Risk Management puzzles, solved step by step

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All topicsCapital and leverage6Compounding and drawdowns8Correlation and diversification8Counterparty exposure and collateral7Credit risk arithmetic10Duration and rates7Liquidity and balance sheet7Logic, estimation and brainteasers7Operational loss and fraud7Options and Greeks7Probability and base rates8Statistics and estimation10VaR and expected shortfall8
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  1. 047A desk makes or loses Rs 1 crore each day, winning with probability 0.52. It stops as soon as it is up Rs 10 crore or down Rs 10 crore. What is the probability it reaches the profit target first?Probability and base ratesHardBank market riskQuant risk

    Try it first

    The daily edge is 52 against 48. What is the chance of hitting plus 10 before minus 10?

    Show the worked solution

    About 69%. This is the gambler's ruin. With r = 0.48 / 0.52, starting 10 steps from each barrier, the chance of hitting the top first is 1 / (1 + r to the power 10). r to the 10 is about 0.449, so the answer is 1 / 1.449, or 69.0%. A 2-point daily edge becomes a 19-point edge because the race takes many days.

    Why does a small daily edge grow so much?

    A casino wins only a couple of points more than half its bets at roulette, yet over thousands of spins it almost never ends a month behind. An edge that barely shows in one step accumulates over many, because the random part grows with the square root of the number of steps while the edge grows in proportion to it. Reaching plus or minus 10 takes around 95 days on average here, long enough for the edge of 0.04 a day to push the walk decisively one way.

    The relationship
    P(+10 first)=1−r101−r20=11+r10,r=0.480.52=0.923,  r10=0.449P(\text{+10 first}) = \frac{1 - r^{10}}{1 - r^{20}} = \frac{1}{1 + r^{10}}, \qquad r = \frac{0.48}{0.52} = 0.923, \; r^{10} = 0.449
    rthe ratio of losing to winning odds on one day
    10the distance in steps from the start to each barrier
    What it says in wordsThe gambler's ruin formula gives the chance of reaching one barrier before the other; starting in the middle it simplifies to one over one plus r to the power of the distance.
    A 2-point daily edge becomes a 19-point edge over the race50%60%70%80%90%100%0.500.520.540.56Chance of winning each daydashed: race odds no better than the daily odds0.52 gives 69.0%91.8%Expected lengthabout 95 daysat p = 0.52
    The chance of reaching plus Rs 10 crore before minus Rs 10 crore rises steeply with the daily win probability, from 50% for a fair coin to 69.0% at 0.52 and 91.8% at 0.56, far above the dashed line where the race would only match the daily edge.

    How do you check the answer, and what does it say about stop-losses?

    Two sanity checks. At 0.50, r is 1 and the formula gives 50%, as symmetry demands. Widen both barriers to 20 and the answer at 0.52 rises to 83.2%. The wider the barriers relative to the daily step, the more the edge dominates the noise. That is the risk manager's reading: a tight stop-loss relative to daily volatility turns a trader with a genuine edge into a near coin flip, while a wide one lets the edge show but exposes more capital to a trader without one.

    State the limits. The model assumes a constant edge, equal step sizes and independent days. Real P&L has fat tails, the edge decays, and a losing streak may itself be evidence that the edge was never there. The formula tells you how an edge would play out, not whether you have one; with drawdownThe fall from a running peak in cumulative profit, measured in rupees or as a percentage. limits, that second question is usually the harder one.

    Where candidates lose it

    The most common answer is 52%, treating the race as one big coin toss with the same odds as a single day. The whole puzzle is about how a small edge compounds over many steps.

    The second trap is knowing the formula but fumbling it. Say the general form, simplify it for the symmetric start, and check it at 0.50; the check earns as much as the number.

    What the interviewer asks next

    • What is the expected number of days before the desk stops?
    • The loss limit is Rs 5 crore and the target stays at Rs 10 crore. What is the answer now?
    • How would you test whether a trader really has a 0.52 edge from their P&L history?
  2. 072A 99% one-day VaR model is exceeded independently with probability 1% each trading day. On average, how many trading days pass before you first see exceptions on two consecutive days?Probability and base ratesHardBank market riskQuant risk

    Try it first

    Before setting up equations: which is it?

    Show the worked solution

    10,100 trading days on average. Let E0 be the expected wait from no current run and E1 the wait after one exception. Then E0 = 1 + 0.99 E0 + 0.01 E1 and E1 = 1 + 0.99 E0. Solving gives E0 = (1 + p) / p squared = 1.01 / 0.0001 = 10,100, about 40 years of trading days if exceptions really are independent.

    Why is the answer not simply one over p squared?

    Waiting for two heads in a row with a fair coin takes six tosses on average, not four, because each time you get one head and then a tail, you are back to the start and the first head was wasted. A run has memory: after one exception you are one step from the finish, and a miss throws you back, so the waiting time needs one equation per state, not one probability. For VaR exceptions at 1% the extra is 100 days, the time spent waiting for each first exception.

    Waiting for a run needs a state you can fall back fromState 0no run yetState 1one exceptionState 2two in a row0.010.010.99: no exception, back to the start0.99: stayE0 = 1 + 0.99 E0 + 0.01 E1E1 = 1 + 0.99 E0E0 = (1 + p) / p squared= 1.01 / 0.0001 = 10,100 daysAbout 40 years of trading days, if exceptions are truly independent
    From no run, an exception moves you to state 1 with probability 0.01; from state 1 another exception ends the wait, but a quiet day with probability 0.99 sends you back to the start. Solving the two waiting-time equations gives 10,100 days, not 10,000.

