that is rare for one person is expected somewhere in a large enough crowd, so the size of the starting group matters as much as the length of the streak. The five-year count is just 1,000 halved five times.
Starting from 1,000 managers with no skill, halving each year leaves about 31 with a perfect five-year record, and the chance that at least one of the 1,000 posts a ten-year streak is about 62%.
How do you get the ten-year figure without a calculator?
Go through the complement: work out the chance that nobody does it. Each manager fails with probability 1023 over 1024. For many small independent chances, (1 minus 1/n) to the power n is close to 1 over e, about 0.37, and here the power is 1,000 against 1,024, so the chance nobody does it is about 0.38. That leaves about 62% for at least one ten-year streak.
The relationship
$P(\text{at least one}) = 1 - \left(1 - \tfrac{1}{1024}\right)^{1000} \approx 1 - e^{-0.977} \approx 62\%$
1/1024 one manager's chance of ten wins in a row, one half to the tenth
1000 the number of managers trying
What it says in wordsTake the chance that every manager fails, and one minus that is the chance at least one succeeds.
Say what it means for fund selection. A long record of beating the index is evidence, but weaker evidence than it looks when it is picked from a large universe after the fact. The limitation of the model is that real returns are not coin tosses and some managers do have skill; the puzzle only shows how much luck alone can produce.
Where candidates lose it
The trap in the second part is answering 1 in 1,024 or 0.1%, the chance for one named manager, when the question asks about anyone in the group. It is the same slip as being amazed that someone at a party shares your birthday.
The other slip is adding the chances, 1,000 times 1 in 1,024, to get 98%. That double counts the cases where two or more managers succeed; the complement avoids it.
What the interviewer asks next
How many managers would you expect with exactly four wins out of five years?
If one manager in the group truly beats the index 60% of years, how likely is a ten-year streak for them?
Why does survivorship, funds closing after bad years, make published track records look better still?
020Ten per cent of fund managers are skilled and beat their index in 70% of years; the rest have no skill and beat it in 50% of years. A manager has beaten the index three years running. What is the probability that this manager is skilled?Probability and expected valueHardFund research and ratingsIndian AMCs
Try it first
After a three-year streak, how likely is the manager to be skilled?
Show the worked solution
About 23%. Picture 1,000 managers. The 100 skilled ones produce 100 times 0.7 cubed, 34.3 three-year streaks. The 900 unskilled produce 900 times 0.5 cubed, 112.5. Of 146.8 streak holders, 34.3 are skilled, 23.4%. The streak more than doubles the odds of skill, from 10%, yet most streak holders are still lucky.
Why does a three-year streak prove so little?
Think of a test for a rare condition that catches most real cases but also flags plenty of healthy people. If the condition is rare, most positive results are false alarms. When the thing you are looking for is rare, even good evidence leaves most positives coming from the larger group, so the starting proportion matters as much as the evidence. Here skill is the rare condition, at 10%, and a three-year streak is a test that unskilled managers pass one time in eight.
Of 1,000 managers, the 100 skilled produce 34.3 three-year streaks and the 900 unskilled produce 112.5, so only about 23% of managers with a streak are actually skilled.
How do you set it up fast in the room?
Use counts, not formulas. Pick 1,000 managers, split them by skill, then split each group by whether it produced the streak, and the answer is one box over the sum of two boxes. Skilled: 100 times 0.343 is 34.3. Unskilled: 900 times 0.125 is 112.5. The answer is 34.3 over 146.8, 23.4%. The same result in odds form: prior odds of 1 to 9, times a likelihood ratiohow many times more likely the evidence is if the manager is skilled than if not of 0.343 over 0.125, about 2.74, gives 2.74 to 9.
The relationship
$P(S\mid \text{streak}) = \frac{0.1 \times 0.7^3}{0.1 \times 0.7^3 + 0.9 \times 0.5^3} = \frac{0.0343}{0.0343 + 0.1125} \approx 23.4\%$
S the manager is skilled
0.1 and 0.9 the shares of skilled and unskilled managers
0.7^3 and 0.5^3 each group's chance of three wins in a row
What it says in wordsThe chance of skill given a streak is the skilled managers' share of all the streaks.
Push it once to show judgement. A ten-year streak changes the picture: 100 times 0.7 to the tenth against 900 times 0.5 to the tenth gives about 76% skilled, because the unskilled pass a ten-year test only about once in a thousand. The limitation is that the 10% and 70% are assumptions, and real years are not independent coin tosses; managers with a style that suits a market phase can string wins together without skill.
Where candidates lose it
The common slip is answering 70%, mixing up the chance a skilled manager wins with the chance a winner is skilled. It is the same inversion that makes people overrate a positive result on a test for a rare condition.
The other slip is ignoring the 10% starting share and answering from the hit rates alone, 0.343 over 0.343 plus 0.125, about 73%. That treats skilled and unskilled managers as equally common, which is the one thing the question told you they are not.
What the interviewer asks next
How many consecutive winning years would push the probability of skill above 50%?
How would the answer change if skilled managers made up 30% of the population?
Why does survivorship in fund databases make this problem worse in practice?
032A corporate bond pays 11% for the year, with a 4% chance of default and 40% recovery if it defaults. A AAA bond pays 7.5% for certain. Which has the higher expected return, and why might a debt fund still choose the AAA bond?Probability and expected valueCoreFixed income desksIndian AMCs
Try it first
What is the risky bond's expected return for the year?
Show the worked solution
