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  1. 003A company has 18 months of cash and starts raising its next round once 9 months have passed. Each month of raising it has an independent 10% chance of closing. What is the chance it closes before the cash runs out, and by which month must it start raising for an 80% chance?Probability and expected valueCoreSeed and early-stage VCSeries A to C VC

    Try it first

    With 9 months of raising at 10% a month, what is the chance of closing?

    Show the worked solution

    About 61%, and for an 80% chance it must start raising after month 2. The chance of failing all nine months is 0.9 to the power 9, about 39%, so it closes about 61% of the time. An 80% chance needs 0.9 to the power n below 20%, which first happens at 16 months (81.5%). Eighteen months of cash less 16 means raising from month 2.

    Why not just add 10% a month?

    Think of a friend who picks up the phone one time in ten. Call nine times and you will not reach them 90% of the time, because every call after the one they answer is wasted. With repeated independent tries, count the chance of failing every time and subtract it from one; adding the chances double counts the months after a close. Adding 10% a month would give 100% at ten months and more than 100% after that, which is the tell that the method is wrong.

    The relationship
    P(close within n)=1−0.9n1−0.99=0.613P(\text{close within } n) = 1 - 0.9^{n} \qquad 1 - 0.9^{9} = 0.613
    0.9the chance of not closing in any one month
    nmonths of raising before the cash runs out
    What it says in wordsThe round closes unless every single month fails, so subtract the chance of all-failure from one.
    Each month adds 10% of what is still open, so the curve flattens20%40%60%80%100%80% targetWrong: 10% x monthsStart after month 9: 9 months, 61.3%16 months: 81.5%0369121518Months spent raising before the cash runs out
    At a 10% chance each month, nine months of raising gives a 61% chance of closing, and the curve first clears 80% at sixteen months, 81.5%, while simply adding 10% a month would wrongly reach certainty at ten.

    How do you find the month it must start?

    Set 0.9 to the power n below 0.2 and solve. Taking logs, n is at least ln 0.2 over ln 0.9, about 15.3, so round up to 16 whole months. Fifteen months gives 79.4%, just short of 80%, and sixteen gives 81.5%. With 18 months of cash, sixteen months of raising means starting after month 2. Doubling the time spent raising only lifts the odds from 61% to about 81%, because each extra month adds 10% of a shrinking remainder.

    What would you say to a founder about this?

    The model is simple and its lesson is not: fundraising odds climb slowly, so a company that waits until half its runway is gone has already given up a large share of its chances. The limitation is worth naming. Real months are not independent; a round that has not closed in six months usually gets harder, not equally likely, so the true curve is flatter still.

    Where candidates lose it

    The fast answer is 90%, nine months at 10% each. It is wrong in a way the interviewer can prove in one line: at eleven months the same method gives 110%.

    The second loss is rounding the 80% month down. 15.3 months must become 16, and fifteen gives 79.4%, which misses the target.

    What the interviewer asks next

    • If the monthly chance falls by one point every month the company has been raising, how does the answer change?
    • What is the expected number of months to close, starting from day one?
    • How much extra runway, in months, buys the last 10 points of probability from 80% to 90%?
  2. 015You can commit Rs 10 crore to a startup now, or put in Rs 2 crore now and the other Rs 8 crore only if it hits a milestone, which it does 30% of the time. Once past the milestone, the company has a 40% chance of paying you Rs 100 crore for the full Rs 10 crore, and zero otherwise; if the milestone is missed, it pays nothing. What is the expected profit of each route, and what is the option to stop worth?Probability and expected valueHardSeed and early-stage VCDeep tech VC

    Try it first

    What is the option to stop after Rs 2 crore worth?

    Show the worked solution

    Committing now earns an expected Rs 2 crore; staging earns Rs 7.6 crore, so the option to stop is worth Rs 5.6 crore. Committing pays Rs 100 crore only 12% of the time, worth Rs 12 crore, against Rs 10 crore invested. Staging risks Rs 2 crore first and adds Rs 8 crore only after the milestone, so in the 70% of cases that fail it saves Rs 8 crore: 0.7 x 8 is Rs 5.6 crore.

    Why is waiting worth anything if the payout is the same?

    Think of booking a wedding venue with a small refundable deposit rather than paying in full a year out. If the engagement is called off, you lose the deposit, not the whole fee. Staging a cheque does not change what you win; it changes what you lose in the worlds where things go wrong, because you stop paying once you learn the milestone was missed. Here the milestone fails 70% of the time, and each of those times the staged route has spent Rs 2 crore instead of Rs 10 crore.

    How do you work each route's expected profit?