    Why would a risk manager care about a 40-year wait?

    Because it turns an observation into a test. If the model is right and exceptions are independent, two in a row should take about forty years of trading to appear; seeing them twice in one year says the exceptions cluster, and clustering means the model misses changes in volatility. Backtesting frameworks test independence for exactly this reason, alongside the plain count of exceptions.

    The relationship
    E0=1+qE0+pE1,E1=1+qE0  ⇒  E0=1+pp2=1.010.0001=10,100E_0 = 1 + qE_0 + pE_1,\quad E_1 = 1 + qE_0 \;\Rightarrow\; E_0 = \frac{1+p}{p^2} = \frac{1.01}{0.0001} = 10{,}100
    E_0expected days to finish from no current run
    E_1expected days to finish just after one exception
    p, qthe daily exception probability 0.01 and its complement 0.99
    What it says in wordsWrite the wait from each state in terms of where the next day takes you, then solve the two equations.

    The limitation is the independence assumption, which is the thing being tested. Real exceptions tend to arrive in bursts when volatility jumps, so the observed wait is usually far shorter, and that gap is evidence against the model rather than bad luck.

    Where candidates lose it

    The common wrong answer is 10,000, one over p squared. It treats every pair of days as a fresh independent trial, ignoring that the pairs overlap and that a failed attempt costs the wait for a new first exception.

    The other trap is 200 days, the wait for any two exceptions. Consecutive is the whole point. Set up the states before you calculate, and the 100 extra days explain themselves.

    What the interviewer asks next

    • How many days on average before three exceptions in a row?
    • How many exceptions would you expect in 250 days, and what is the chance of seeing 5 or more?
    • What does it tell you if exceptions cluster in the same weeks?
  3. 097A client offers you a game: a fair coin is flipped until the first tail, and you are paid Rs 2 raised to the number of flips. The client can pay at most Rs 1 crore. What is the fair price of the game?Probability and base ratesHardQuant riskCounterparty risk

    Try it first

    Roughly what is the capped game worth?

    Show the worked solution

    About Rs 24.19. A game that ends on flip n pays 2 to the n with probability one half to the n, so each flip count adds exactly Rs 1. Uncapped, that sum is infinite. The client can pay only Rs 1 crore, which 2 to the 23 stays below, so 23 flip counts add Rs 23 and every longer game pays Rs 1 crore, adding about Rs 1.19.

    Why is the uncapped game worth an infinite amount?

    Picture a prize that doubles every time you survive another round, while the chance of surviving halves. Each round's prize times its chance is always the same Rs 1. In this game every possible length contributes exactly Rs 1 to the expected value, one Rs 1 for each flip count, forever, so the sum never stops. This is the St Petersburg paradox: the arithmetic says pay anything, yet nobody would pay even Rs 100.

    Each flip count adds Rs 1 until the payer runs out of money1510152023242530Rs 10payer's cap: Rs 1 crore2 to the 23 = Rs 83,88,60823 flip counts x Rs 1 each = Rs 23capped tailadds Rs 1.19Number of flips until the first tailFair priceRs 24.1923 + 1.19
    Each flip count from 1 to 23 adds exactly Rs 1 to expected value, but beyond the payer's Rs 1 crore cap the contributions halve each flip and add only Rs 1.19 in total, so the capped game is worth about Rs 24.19.

    How does the cap turn infinity into Rs 24?

    Find where the cap bites: 2 to the 23 is Rs 83,88,608 and 2 to the 24 is over Rs 1 crore, so the first 23 flip counts pay in full. Those contribute Rs 23; every longer game pays the capped Rs 1 crore, and the chance of lasting past 23 flips is one half to the 23, so the tail adds Rs 1 crore divided by 2 to the 23, about Rs 1.19. The fair price is about Rs 24.19.

    The relationship
    EV=∑n=1232−n⋅2n+107⋅2−23=23+1.19=24.19EV = \sum_{n=1}^{23} 2^{-n} \cdot 2^{n} + 10^{7} \cdot 2^{-23} = 23 + 1.19 = 24.19
    2^{-n}the chance the first tail comes on flip n
    2^nthe payout if it does
    10^7the payer's cap, Rs 1 crore
    What it says in wordsEvery flip count below the cap adds one rupee; the capped tail adds the cap times the chance of reaching it.

    Why is this a counterparty risk question?

    Because the value came entirely from payouts the client could never make. A promised payoff is worth only what the payer can actually pay, and most of this game's theoretical value sat in states where the payer would default. Raising the cap a thousandfold to Rs 1,000 crore only lifts the value to about Rs 34.16, because each doubling of capacity adds just one rupee. The same logic prices protection bought from a seller who could not survive the event it insures.

    Where candidates lose it

    The tempting answer is infinity, or a large number near the cap, because the uncapped maths is famous. The question gives the cap precisely to see whether you use it.

    The other slip is stopping at Rs 23 and forgetting the capped tail, or adding Rs 1 crore for every long game rather than weighting it by the chance of getting there.

    What the interviewer asks next

    • What would you pay if the client could pay at most Rs 1,000 crore?
    • Why might a risk-averse person pay far less than the expected value even with the cap?
    • Where do you see payoffs that are only as good as the payer's capacity in real markets?