The risky bond, 8.16% against 7.5%. Ninety six times in a hundred it pays 11%; four times it returns 40 of 100, a 60% loss. Weighted, 0.96 x 11 plus 0.04 x (minus 60) is 8.16%. A debt fund may still prefer the AAA because its investors treat it like a deposit: a single default is a sudden loss, triggers redemptions, and in a fund of twenty such bonds, one default a year is more likely than not.
How do you set up the expected return?
Think of lending Rs 100 to twenty-five friends at a good rate, knowing that one of them, on average, will repay only Rs 40. You have to count that friend before you celebrate the rate. Expected return weights every outcome by its chance, and the default outcome is a large loss, not a zero. Recovery of 40% means you lose 60 of your 100, and for simplicity the coupon is also lost in default.
The risky bond pays 11% with 96% probability and loses 60% with 4% probability, an expected 8.16%, above the AAA bond's certain 7.5%. The whole case against it sits in the loss branch, which a debt fund's investors are not expecting to see.
The relationship
$E[r] = (1-p)\,c + p\,(R - 1) = 0.96 \times 11\% + 0.04 \times (-60\%) = 8.16\%$
p probability of default in the year, 4%
c the coupon, 11%
R recovery, 40% of the money lent
What it says in wordsExpected return is the coupon when paid, weighted by its chance, plus the default loss, weighted by its chance.
If the risky bond pays more on average, why hold the AAA?
Because an average is not what a debt fund investor experiences. A debt fund is bought as a place for money that must not fall, so the question is not only the average but the chance and size of a loss. On one bond, the return has a standard deviation of about 13.9%, against zero for the AAA. Spread across twenty such bonds, the chance that at least one defaults in a year is 1 minus 0.96 to the twentieth, about 56%. Each default cuts the NAV overnight and can set off redemptions that force the fund to sell its better bonds.
There is also a margin-of-safety check: the risky bond only matches the AAA if the default chance rises to (11 minus 7.5) over 71, about 4.9%. A small error in the 4% estimate wipes out the advantage. Default probabilities are estimates, not facts, and they tend to rise together in a downturn, which is exactly when investors redeem.
Where candidates lose it
Candidates often forget that recovery is 40, not zero, and compute 0.96 x 11 = 10.56% or subtract only the coupon. Others treat default as a zero return and get 10.56% as well. The loss in default is the principal not recovered, 60%.
The bigger miss is stopping at 8.16% and declaring the risky bond better. The interviewer asked why a fund might still choose the AAA; the answer is about who holds the fund and what a loss does to them, not about the average.
What the interviewer asks next
What default probability makes the two bonds equal on expected return?
How does holding 50 such bonds instead of one change the picture?
Why would the default probability of these bonds be correlated, and why does that matter?
047An equity fund has a 25% chance of a negative year, independently each year. What is the chance that an investor who holds it for five years sees at least one losing year?Probability and expected valueWarm upFund research and ratingsIndian AMCs
Try it first
Your first guess: the chance of at least one losing year in five?
Show the worked solution
About 76%. The easy route is the opposite event. Five positive years in a row needs a 75% chance to come up five times: 0.75 to the fifth is about 23.7%. Every other outcome includes at least one losing year, so the chance is 1 minus 0.237, about 76.3%. A small yearly chance of loss becomes close to a sure thing over an ordinary holding period.
Why work with the opposite event?
Ask a cricket fan the chance that a bowler concedes at least one boundary in an over, and the quick way is to ask the chance he concedes none. At least one is messy to count directly, because it covers one loss, two losses and every combination; none is a single clean path, so compute none and subtract from one. Here none means five positive years, each with a 75% chance, independent of the others: 0.75 times itself five times is 23.7%.
The chance that every year so far has been positive falls by a quarter each year, from 75% after one year to 23.7% after five, so the chance of at least one losing year in five is about 76.3%.
The relationship
$P(\text{at least one loss}) = 1 - (1-p)^n = 1 - 0.75^5 \approx 0.763$
p the chance of a losing year, 25%
n years held, 5
What it says in wordsThe chance of at least one losing year is one minus the chance that every year is positive.
Why does an adviser care about this number?
Because it sets the client's expectation before the first bad year arrives. Over five years a losing year is the likely case, not the unlucky one, so a client who has not been told this reads the first negative year as something gone wrong. On the same assumptions the expected number of losing years in five is 1.25, the chance of exactly one is about 40%, and over ten years the chance of at least one rises to about 94%. A losing year is a calendar-year label; it says nothing about whether the five-year return was positive.
State the two assumptions. The 25% is an illustration, not a measured figure for any market, and real yearly returns are not fully independent: bad years cluster in some periods. Clustering lowers the chance of at least one losing year slightly while making the losing stretches longer. The method survives both caveats; the exact figure does not.
Where candidates lose it
The trap is answering 25%, as if holding longer did not create more chances to see a loss, or adding 25% five times and reaching 125%. Both come from treating at least one as a simple sum.
The quieter loss is forgetting the independence assumption. Say it in one breath with the answer; it is what makes 0.75 to the fifth valid.
What the interviewer asks next
What is the chance of exactly two losing years in five?
How many years would you have to hold for the chance of at least one losing year to pass 90%?
Why might the chance of a losing five-year period be far lower than the chance of a losing year?