    Committing now: the company pays Rs 100 crore only if it passes both hurdles, 0.3 x 0.4, which is 12% of the time. That is worth Rs 12 crore against Rs 10 crore paid, an expected profit of Rs 2 crore. Staging: pay Rs 2 crore for sure, then in the 30% of worlds that hit the milestone, pay Rs 8 crore for a 40% shot at Rs 100 crore, which is worth Rs 32 crore at that point. Minus 2, plus 0.3 x 32, gives Rs 7.6 crore.

    The relationship
    EV1=0.3×0.4×100−10=2EV2=−2+0.3 (0.4×100−8)=7.6EV_1 = 0.3 \times 0.4 \times 100 - 10 = 2 \qquad EV_2 = -2 + 0.3\,(0.4 \times 100 - 8) = 7.6
    0.3chance the milestone is hit
    0.4chance of the Rs 100 crore payout once past the milestone
    2, 8the first and second tranches, Rs crore
    What it says in wordsAverage every path's profit by its probability; the staged route only pays the Rs 8 crore on the paths where the milestone was hit.
    Staging lets you skip the Rs 8 crore in the 70% of worlds that failRoute 1: commit Rs 10 crore nowPay now-1030%70%Milestone hitno extra cashMilestone missedpayout 040%60%Successpayout +100Failurepayout 0Route 2: Rs 2 crore now, Rs 8 crore only after the milestonePay now-230%70%Milestone hitpay -8Milestone missedstop: lose only 240%60%Successpayout +100Failurepayout 00.3 x 0.4 x 100 - 10EV = +2.0-2 + 0.3 x (40 - 8)EV = +7.6Option to stop = 7.6 - 2.0 = Rs 5.6 crore, which is 70% x the Rs 8 crore you no longer risk. Rs crore throughout.
    Committing Rs 10 crore up front earns an expected Rs 2 crore, while paying Rs 2 crore first and Rs 8 crore only after the milestone earns Rs 7.6 crore, so the right to stop is worth Rs 5.6 crore, the Rs 8 crore saved in the 70% of cases that fail.

    What would a founder say about this, and what is the limit?

    A founder will rarely give you the same price for both tranches, because staging moves risk onto the company. The Rs 5.6 crore is the most it is worth paying for the right to stage, in a higher second-tranche price or a smaller stake. Staging also has costs the model ignores: a company that has to hit a milestone to get its money may be run for the milestone rather than for the business, and the uncertainty can scare off other investors.

    Where candidates lose it

    The common slip is to say both routes are the same because the total cheque and the payout are the same. The interviewer is testing whether you see that information arrives between the tranches, and that you can act on it.

    The second loss is getting Rs 7.6 crore and not explaining it. The cleanest check is 0.7 x Rs 8 crore: the option is worth exactly the money you no longer risk in the failure cases.

    What the interviewer asks next

    • The founder insists the second tranche is priced 50% higher. Is staging still better?
    • At what milestone probability does the option to stop become worthless?
    • Why do deep tech investors use milestone tranches more than consumer investors?
  3. 024A startup survives only if at least two of its three enterprise pilots convert to paid contracts. The pilots convert independently with probabilities of 60%, 50% and 30%. What is the chance the startup survives?Probability and expected valueCoreSeries A to C VCSaaS-focused VC

    Try it first

    What is the chance at least two pilots convert?

    Show the worked solution

    About 45%, below a coin flip. Exactly two pilots convert in three ways: pilots 1 and 2 only, 0.6 x 0.5 x 0.7 = 0.21; pilots 1 and 3 only, 0.6 x 0.5 x 0.3 = 0.09; pilots 2 and 3 only, 0.4 x 0.5 x 0.3 = 0.06. All three convert with probability 0.09. Adding the four gives 0.45.

    How do you make sure you count every way to survive?

    Think of a cricket team that needs two of its three openers to score fifty. You would list who scores and who does not, because each opener's failure matters to the case where the other two succeed. List the outcomes that satisfy the condition, write each as a product of the chance each pilot converts or does not, and add them. At least two out of three means exactly two, which can happen three ways, or all three.

    Add the four surviving leaves: 0.09 + 0.21 + 0.09 + 0.06 = 0.45Pilot 1 (60%)Pilot 2 (50%)Pilot 3 (30%)Y Y Y0.090survivesY Y N0.210survivesY N Y0.090survivesY N N0.210failsN Y Y0.060survivesN Y N0.140failsN N Y0.060failsN N N0.140failsY = convertsSurvives45%Best two only30%Pairs added63%Solid green: converts. Dashed red: does not. Leaves read pilot 1, 2, 3.Pairs added count the all-three case three times: 0.63 - 2 x 0.09 = 0.45.
    Of the eight possible outcomes, the four with at least two conversions carry probabilities of 0.09, 0.21, 0.09 and 0.06, which add to 0.45, while using only the best two pilots gives 30% and adding the pair products gives an overcounted 63%.

    Why do the quick shortcuts fail?

