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  1. 001You roll a fair die until each even number, 2, 4 and 6, has appeared at least once. You are told the game ended on a 2. What is the probability that the first roll was a 1, and why is it not 1/5?Probability and brainteasersHardSquarepoint CapitalLondon · 2026

    Try it first

    Commit before you work it: given the game ended on a 2, what is the chance the first roll was a 1?

    Show the worked solution

    The answer is 1/6. After a first roll of 1, 3 or 5 the game ends on 2 with chance 1/3; after a 2 it cannot; after a 4 or 6 it ends on 2 with chance 1/2. Overall the game ends on 2 with chance 1/3, so a first roll of 1 carries (1/6 x 1/3) / (1/3) = 1/6. The 4 and 6 absorb the share the 2 lost.

    Why does 1/5 feel right, and where does it go wrong?

    Think of a cricket team told that the match was won off the last ball. That news does not just rule out one scenario; it makes the close games far more likely than the easy ones. Conditioning works the same way here. Being told the game ended on 2 is evidence, and evidence reweights every starting point by how well it explains what you saw. The 1/5 answer treats the news as if it only deleted the 2 and left the other five faces equally likely.

    So ask, for each first roll, how likely it is that 2 ends up the last even number. After a 1, 3 or 5, all three evens are still missing and each is equally likely to come last, so 1/3. After a 2, the 2 is already seen and cannot come last, so 0. After a 4, only 2 and 6 are missing, and 2 comes last exactly when 6 shows first, so 1/2. The same holds after a 6.

    Learning the game ended on 2 reweights the first roll, it does not just delete the 2Chance the gamethen ends on a 21/3First roll 102 already seenFirst roll 21/3First roll 31/2First roll 41/3First roll 51/2First roll 6Weight on each firstroll, given it ended on 21/601/61/41/61/4The naive answer spreads 1/5 evenly over five faces. The 4 and the 6 each earn 1/4, so 1, 3 and 5 get only 1/6.
    The chance the game ends on 2 is 1/3 after a first roll of 1, 3 or 5, zero after a 2, and 1/2 after a 4 or 6, so once you learn it ended on 2 the first roll carries weight 1/6 for each odd face and 1/4 each for the 4 and the 6.

    How do you turn those chances into the answer?

    Apply Bayes ruleThe chance of a cause given what you saw equals its prior chance times how likely it made the observation, divided by the total chance of the observation.. Each face starts at 1/6. Multiply by its chance of ending on 2 and add them up: three faces at 1/18, one at 0, two at 1/12, which totals 1/3. The 1 contributes 1/18 of that 1/3, which is 1/6, exactly its starting weight. The 4 and the 6 rise from 1/6 to 1/4 each, and the 2 falls to zero.

    The relationship
    P(1∣end on 2)=16⋅1336⋅13+16⋅0+26⋅12=1/181/3=16P(1 \mid \text{end on }2) = \frac{\tfrac{1}{6}\cdot\tfrac{1}{3}}{\tfrac{3}{6}\cdot\tfrac{1}{3} + \tfrac{1}{6}\cdot 0 + \tfrac{2}{6}\cdot\tfrac{1}{2}} = \frac{1/18}{1/3} = \frac{1}{6}
    1/6the chance of each first roll before you know anything
    1/3the chance of ending on 2 after an odd first roll
    1/2the chance of ending on 2 after a first roll of 4 or 6
    What it says in wordsWeight each first roll by how likely it makes ending on 2, then divide by the total chance of ending on 2.

    The research version of this is reading a data point. A company that beats estimates is more likely to be one that guided low, not just one that is doing well. Before you update, ask which starting stories make the thing you observed more likely, and shift weight towards them.

    Where candidates lose it

    The trap is saying 1/5 fast, because it sounds like careful conditioning: remove the impossible case and spread the rest evenly. The interviewer asked why it is not 1/5 precisely because that answer throws away how strongly each start predicts the ending.

    The second loss is getting 1/6 by luck and being unable to explain it. Say the three conditional chances, 1/3, 0 and 1/2, out loud; that list is the whole argument.

    What the interviewer asks next

    • What is the probability the first roll was a 4, given the game ended on 2?
    • What is the expected number of rolls until all three evens have appeared?
    • If the game instead ends when any two evens have appeared, how does the answer change?

    Asked at Squarepoint Capital, Quant Research Intern Interview, London, 2026 (Wall Street Oasis): why is the probability of seeing a 1 on our first roll, given that we end on a 2, not 1/5

  2. 026You start with Rs 2. Each round you win Rs 1 with probability 0.6 or lose Rs 1 with probability 0.4. You stop when you reach Rs 5 or go broke at Rs 0. What is the chance you reach Rs 5?Probability and brainteasersHardTwo SigmaNew York · 2023

    Try it first

    Before any algebra: roughly where does the answer sit?