058A fund's yearly returns average 12% with 18% volatility, and are roughly normal and independent from year to year. What is the chance of losing money in any one year, and what is the chance that its average return over ten years is negative?Probability and expected valueHardFund research and ratingsIndian AMCs

Try it first

Which pair is closest?

Show the worked solution

About 25% in any one year, and about 2% over ten years. One year: zero sits 12/18 = 0.67 standard deviations below the mean, which leaves 25.2% of outcomes below it. The ten-year average keeps the 12% mean but its spread falls to 18 divided by the square root of 10, 5.7%, so zero is 2.11 standard deviations away and the chance is 1.8%.

How likely is a losing year?

Picture the daily commute. Any one day might be twenty minutes late because of rain or a breakdown; your average over a month is almost never more than a few minutes off. Single outcomes are noisy; averages of many independent outcomes are much less noisy, because the bad days and the good days partly cancel. For one year, the question is how far zero sits below a 12% mean when the spread is 18%. That distance is 12/18 = 0.67 standard deviations, and the normal table puts 25.2% of the curve below it. One year in four is a loss, which matches what investors in equity funds actually live through.

The one-year return curve, centred on 12% with an 18% spread, has 25% of its area below zero, while the ten-year average curve has the same centre but a 5.7% spread and only 1.8% of its area below zero.

Why does the ten-year chance collapse to about two percent?

Averaging ten independent years keeps the centre at 12% but divides the spread by the square root of ten. The standard errorThe spread of an average. For independent draws it equals the spread of one draw divided by the square root of the number of draws. of the ten-year average is 18 / 3.16 = 5.7%, so zero is now 2.11 standard deviations below the mean instead of 0.67. The tail beyond 2.1 standard deviations is 1.8%. Five years sits in between: a spread of 8.0% and a chance of about 7%.

