Private Wealth Management puzzles, solved step by step

Puzzles: 100
Traced to a firm: 3
Topics: 13
Hard: 30

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018One fund in ten is truly skilled and beats its benchmark in 70% of years; the rest are unskilled and beat it in 50% of years. A fund has just beaten its benchmark three years running. What is the chance it is skilled?Probability and risk of lossHardWealth management
Try it first
After three straight wins, how likely is the fund to be skilled?
Show the worked solution
About 23%. Take 1,000 funds: 100 skilled and 900 unskilled. A skilled fund wins three years running with chance 0.7 cubed, 34.3%, so 34.3 funds. An unskilled fund does it with chance 0.5 cubed, 12.5%, so 112.5 funds. Of the 146.8 funds with a streak, 34.3 are skilled: 23.4%. The streak moves the odds from 10% to 23%, but most streak funds are still unskilled.
Why is the answer not close to 70%?
Think of a medical test that is good but not perfect, used for a rare condition. Most positive results come from the many healthy people, simply because there are so many more of them. How common skill is to begin with matters as much as how well skill shows up in results. Skilled funds are one in ten, so even though they streak more often, unskilled funds produce most of the streaks by sheer numbers.
Of 1,000 funds, 34.3 skilled and 112.5 unskilled funds post a three-year winning streak, so a fund with a streak is skilled only 23.4% of the time, up from a starting 10%.
How do you set it up without a formula sheet?
Use natural frequencies. Pick a round population, 1,000 funds, and count how many land in each branch; the answer is one count over the total count. That is Bayes ruleA way of updating a starting probability with new evidence, by weighing how likely the evidence is under each possible explanation. without the notation, and it is far harder to get wrong out loud. The formula version gives the same 23.4%.
The relationship
$P(S\mid W^3) = \frac{0.1 \times 0.7^3}{0.1 \times 0.7^3 + 0.9 \times 0.5^3} = \frac{0.0343}{0.1468} \approx 23\%$
S the fund is skilled
W^3 three wins in a row
0.1, 0.9 the share of skilled and unskilled funds before any results
0.7^3, 0.5^3 the chance of a three-year streak for each type
What it says in wordsThe chance of skill given a streak is the skilled streaks divided by all streaks.
The client version is one sentence: a three-year record is weak evidence on its own. It is also worth naming the limitation of the model: real skill is not a clean 70% and fund returns are not independent year to year, but the direction of the answer survives both.
Where candidates lose it
The trap is answering 70% or 34%, the numbers attached to skilled funds. Both describe how a skilled fund behaves, not how many streak funds are skilled, and confusing the two is the most common error in probability questions.
The other loss is fumbling the formula. Counting 1,000 funds through the tree is faster, is easier to say, and checks itself.
What the interviewer asks next
What if the fund has won five years running?
If one fund in four were skilled, what would three wins imply?
How would you use this when a client wants to buy last year's top fund?
051A client stakes his whole portfolio on a fair coin, again and again. Heads, the portfolio rises 60%; tails, it falls 40%. The expected return per flip is plus 10%. What happens to the typical client after many flips?Probability and risk of lossHardPrivate banking
Try it first
After 40 flips, where does the typical client, the one in the middle of the pack, stand?
Show the worked solution
The typical client shrinks towards zero, even though the average grows 10% a flip. A head and a tail together multiply wealth by 1.6 x 0.6 = 0.96, so the client in the middle loses about 2% a flip. After 40 flips the average stands at 45 times the start, but the median client holds 0.44 of it and only 44% of clients are ahead. A handful of lucky paths carry the average.
Why does a positive average not help the client in the middle?
Think of a shop that raises a price 60% and then cuts it 40%. A Rs 100 item goes to Rs 160 and then to Rs 96, not back to Rs 100, because the cut is taken on the bigger number. When the whole portfolio is staked every time, returns multiply, and the growth one client actually lives through is the geometric average, not the arithmetic one. Here that is the square root of 1.6 x 0.6, which is 0.9798: a loss of about 2.0% a flip, while the arithmetic average says plus 10%.
The relationship
$\underbrace{\tfrac12(1.6)+\tfrac12(0.6)}_{\text{average}} = 1.10 \qquad \underbrace{\sqrt{1.6\times0.6}}_{\text{typical}} = 0.98$
1.6 the wealth multiple after a head
0.6 the wealth multiple after a tail
1/2 the chance of each outcome
What it says in wordsThe average across many clients grows 10% a flip; the path one client lives through shrinks about 2% a flip.
Over 40 flips the average across all clients rises 10% a flip to 45 times the start, while the typical client, with as many heads as tails, zigzags down to 0.44 times, because every head and tail pair leaves 0.96 of the money.
Where has the average gone, then?
It sits with a very small number of clients who threw long runs of heads. The average is real, but it is an average across people, and no single client can collect it by waiting. To finish ahead after 40 flips a client needs at least 21 heads, which happens 43.7% of the time. Stretch it to 100 flips and the average reaches about 13,781 times while the middle client holds 0.13, about an eighth of what he began with.