    Multiplying the two best pilots, 0.6 x 0.5, gives 30% and forgets that pilots 1 and 3, or 2 and 3, also keep the company alive. Adding the three pair products, 0.30 + 0.18 + 0.15, gives 63% and makes the opposite error. Each pair product already includes the case where the third pilot converts too, so the all-three outcome is counted three times and must be taken off twice: 0.63 minus 2 x 0.09 is 0.45. That correction gives the same answer as the list, which is a useful check in the room.

    The relationship
    P=∑i<jpipj−2 p1p2p3=0.63−2(0.09)=0.45P = \sum_{i<j} p_i p_j - 2\,p_1p_2p_3 = 0.63 - 2(0.09) = 0.45
    p_i p_jthe chance that a given pair of pilots both convert, whatever the third does
    p_1 p_2 p_3the chance all three convert, 0.09
    What it says in wordsAdd the pair chances, then remove the two extra counts of the all-three case.

    What would you tell the investment committee?

    That three pilots which each sound promising still leave the company more likely to fail than survive. The answer also depends heavily on independence: if the pilots share a buyer type or a product gap, they tend to succeed or fail together, and the real survival chance could be noticeably higher or lower. Ask what the pilots have in common before trusting the 45%.

    Where candidates lose it

    The fast wrong answers are 30%, the two best pilots, and 63%, the pair products added up. Both come from skipping the list of outcomes, and the interviewer can see exactly which counting error you made.

    The second loss is forgetting the all-three case, which gives 36%. At least two includes three; say so before you add.

    What the interviewer asks next

    • What is the chance exactly one pilot converts?
    • If the company could add a fourth pilot at 40%, what would its survival chance become under a two-of-four rule?
    • How would correlation between the pilots change your answer?
  4. 026A startup has a 25% chance of dying in any given year, independently of what happened the year before. What is the chance it is still alive after five years, and after ten?Probability and expected valueWarm upSeed and early-stage VCIndia VC

    Try it first

    Quick instinct: what is the chance the startup is still alive after five years?

    Show the worked solution

    About 23.7% after five years and about 5.6% after ten. Each year the company survives with probability 0.75, and the years are independent, so the chances multiply: 0.75 to the fifth is 0.237. Ten years is that number squared, 0.237 x 0.237, about 0.056. Half the companies are gone before the end of year three.

    Why do you multiply survival rather than add up the deaths?

    Think of a batch of 100 phones, each with a one in four chance of breaking in every year you own it. In year one about 25 break and 75 are left. In year two the 25% applies to those 75, not to the original 100, so about 19 break and 56 are left. A yearly death rate only acts on the companies still alive, so survival shrinks by the same factor each year instead of falling by the same amount. Adding 25% five times gives 125%, and a probability above 100% is the sign you are using the wrong operation.

    Share of startups still alive, 25% chance of dying each yearone in four100%Yr 075%Yr 156%Yr 242%Yr 332%Yr 423.7%Yr 518%Yr 613%Yr 710%Yr 88%Yr 95.6%Yr 10Each year keeps 0.75 of the survivorsYear 5: 0.75^5 = 0.237Year 10: 0.237 x 0.237 = 0.056Year ten is year five squared, because the same five-year factor applies twice
    With a 25% chance of dying every year, the share of startups alive falls to 23.7% at year five, below the one in four line, and to 5.6% at year ten, which is the five-year figure squared.
    The relationship
    S(n)=(1−d)nS(5)=0.755≈0.237S(10)=0.2372≈0.056S(n) = (1-d)^n \qquad S(5) = 0.75^5 \approx 0.237 \qquad S(10) = 0.237^2 \approx 0.056
    dthe chance of dying in any one year, 0.25
    nthe number of years
    S(n)the chance of still being alive after n years
    What it says in wordsSurvival after n years is the one-year survival chance multiplied by itself n times.

    How do you get 0.75 to the fifth in your head?

    Build it from squares. 0.75 squared is 0.5625, call it 0.56. Squared again, 0.56 x 0.56 is about 0.316, which is year four. One more 0.75 takes 0.316 to 0.237 for year five. Year ten is year five squared, so once you have 0.237 the second answer is one step: 0.237 x 0.237 is about 0.056. Saying the squaring route out loud shows the interviewer a method rather than a memorised number.

    What does a flat death rate mean for a seed portfolio?

    Two more numbers fall out of the same 25%. The average company lives four years, one over the yearly death rate, and half are gone within about 2.4 years, where 0.75 to the n crosses one half. A seed fund of 30 companies with this death rate expects only about 1.7 of them alive at year ten, which is why seed funds are sized for most companies failing. Say the limitation too: real death rates are not flat. They are highest in the first two years and fall for companies that find a market, so a flat 25% overstates late deaths and understates early ones.