The relationship

P(\bar r_{10} < 0) = \Phi\!\left(-\frac{12}{18/\sqrt{10}}\right) = \Phi(-2.11) \approx 1.8\%

r bar 10	the average yearly return over ten years
12	the mean yearly return, per cent
18 / sqrt(10)	the spread of the ten-year average
Phi	the share of a normal curve below a given number of standard deviations

What it says in wordsDivide the distance to zero by the spread of the average, not by the spread of one year, then read the tail.

Now the limits, because a sharp interviewer will push. A negative arithmetic average is not quite the same as losing money: compounding knocks roughly half the variance off growth, so the fund compounds nearer 10.4% than 12%, and the chance of ending ten years below the starting amount is a little higher than 1.8%. Real returns also have fatter tails than a normal curve and are not fully independent, since bad years cluster. The direction survives all of that: time narrows the spread of the average, not the risk of a bad single year.

Where candidates lose it

The first trap is saying the risk of loss is the same at every horizon, or that it falls in proportion to time. It falls with the square root of time, which is why ten years cuts 25% to about 2%, not to zero and not to 2.5%.

The second is overselling the answer. Say what the two percent assumes: normal returns, independent years and a fixed mean, and that compounding and fat tails push the true figure somewhat higher.

What the interviewer asks next

What volatility would make the one-year chance of loss exactly one in three?
Why is the chance of ending below your starting value higher than the chance of a negative arithmetic average?
Over how many years does the chance of a negative average fall below 1%?

