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  1. 001A book holds 20 independent positions of 5% each, and each has a 10% chance of going to zero over the year. What is the probability that you lose 15% or more of the book?Betting and sizingCoreMulti-manager platformsProp and quant trading firms

    Try it first

    Before calculating: roughly how likely is a loss of 15% or more?

    Show the worked solution

    About 32%. A 15% loss means three or more of the 20 positions go to zero. The count of zeros is binomial with 20 tries at 10%, so the chance of zero, one or two is 12.2% + 27.0% + 28.5% = 67.7%, and the chance of three or more is 32.3%. The book expects two blow-ups a year, so three is not a tail event.

    Why is a rare event per name a common event per book?

    Think of a wedding with twenty guests, each with a one in ten chance of arriving late. Any single guest is almost certainly on time, but a host who plans for nobody being late is planning badly: on average two will be. When you hold many independent risks, the question is not whether one fails but how many do, and the expected count here is 20 x 10% = 2. A loss of 15% needs three failures, which is one more than an average year.

    How many of 20 positions go to zero in a year, each at a 10% chance12.2%00%27.0%1-5%28.5%2-10%19.0%3-15%9.0%4-20%3.2%5-25%0.9%6-30%0.2%7-35%<0.1%8-40%LostBook15% or worseThree or more zeros32.3%of yearsExpected zeros = 20 x 10% = 2, so three is only one more than the average year
    With 20 positions each carrying a 10% chance of going to zero, the most likely outcomes are one or two zeros; three or more zeros, a loss of 15% or more, happen in 32.3% of years.

    How do you count the ways to lose three or more?

    Count the outcomes you can live with and subtract. It is faster to add up zero, one and two blow-ups and take them from one than to add up three through twenty. Zero needs all twenty to survive: 0.9 to the twentieth, 12.2%. One needs a single failure, with twenty choices of which name: 20 x 0.1 x 0.9 to the nineteenth, 27.0%. Two has 190 possible pairs: 190 x 0.01 x 0.9 to the eighteenth, 28.5%.

    The relationship
    P(X≥3)=1−∑k=02(20k)(0.1)k(0.9)20−k≈1−0.677=0.323P(X \ge 3) = 1 - \sum_{k=0}^{2} \binom{20}{k} (0.1)^k (0.9)^{20-k} \approx 1 - 0.677 = 0.323
    Xthe number of positions that go to zero
    \binom{20}{k}the number of ways to choose which k names fail
    0.1 and 0.9the chance one name fails, and survives
    What it says in wordsThe chance of three or more failures is one minus the chance of zero, one or two.

    What does a risk manager take from this?

    Sizing each position so a single wipe-out is survivable does not make the book survivable. Five per cent a name feels small, yet a 15% drawdown is roughly a one in three year event on these odds, and a platform with a 10% drawdown limit would see it breached in 60.8% of years, because two zeros, the average outcome, already cost 10%. Say the limitation as well: the positions are assumed independent. In a sell-off failures cluster, and correlation fattens exactly the tail you have just computed.

    Where candidates lose it

    The fast wrong answer multiplies: 10% cubed is 0.1%, so three blow-ups look like a freak. That ignores the 1,140 different ways to choose which three names fail, and it ignores four, five and more.

    The second loss is stopping at exactly three. The question says 15% or more, so either sum three through twenty or, far faster, take the complement of zero, one and two. Say which route you are taking before you start.

    What the interviewer asks next

    • What is the chance of losing 10% or more?
    • If the 20 names are positively correlated, does the chance of a 15% loss rise or fall, and why?
    • How would you resize the book so a 15% loss happens less than one year in ten?