    Where candidates lose it

    The fast wrong answer is zero, or some version of five times 25%, because the candidate adds the yearly chances. That treats a dead company as able to die again. The interviewer is checking whether you know that independent yearly chances multiply.

    The second loss is getting 23.7% and then working ten years from scratch, slowly and aloud. Square the five-year figure; it is quicker and it shows you see the structure.

    What the interviewer asks next

    • What yearly death rate leaves exactly half the companies alive after five years?
    • If the death rate is 40% in year one and 15% every year after, what is five-year survival?
    • A fund wants at least three companies alive at year ten. How many should it back at a 25% yearly death rate?
  5. 038A startup has 3 months of cash. Each month it either signs a contract worth one extra month of runway, with probability 0.4, or burns a month, with probability 0.6. What is the chance it reaches 6 months of cash, the level at which it can raise, before it reaches zero?Probability and expected valueHardSeed and early-stage VCIndia VC

    Try it first

    Starting halfway between zero and the raise, what is the chance of reaching 6 months first?

    Show the worked solution

    About 23%. This is a random walk between two walls, zero and six months, starting at three. With r the ratio of down to up probabilities, 0.6 over 0.4 or 1.5, the chance of hitting the top first from rung i is (1 minus r to the i) over (1 minus r to the 6). From three that is 2.375 over 10.39, or 22.9%, against 50% if the odds were even.

    Why is the answer not 50% when the company starts halfway?

    Picture someone walking along a narrow wall in the wind, three steps from either end, and the wind pushes them back towards the start a little more often than forward. Each single step is only slightly unfair, but the walk ends at whichever end comes first, and over many steps the small push decides it. A 60/40 tilt per month sounds mild, yet over the many months the walk can last it compounds into odds of roughly three to one against reaching the raise. Only a perfectly fair walk gives 50% from the midpoint.

    Months of cash: up one with 0.4, down one with 0.6, until 0 or 60%04.8%112.0%222.9%339.1%463.5%5100%6out of cashcan raisestarts hereup 0.4down 0.6dashes: chance of reaching 6 with a fair 50-50 walkFrom 3 months: 22.9%a fair walk would give 50%
    With each month 60/40 against it, the company's chance of reaching six months before zero is 22.9% from a start of three, far below the 50% a fair walk would give from the same midpoint.

    How do you solve it on a whiteboard?

    Let h(i) be the chance of reaching 6 from rung i. One month from now the company is at i plus 1 with chance 0.4 or i minus 1 with chance 0.6, so h(i) is 0.4 h(i + 1) plus 0.6 h(i - 1), with h(0) = 0 and h(6) = 1. The solution of that recurrence is the gambler’s ruinThe classic problem of a player betting one unit at a time until reaching a target or losing everything. It gives the chance of hitting either wall of a random walk first. formula, (1 - r to the i) over (1 - r to the N), with r = q over p = 1.5. Then it is arithmetic: 1.5 cubed is 3.375 and 1.5 to the sixth is 3.375 squared, about 11.39, so h(3) = 2.375 / 10.39, about 0.229.

    The relationship
    h(i)=1−ri1−rN,r=qp=0.60.4=1.5h(3)=1−3.3751−11.39=2.37510.39≈0.229h(i) = \frac{1 - r^{i}}{1 - r^{N}}, \quad r = \frac{q}{p} = \frac{0.6}{0.4} = 1.5 \qquad h(3) = \frac{1 - 3.375}{1 - 11.39} = \frac{2.375}{10.39} \approx 0.229
    h(i)the chance of reaching the raise from i months of cash
    p, qthe chances of a good month, 0.4, and a bad one, 0.6
    Nthe runway at which the company can raise, 6 months
    What it says in wordsThe chance of reaching the top wall first depends on the start, the distance to each wall and how tilted each step is.

    What would change the company's odds most?

    The formula shows two levers. Moving the bar closer helps a lot: if the company could raise at 4 months instead of 6, the chance from 3 rises to 58.5%. Cutting the tilt helps even more, because the damage comes from compounding a bad ratio over many steps, which is why investors push early teams to cut burn rather than hope for a run of contracts. The limit of the model is that months are not independent coin flips: contracts cluster, and a founder can change the step size by cutting costs. It is a way to see the shape of the risk, not a forecast.

    Where candidates lose it

    The common loss is answering 50% because the company starts in the middle, or 40% because that is the chance of a good month. Neither accounts for the walk ending at whichever wall comes first after many tilted steps.

    The second loss is computing only the straight path, 0.4 cubed or 6.4%, and missing all the paths that wander down and back up. The recurrence, or the gambler's ruin formula, counts every path at once.