072A toll-road InvIT unit pays Rs 12 in a good year (40% chance), Rs 8 in a normal year (45%) and Rs 2 in a bad year (15%). What is the expected payout, and what is the unit worth at a 10% required return if this pattern continues forever?Probability and expected valueCoreNuveenChicago · 2023
Try it first
What is the unit worth at a 10% required return?
Show the worked solution
The expected payout is Rs 8.70 and the unit is worth about Rs 87. Weight each year by its chance: 0.40 x 12 = 4.80, 0.45 x 8 = 3.60, 0.15 x 2 = 0.30, which sum to Rs 8.70. A payment expected every year forever is worth that payment divided by the required return, so 8.70 / 0.10 = Rs 87. Valuing the likeliest year alone would give Rs 80.
How do you value income that changes every year?
A farmer whose crop is good four years in ten, ordinary in about half, and poor in the rest does not plan around a normal year. He plans around what the land produces on average over many years. When income is uncertain, you value the expected valueThe probability-weighted average of all possible outcomes: each outcome times its chance, added up. of the income, not the likeliest outcome and not the plain average of the outcomes. For this unit: 0.40 x Rs 12 = 4.80, 0.45 x Rs 8 = 3.60, 0.15 x Rs 2 = 0.30, a total of Rs 8.70 a year.
Weighting Rs 12, Rs 8 and Rs 2 by their chances gives an expected payout of Rs 8.70, which at a 10% required return forever values the unit at Rs 87, while valuing only the likeliest year would give Rs 80.
The relationship
$V = \frac{E[D]}{r} = \frac{0.40(12) + 0.45(8) + 0.15(2)}{0.10} = \frac{8.70}{0.10} = 87$
E[D] the expected yearly distribution
r the required return, 10%
V the value of the unit, assuming the pattern repeats forever with no growth
What it says in wordsAverage the payouts by their chances, then treat the average as a level payment forever.
Where do the two common wrong answers come from?
The likeliest year pays Rs 8, so some candidates value the unit at Rs 80. That throws away the 40% chance of Rs 12, which is worth Rs 7 of value here. Others average 12, 8 and 2 to get Rs 7.33, as if each year were equally likely, which gives Rs 73 and punishes the unit for a bad year that happens only 15% of the time. The weighting is the whole answer.
Now the honest limits, because the follow-up is usually how you would assess such an asset in practice. The expected value hides the spread: the payout's standard deviation here is about Rs 3.36, large against Rs 8.70, and an investor needing steady income cares about that. Uncertainty usually shows up in a higher required return, not in a lower expected payout, so the 10% must be chosen with the spread in mind. A toll road's traffic also trends and its concession ends, so the forever assumption is a simplification to state openly.
Where candidates lose it
The trap is valuing the most likely year, Rs 8, which gives Rs 80. It feels prudent but ignores that good years happen 40% of the time.
The second miss is double counting risk: weighting the payouts down for the bad year and then also using a high required return for the same risk. Say that probabilities go in the cash flow and the price of uncertainty goes in the rate, once each.
What the interviewer asks next
The bad-year chance rises to 30%, taken from the good years. What is the unit worth now?
Why might two investors pay different prices for this unit with the same expected payout?
The concession ends after 20 years with nothing left. Roughly what is the unit worth then?
Asked at Nuveen, Multifamily, Chicago, 2023 (Wall Street Oasis): Walk me through how you would assess the value of a property if the income stream is unpredictable?
084Two diversified funds each pick 50 stocks at random from the same universe of 100 stocks. How many stocks do you expect them to hold in common, and what does that suggest about a client who holds three large cap funds?Probability and expected valueHardIndian AMCsDistribution and sales
Try it first
How many stocks do you expect the two funds to share?
Show the worked solution
25 stocks, half of each fund, give or take about 2.5. Each of fund B's 50 picks has a 50 in 100 chance of being among fund A's, so the expected overlap is 50 x 0.5 = 25. Two funds fishing in the same pond share half their holdings by chance alone. With three such funds, about 12.5 stocks sit in all three, and together they hold only about 87.5 different names.
Why is the overlap so large with no coordination?
Two friends each order five dishes from a menu of ten without talking to each other. Each dish you pick has a 50% chance of being on your friend's list, so you expect to share about two and a half dishes. When each fund holds a large share of the same universe, overlap is not a coincidence to be explained; it is the default the arithmetic produces. The tool is linearity of expectation: go stock by stock, add up the chance that each one is in both funds, and the dependence between picks never needs to be modelled.
The relationship
$E[\text{shared}] = N \times \frac{k}{N} \times \frac{k}{N} = 100 \times 0.5 \times 0.5 = 25$
N stocks in the universe, 100
k stocks each fund holds, 50
k/N the chance a given stock is in one fund
What it says in wordsEach stock is in both funds with probability one half times one half, and there are 100 stocks, so expect 25 shared.
In one random draw of two 50-stock funds from 100 stocks, 25 stocks land in both, matching the expected 50 x 50/100, and a third such fund would leave only about 87.5 distinct names across 150 holdings.
How firm is 25, and what changes with three funds?
The count follows a hypergeometric distributionThe distribution of how many marked items you get when you draw a fixed number without replacement from a pool that holds a fixed number of marked items. with a standard deviation of about 2.5, so most random pairs share between 20 and 30 stocks. With three funds each holding half the universe, a stock is in all three with probability one in eight, about 12.5 stocks, and in none with probability one in eight too. Three such funds do not give three times the diversification: 150 holdings collapse to about 87.5 distinct stocks, and 50 of those are held by two or three of the funds at once.
Real funds are not random, and the difference cuts one way. A large cap mandate points every manager at the same biggest companies, and benchmark weights pull portfolios further together, so real overlap between large cap funds tends to sit above this random baseline rather than below it. Use the random case as the floor: if two funds overlap much more than chance would give, the second fund adds little except a second fee.
Where candidates lose it
The instinct is that two independent managers should have little in common, so candidates guess 5 or 10. They forget that each fund covers half the universe, which makes sharing the norm, not the exception.
The second loss is stopping at 25 without the portfolio point. The question is really about clients holding three or four funds from the same category who believe they are diversified; say what the overlap does to that belief, with the 87.5 distinct names as the number.
What the interviewer asks next
If each fund picks only 20 of the 100 stocks, what overlap do you expect?
How would you measure overlap by portfolio weight rather than by count?
A client holds four large cap funds. What would you look at first?
098You roll a fair die and are paid the face value in thousands of rupees. After seeing the first roll you may reroll once, but then you must take the second roll. When should you reroll, and what is the game worth?Probability and expected valueCoreIndian AMCsGlobal asset managers
Try it first
Which first rolls should you reroll?
Show the worked solution
Reroll on 1, 2 or 3; keep 4, 5 or 6. The game is worth 4.25 thousand, Rs 4,250. A reroll is a new die worth 3.5 on average, so keep any face above 3.5. Half the time you reroll and expect 3.5; the other half you keep 4, 5 or 6, worth 5 on average. So the game is 0.5 x 3.5 + 0.5 x 5, or 4.25. The right to reroll adds Rs 750 to a plain roll's Rs 3,500.
How do you decide whether to keep a roll?
A friend offers you a sealed envelope known to hold Rs 350 on average, in exchange for the Rs 300 note in your hand. You swap; if the note were Rs 500, you would keep it. Keep any outcome that beats what the alternative is expected to give, and the alternative here is a fresh roll worth 3.5 on average. So 4, 5 and 6 are kept and 1, 2 and 3 are thrown back. The decision depends only on comparing the face in front of you with 3.5.
Each first roll below the reroll's expected 3.5 is thrown back and each roll above it is kept, which makes the game worth 4.25, or Rs 4,250, against Rs 3,500 for a single roll with no option.
The relationship
$V = \frac{3}{6} \times 3.5 + \frac{4 + 5 + 6}{6} = 1.75 + 2.5 = 4.25$
3/6 the chance the first roll is 1, 2 or 3 and you reroll
3.5 the expected value of the reroll
(4+5+6)/6 the expected value from the faces you keep
What it says in wordsWeight each branch by its chance: rerolled faces are worth 3.5, kept faces are worth themselves.
Why is the option worth Rs 750, and what if there are more rerolls?
Without the reroll you get 3.5 on average. The option lets you throw away the low outcomes and replace them with an average one, so it lifts the value to 4.25: an option is worth something because you choose after seeing the outcome, and its value comes entirely from the bad outcomes it lets you escape. With two rerolls, work backwards: the last reroll is worth 3.5, so the middle roll is kept on 4 or more and is worth 4.25; the first roll is then kept only on 5 or 6, because 4 is below 4.25. The game rises to 4.67.
The fund-desk parallel is any decision with a later choice built in: a redemption right, a switch option, a stop-loss. Each one is worth what it lets you avoid, and each has a threshold set by what the alternative is expected to give. Saying the general rule, keep what beats the continuation value, is what lifts the answer above a piece of arithmetic.
Where candidates lose it
The common slip is setting the threshold by feel, keeping 3 because it is close to average or rerolling 4 because it feels low. The cutoff is exactly the reroll's expected value, 3.5, and a whole number on a die is either above it or below it.
The second loss is computing the value as the simple average of the best choices without weighting by probability. Say the threshold, then weight each branch: half the time 3.5, half the time 5.
What the interviewer asks next
With two rerolls allowed, what is the game worth? (4.67)
What would you pay to play if each reroll cost Rs 500?
If the payout were the face value squared, would the threshold change?