What changes if he stakes only part of the portfolio?
Staking less cuts the damage of a tail more than it cuts the gain of a head. The Kelly fractionThe share of wealth to stake on a repeated favourable bet that maximises the long-run growth rate. Named after John Kelly, who derived it in 1956. for this coin is 0.5/0.4 minus 0.5/0.6, about 42% of the portfolio. At that stake a head and a tail multiply wealth by 1.25 x 0.833, which is 1.0417, so the typical client grows about 2.1% a flip instead of shrinking. The same favourable bet turns from ruinous to useful purely through position size. That is the point a wealth interviewer wants: volatility costs compound wealth even when the expected return is positive.
Where candidates lose it
The trap is answering with the expected value. Candidates hear plus 10% a flip, compound it, and describe a client who gets rich. They have averaged across clients when the question asked about one client through time.
The second loss is getting the geometric average right but not being able to say what to do about it. Have the sizing answer ready: stake part of the portfolio and the same coin grows the typical client.
What the interviewer asks next
What stake on this coin maximises the typical client's growth, and why is staking more than that worse?
How does this link to the gap between a fund's average annual return and its compound annual return?
The coin pays plus 50% or minus 40%. Does the typical client now grow if he stakes everything?
095A portfolio has an expected return of 10% a year and a volatility of 20%, with independent, normally distributed years. What is the chance that its average annual return over ten years is negative?Probability and risk of lossHardWealth management
Try it first
Your estimate of the chance of a negative ten-year average.
Show the worked solution
About 6%. The average of ten independent years has the same 10% mean but a volatility of 20 divided by the square root of 10, which is 6.3. Zero is then 10 / 6.3, about 1.58 standard deviations below the mean, and the normal table puts 5.7% of outcomes below that, against 31% for a single year.
Why does averaging over ten years narrow the range?
One cricket innings can be anything from a duck to a century; a batter's average over ten innings is far more stable. Good and bad innings cancel partly. Averaging independent years cancels part of their randomness, so the volatility of the average falls with the square root of the number of years, while the expected return stays the same. Ten years divide 20 by about 3.16.
The relationship
$\sigma_{\bar{R}} = \frac{20\%}{\sqrt{10}} = 6.32\% \qquad z = \frac{0 - 10}{6.32} = -1.58 \qquad \Phi(-1.58) \approx 5.7\%$
\sigma_{\bar{R}} the volatility of the ten-year average return
\sqrt{10} the square root of the number of independent years
\Phi the share of a normal distribution below a given z
What it says in wordsShrink the volatility by root ten, then read off how far below the mean zero now sits.
Both bells centre on 10%, but the ten-year average has a volatility of 6.3 instead of 20, so the red tail below zero shrinks from 31% of outcomes for one year to 5.7% for the ten-year average.
Does this mean ten years makes the portfolio safe?
No, and say why. Time narrows the range of the average return, but it widens the range of rupee outcomes, because each year's result compounds on a bigger pot. The chance of a bad decade is smaller than the chance of a bad year, and a bad decade still costs far more money when it comes.
Two other limits belong in the answer. Years are not fully independent, and real returns have fatter tails than the normal curve, so the true chance is likely higher than 5.7%. And the average return is not the compound return: using the roughly 8% compound rate from volatility drag, the chance of ending the decade with less money than you started with is closer to 10% on the same rough method.
Where candidates lose it
Candidates either keep the one-year 31%, missing that averaging narrows the spread, or say zero because ten years sounds long. Both skip the square root of ten, which is the whole point.
The second loss is claiming that time removes risk. Give the 6%, then say the rupee range widens and the tails are fatter than normal: the interviewer wants the number and its limit.
What the interviewer asks next
What is the chance over 25 years?
How many years until the chance of a negative average falls below 1%?
Why does the chance of ending with less money differ from the chance of a negative average return?

S	the fund is skilled
W^3	three wins in a row
0.1, 0.9	the share of skilled and unskilled funds before any results
0.7^3, 0.5^3	the chance of a three-year streak for each type

1.6	the wealth multiple after a head
0.6	the wealth multiple after a tail
1/2	the chance of each outcome

\sigma_{\bar{R}}	the volatility of the ten-year average return
\sqrt{10}	the square root of the number of independent years
\Phi	the share of a normal distribution below a given z

Try it first

Why is the answer not close to 70%?

How do you set it up without a formula sheet?

Where candidates lose it

What the interviewer asks next

Try it first

Why does a positive average not help the client in the middle?

Where has the average gone, then?

What changes if he stakes only part of the portfolio?

Where candidates lose it

What the interviewer asks next

Try it first

Why does averaging over ten years narrow the range?

Does this mean ten years makes the portfolio safe?

Where candidates lose it

What the interviewer asks next