    What the interviewer asks next

    • What is the chance of reaching 6 months if the odds of a good month rise to 0.5?
    • With the same 60/40 odds, from what starting runway does the company have a better than even chance of reaching 6 months?
    • If each contract added two months of runway instead of one, how would you set the problem up?
  6. 050Two portfolio companies each have a 20% chance of failing this year. What is the chance that at least one fails if the failures are independent, if they are perfectly correlated, and if the chance both fail is 10%?Probability and expected valueCoreMulti-stage VCFund of funds and LPs

    Try it first

    If the failures are independent, what is the chance at least one company fails?

    Show the worked solution

    36% if independent, 20% if perfectly correlated, and 30% if the chance both fail is 10%. In each case the chance of at least one failure is 20% plus 20% minus the chance both fail. Independent failures overlap 0.2 x 0.2 = 4%, giving 36%. Perfectly correlated failures overlap completely, 20%, giving 20%. With a 10% overlap the answer is 30%. Correlation makes any failure less likely but both failing far more likely.

    Why can you not just add the two chances?

    Think of two friends who each forget your birthday one year in five. If you add the chances you get two in five, but that counts the years when both forget twice, once for each friend. The chance of at least one event is the sum of the two chances minus the chance both happen, so the overlap decides the answer. For two independent events the overlap is the product, 0.2 x 0.2 = 4%, and the answer is 36%. The complement route checks it: neither fails with chance 0.8 x 0.8 = 64%.

    The relationship
    P(A∪B)=P(A)+P(B)−P(A∩B)0.2+0.2−0.04=0.360.4−0.10=0.300.4−0.20=0.20P(A \cup B) = P(A) + P(B) - P(A \cap B) \qquad 0.2 + 0.2 - 0.04 = 0.36 \qquad 0.4 - 0.10 = 0.30 \qquad 0.4 - 0.20 = 0.20
    P(A), P(B)each company's chance of failing, 20%
    P(A and B)the chance both fail, which depends on how the failures are linked
    P(A or B)the chance at least one fails
    What it says in wordsAdd the two chances and subtract the overlap once, because the overlap was counted in both.
    Two companies, each 20% likely to fail: overlap is what changesIndependentABboth fail: 4%at least one fails20 + 20 - 4 = 36%Partly correlatedABboth fail: 10%at least one fails20 + 20 - 10 = 30%Perfectly correlatedA = Bboth fail: 20%at least one fails20 + 20 - 20 = 20%Higher correlation: fewer chances of any failure, a much higher chance of both
    As the overlap between two 20% failure chances grows from 4% to 10% to 20%, the chance that at least one company fails falls from 36% to 30% to 20%, while the chance that both fail rises fivefold.

    What does correlation do to a venture portfolio?

    It trades many small surprises for fewer, larger ones. As failures become more correlated, the chance of at least one failure falls from 36% to 20%, but the chance both fail rises from 4% to 20%, five times higher. The 10% case corresponds to a correlation of about 0.37 between the two failure events. A fund full of companies selling to the same customers, or depending on the same funding market, looks diversified by count but behaves like fewer, bigger bets.

    Why would an LP care about this more than a single fund manager?

    An LP holding several funds cares about how many bad outcomes can arrive at once. Correlated failures are what turn an ordinary bad year into a vintage where most companies struggle together, and an average-case model that treats failures as independent understates that risk. The limitation is that correlation is hard to measure for private companies with few data points; in practice investors judge it from shared exposures, such as the same sector, the same customers or the same reliance on new funding, rather than from a computed number.

    Where candidates lose it

    The common loss is answering 40% for the independent case by adding the two chances. That double-counts the 4% where both fail, and the interviewer is listening for the subtraction or the complement.

    The second loss is thinking correlation makes failure more likely across the board. It makes at least one failure less likely and both failing more likely, and the interesting answer says both halves.

    What the interviewer asks next

    • With ten independent companies at 20% each, what is the chance at least one fails?
    • What is the largest possible chance that at least one of the two fails, and what overlap gives it?
    • How would you estimate correlation between two private portfolio companies with no price data?
  7. 056Your pro rata right lets you invest Rs 5 crore in the next round, and you decide after you see how it is priced. 40% of rounds turn out good and return 5x the round price; 60% are bad and return 0.5x. Assume the pricing tells you which kind it is. What is the right worth, compared with an obligation to invest Rs 5 crore in every round?Probability and expected valueCoreSeed and early-stage VCMulti-stage VC

    Try it first

    How much more is the right worth than the obligation, in expected profit?

    Show the worked solution

    Rs 8 crore of expected profit against Rs 6.5 crore, so the choice is worth Rs 1.5 crore. A good round turns Rs 5 crore into Rs 25 crore, a Rs 20 crore profit; a bad one leaves Rs 2.5 crore, a Rs 2.5 crore loss. Forced to invest, you expect 0.4 x 20 minus 0.6 x 2.5, or Rs 6.5 crore. With the right you pass on bad rounds, so you expect 0.4 x 20, or Rs 8 crore. The Rs 1.5 crore gap is the price of the bad branch you no longer have to take.