1/1024	one manager's chance of ten wins in a row, one half to the tenth
1000	the number of managers trying

S	the manager is skilled
0.1 and 0.9	the shares of skilled and unskilled managers
0.7^3 and 0.5^3	each group's chance of three wins in a row

p	probability of default in the year, 4%
c	the coupon, 11%
R	recovery, 40% of the money lent

E[D]	the expected yearly distribution
r	the required return, 10%
V	the value of the unit, assuming the pattern repeats forever with no growth

Try it first

Why do perfect records appear even when nobody is skilled?

How do you get the ten-year figure without a calculator?

Where candidates lose it

What the interviewer asks next

Try it first

Why does a three-year streak prove so little?

How do you set it up fast in the room?

Where candidates lose it

What the interviewer asks next

Try it first

How do you set up the expected return?

If the risky bond pays more on average, why hold the AAA?

Where candidates lose it

What the interviewer asks next

Try it first

Why work with the opposite event?

Why does an adviser care about this number?

Where candidates lose it

What the interviewer asks next

Try it first

How likely is a losing year?

Why does the ten-year chance collapse to about two percent?

Where candidates lose it

What the interviewer asks next

Try it first

How do you value income that changes every year?

Where do the two common wrong answers come from?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the overlap so large with no coordination?

How firm is 25, and what changes with three funds?

Where candidates lose it

What the interviewer asks next

Try it first

How do you decide whether to keep a roll?

Why is the option worth Rs 750, and what if there are more rerolls?

Where candidates lose it

What the interviewer asks next