    Why is a right worth more than an obligation to do the same thing?

    A season ticket that lets you skip any match you like is worth more than one that forces you to sit through the washed-out ones, even though the matches are the same. A right is worth exactly the losses it lets you refuse, so its value is the expected loss on the branches you would walk away from. Here the bad branch costs Rs 2.5 crore 60% of the time, an expected Rs 1.5 crore, and the right lets you skip all of it. The good branch is the same Rs 20 crore profit either way, so it adds nothing to the difference.

    The right lets you skip the bad branch; the obligation does notPro rata right: decide after seeing the priceGood 40%Bad 60%Invest: +Rs 20 crPass: Rs 0Invest: -Rs 2.5 crPass: Rs 0Expected profit = 0.4 x 20 = Rs 8 croreObligation: invest in every roundGood 40%Bad 60%+Rs 20 cr (5x)-Rs 2.5 cr (0.5x)no way out0.4 x 20 - 0.6 x 2.5 = Rs 6.5 croreGap: Rs 1.5 crorethe value of saying no
    In the left tree you pass after a bad round and keep zero, so the expected profit is Rs 8 crore; in the right tree you must also take the Rs 2.5 crore loss 60% of the time, so the expected profit falls to Rs 6.5 crore, and the Rs 1.5 crore gap is the value of the choice.
    The relationship
    Vright−Voblig=p W−[p W+(1−p) L]=−(1−p) L=0.6×2.5=1.5V_{\text{right}} - V_{\text{oblig}} = p\,W - \big[p\,W + (1-p)\,L\big] = -(1-p)\,L = 0.6 \times 2.5 = 1.5
    pchance the round is good, 40%
    Wprofit in a good round, Rs 20 crore
    Lprofit in a bad round, minus Rs 2.5 crore
    What it says in wordsThe right and the obligation share the good branch, so the right's extra value is just the expected loss on the bad branch it lets you skip.

    Where does this simple answer overstate the right in real life?

    The question lets the round's pricing tell you for certain whether it is good. In practice the signal is noisy, so a pro rata right is worth less than Rs 1.5 crore here: you will sometimes pass on a good round and sometimes take a bad one. If your read were no better than a coin, the choice would be worth nothing, because you could not tell the branches apart. Two more limits: a hot round may leave you no allocation even with the right, and a fund must hold reserves to exercise it, and that cash earns nothing while it waits.

    This is why seed funds fight for pro rata rights and why the best later investors try to cut them back. The value sits with whoever gets to look before deciding, and it grows with how far apart the good and bad outcomes are. Say that last point in the room: a wider spread between good and bad rounds makes the right worth more, exactly as a wider spread makes any option worth more.

    Where candidates lose it

    The usual slip is to value the right at the Rs 8 crore it earns and stop there. The question asks for the right compared with the obligation, and the answer is the Rs 1.5 crore difference, the expected loss you get to avoid.

    The second trap is ignoring the assumption that pricing reveals the quality of the round. Say it out loud and add that a noisier signal shrinks the value toward zero; that sentence is what separates a mechanical answer from an investor's one.

    What the interviewer asks next

    • If your read on round quality were right only 75% of the time, what would the right be worth?
    • Why might a lead investor in the next round want to cap your pro rata?
    • How does holding reserves to exercise pro rata rights affect the fund's overall return?
  8. 06810% of the startups you meet are genuinely good. Your screen flags 80% of the good ones as worth pursuing, but also flags 20% of the bad ones. A company passes your screen. What is the chance it is actually good?Probability and expected valueCoreSeed and early-stage VCMulti-stage VC

    Try it first

    A company passes. The chance it is good is about:

    Show the worked solution

    About 31%, not 80%. Picture 1,000 companies. 100 are good and your screen passes 80 of them. 900 are bad and it passes 20% of those, 180 companies. So 260 pass, and only 80 of them are good: 80 divided by 260 is 30.8%. The screen raises the odds from 10% to about 31%, which is useful, but most companies that pass are still bad because bad ones are so common.

    Why is the answer so far below 80%?

    A smoke alarm that rings for every real fire and for one in five burnt toasts will mostly ring for toast, because toast is far more common than fire. When the thing you are looking for is rare, even a small false alarm rate applied to the large crowd of negatives produces more false passes than true ones. Here 20% of 900 bad companies is 180, against 80% of 100 good companies, 80. The screen's 80% describes how it treats good companies; the question asks what a pass tells you, which also depends on how many bad companies get in.

    1,000 companies: of the 260 that pass the screen, only 80 are good260 pass the screen80 good, passed180 bad, passed20 good, missed720 bad, rejectedP(good | passed) = 80 / 260 = 30.8%not 80%Base rate 10%:bad companiesoutnumber good 9 to 1
    Of 1,000 companies, the screen passes 80 good ones and 180 bad ones, so the 260 passes are mostly false alarms and the chance a passing company is good is 80 out of 260, about 31%, not the 80% the screen's hit rate suggests.
    The relationship
    P(G∣pass)=0.8×0.10.8×0.1+0.2×0.9=0.080.26=0.308P(G \mid \text{pass}) = \frac{0.8 \times 0.1}{0.8 \times 0.1 + 0.2 \times 0.9} = \frac{0.08}{0.26} = 0.308
    Gthe company is genuinely good
    0.8share of good companies the screen passes
    0.2share of bad companies the screen also passes
    0.1share of all companies that are good, the base rate
    What it says in wordsOf everything that passes, the good share is the true passes divided by all passes, true and false.

    What would actually improve the screen?

    Run the alternatives. Halving the false pass rate to 10% lifts the answer to 80 out of 170, about 47%, while raising the hit rate from 80% to 95% only takes it to 95 out of 275, about 35%. When good companies are rare, cutting false passes matters more than catching every good one. A second, independent check helps the same way: if a passing company goes through another screen with the same rates, the base rate is now 31%, and a second pass takes it to about 64%.

    The limit is that the two checks must be independent; a second partner who looks at the same deck with the same biases is not a second screen. And the 10% base rate is an assumption about your deal flow; a fund with better sourcing starts from a higher base and every pass means more.

    Where candidates lose it

    The trap is answering 80%, confusing the chance a good company passes with the chance a passing company is good. It is the most common probability error there is, and the interviewer expects you to spot it.

    The second loss is reaching 31% through a formula and fumbling the explanation. Use natural frequencies, 1,000 companies, 80 true passes, 180 false ones; it is faster and harder to get wrong.

    What the interviewer asks next

    • If a company passes two independent screens with the same rates, what is the chance it is good?
    • Which matters more here, raising the hit rate or cutting false passes?
    • How would better sourcing change these numbers?
  9. 082You will meet 20 founders, one at a time, in random order, and can back only one. You must decide yes or no at the end of each meeting and cannot go back to anyone you passed. What rule maximises the chance that you back the single best founder, and what is that chance?Probability and expected valueHardSeed and early-stage VCMulti-stage VC

    Try it first

    Roughly how often does the best possible rule land the single best of the 20?

    Show the worked solution

    Let the first 7 founders go, then back the first one who is better than all of them; you back the single best about 38% of the time. The first 7 set the bar. You win when the best founder comes after them and nobody earlier than the best beats the bar first. Across all cut-offs, 7 gives the highest chance, 38.4%, against 5% for picking at random.

    Why let good founders pass at all?

    Renting a flat in a tight market works the same way: the first few viewings teach you what a good flat looks like, and you commit to the next one that beats them. The early meetings are the price of learning where the bar sits, and the rule trades a small chance that the best founder is among them for a much better read on everyone after. Commit too early and you have no bar; wait too long and the best has probably already gone.

    Let the first 7 go, then back the first founder better than all of them10%20%30%40%1/e = 36.8%, the large-n limit057101519Founders you let pass before you are willing to back oneSkip 7: 38.4%Back the first: 5%Skip 15: 22.2%
    The chance of backing the best of 20 founders rises from 5% if you back the first one to a peak of 38.4% if you let 7 pass, then falls slowly, to 22.2% if you let 15 pass, staying near the 1/e limit of 36.8% across a wide middle range.

    How do you work out the chance for a given cut-off?

    Say you let r founders pass. Suppose the best founder is at position i, after the first r. You back that founder only if nobody between r and i beats the bar first, which happens exactly when the best of the first i minus 1 founders sits inside the first r, a chance of r over i minus 1. Each position is equally likely to hold the best, one in 20, so add up across positions.

    The relationship
    P(r)=rn∑i=r+1n1i−1P(7)=720(17+18+⋯+119)≈0.384P(r) = \frac{r}{n}\sum_{i=r+1}^{n}\frac{1}{i-1} \qquad P(7) = \frac{7}{20}\left(\tfrac{1}{7} + \tfrac{1}{8} + \dots + \tfrac{1}{19}\right) \approx 0.384
    nfounders in total, 20
    rfounders you let pass to set the bar
    1/(i-1)weight for the best founder sitting at position i
    What it says in wordsAverage, over every position the best founder could hold, the chance that nobody earlier stole the pick.

    The curve is flat near the top: letting 6 pass gives 37.9% and 8 gives 38.2%. The general rule is to let about n/e go, 37% of the field, and win about 37% of the time; for 20 founders that is 7.4, which rounds to 7.

    Where does the model stop describing real deal flow?

    It assumes you only value the single best, can only rank founders against each other, see them in random order and never get a second chance. A real investor is happy with a top three founder, has an absolute sense of quality from past deals, and can sometimes return to a founder a week later. Each of those argues for committing earlier. Say that limit after the number; it shows you know what the model is for.

    Where candidates lose it

    Most candidates either say 1 in 20, treating the choice as blind, or propose meeting everyone and then choosing, which the rules forbid. The interviewer is testing whether you see that watching without committing is itself a strategy.

    The second loss is quoting 37% from memory without showing where it comes from. Give the r over i minus 1 argument in one sentence and the 7 for 20 falls out.

    What the interviewer asks next

    • With 100 founders, how many do you let pass and what is your chance?
    • You are happy with either of the two best founders. Should you stop earlier or later?
    • A founder you passed is still available at the end half the time. How does the rule change?
  10. 094A founder holding a term sheet can keep shopping it to other investors, but cannot go back to an offer she turned down. Model each offer as a fair die roll that pays the face value in Rs crore. She may stop after any roll. With at most two rolls, and then with at most three, what stopping rule maximises her expected value, and what is it worth?Probability and expected valueCoreSeed and early-stage VCSeries A to C VC

    Try it first

    With three rolls allowed, which first rolls should she keep?

    Show the worked solution

    With two rolls, keep a 4, 5 or 6 and the game is worth 4.25; with three, keep only a 5 or 6 on the first roll and it is worth about 4.67. Work backwards. The last roll is worth 3.5 on average, so on the roll before it keep anything above 3.5. That makes two rolls worth 4.25, which becomes the bar for the first of three rolls: only a 5 or 6 beats it.

    Why work backwards from the last roll?

    Deciding whether to take a flat today depends on what the next viewing is likely to offer, and that depends on whether there is another after it. The value of rolling again is the value of the game that remains, so solve the last stage first, where there is no choice, and carry its value back as the bar for the stage before. With one roll left she takes whatever comes, worth 3.5 on average. That 3.5 is what she gives up by keeping an earlier roll.

    Keep an offer only if it beats what rolling again is worthOne roll leftno choice: keep it123456Value3.50Two rolls leftroll on is worth 3.5123456bar 3.5Value4.25Three rolls leftroll on is worth 4.25123456bar 4.25Value4.67keep and stoproll again
    With one roll left every face is kept and the game is worth 3.5; with two rolls left she keeps only faces above 3.5, which lifts the value to 4.25; with three rolls left the bar rises to 4.25, so she keeps only a 5 or 6 and the game is worth 4.67.

    How do you compute each value?

    With two rolls: keep 4, 5 or 6 on the first, each with chance one in six, and roll again on 1, 2 or 3, which is worth 3.5. Each stage's value is the average, over the six faces, of the better of keeping that face and rolling on. That gives (4 + 5 + 6) / 6 + (3/6) x 3.5 = 2.5 + 1.75 = 4.25. With three rolls the bar is 4.25: keep 5 or 6, worth (5 + 6) / 6, and roll on otherwise, worth (4/6) x 4.25. Together that is 4.67.

    The relationship
    Vk=16∑f=16max⁡(f, Vk−1)V1=3.5,  V2=4.25,  V3≈4.67V_{k} = \frac{1}{6}\sum_{f=1}^{6}\max(f,\,V_{k-1}) \qquad V_1 = 3.5,\; V_2 = 4.25,\; V_3 \approx 4.67
    V_kvalue of the game with k rolls left, Rs crore
    fthe face showing
    max(f, V_{k-1})keep the face or roll on, whichever is worth more
    What it says in wordsAt each stage, keep the face if it beats the value of continuing, and average over the faces.

    What does this say about shopping a real term sheet?

    Two things. The bar for accepting rises with the number of chances left, so an early offer that is merely average should be declined if the process has room. But each extra roll adds less: the second roll adds 0.75, the third only about 0.42, and in real fundraising each roll costs weeks and the dice are not fair, since an offer turned down rarely comes back and a long process can scare off the next investor. The model is a way to think about the bar, not a reason to keep shopping.

    Where candidates lose it

    The most common slip is using 3.5 as the bar at every stage, so candidates keep a 4 on the first of three rolls. With two rolls still to come, continuing is worth 4.25, and a 4 falls short of that.

    The second is computing the value of keeping 4, 5 or 6 as their average, 5, and forgetting the half of the time she rolls again. Weight every branch by its chance: the answer for two rolls is 4.25, not 5.

    What the interviewer asks next

    • What is the game worth with four rolls, and what does she keep on the first?
    • Each extra roll now costs Rs 0.3 crore. How many rolls should she plan for?
    • How would the rule change if she could return to any offer she turned down?
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