    Show the worked solution

    135 over 211, about 64%. Let P(i) be the chance of reaching Rs 5 from Rs i. Each P(i) is 0.6 times P(i+1) plus 0.4 times P(i-1), with P(0) = 0 and P(5) = 1. The solution is P(i) = (1 - (2/3)^i) / (1 - (2/3)^5), and at i = 2 that is (5/9) divided by (211/243), or 135/211. A fair coin would give 40%.

    How do you set the problem up without drawing every path?

    Think of a cricket team chasing a small target with few wickets in hand. What matters is not the path of each ball but where the game stands now: runs needed and wickets left. The only thing that matters at any moment is how much money you hold, so write one unknown for each amount and one equation linking it to its neighbours. From Rs i you move to Rs i+1 with chance 0.6 and to Rs i-1 with chance 0.4, so P(i) = 0.6 P(i+1) + 0.4 P(i-1). The two ends are fixed: P(0) = 0 because you are broke, and P(5) = 1 because you have won.

    The relationship
    P(i)=1−(q/p)i1−(q/p)NP(2)=1−(2/3)21−(2/3)5=135211P(i) = \frac{1-(q/p)^{i}}{1-(q/p)^{N}} \qquad P(2) = \frac{1-(2/3)^2}{1-(2/3)^5} = \frac{135}{211}
    p, qthe chances of winning and losing one round, 0.6 and 0.4
    Nthe target, Rs 5
    iyour starting capital, Rs 2
    What it says in wordsThe ratio of losing to winning odds, raised to your capital and to the target, gives the chance of hitting the target first.
    Chance of reaching Rs 5 before Rs 0: a 60/40 edge against a fair coin0%50%100%Rs 0Rs 1Rs 2Rs 3Rs 4Rs 5Starting capital38.4%20%64.0%40%81.0%60%92.4%80%Start at Rs 2Fair coin: 2 / 5 = 40%60/40 edge: 135 / 21164.0%Win 60%, lose 40%Fair coin, 50/50Rs 0 and Rs 5 are the twostopping points, 0% and 100%
    With a fair coin the chance of reaching Rs 5 rises in a straight line with starting capital, 40% from Rs 2. A 60/40 edge lifts every point well above that line, to 64.0% from Rs 2 and 92.4% from Rs 4.

    How do you check 64% without trusting the formula?

    Solve the chain by hand, starting from the bottom. Write P(1) = a. Then P(2) = (a - 0.4 x 0) / 0.6 = a/0.6 = 1.667a, and each later step follows the same rule until P(5) = 211/81 x a. Setting P(5) = 1 gives a = 81/211, or 38.4%, and P(2) = 135/211. Two routes landing on the same fraction is the check an interviewer wants to hear. It also tells you the shape: the edge matters most in the middle, where the fair line and the edge curve are furthest apart.

    Say the desk version in one line. A small, repeatable edge does not make ruin impossible when you start thin: from Rs 2 you still go broke about 36 times in 100. More capital, not a bigger edge, is what pushes the ruin chance towards zero.

    Where candidates lose it

    The common loss is answering 40%, the fair coin result, because the candidate remembers that the chance of reaching the target is capital over target. That rule only holds when winning and losing are equally likely.

    The second loss is trying to list paths. The game can run for any number of rounds, so path counting never ends. Name the state, write one equation per state, and use the two boundaries.

    What the interviewer asks next

    • What is the chance of reaching Rs 5 if the win probability drops to 0.5?
    • If the target were Rs 1,000 instead of Rs 5, roughly what is your chance of ever going broke from Rs 2?
    • How many rounds do you expect the game to last from Rs 2?

    Asked at Two Sigma, Research, New York, 2023 (Wall Street Oasis): Biased gamblers ruin problems; Markov Chain problems; sampling uniformly from triangle

  3. 051You have n cars, each with a full tank that lasts exactly 1,000 miles, and fuel can be passed from one car to another in the middle of the journey. What is the farthest one car can get? Give the answer for three cars, and say what happens as n grows without limit.Probability and brainteasersHardMillennium ManagementLondon · 2024

    Try it first

    With three cars, how far can the last car get?

    Show the worked solution

    With three cars the last car reaches about 1,833 miles; with n cars it reaches 1,000 x (1 + 1/2 + ... + 1/n). The cars drive together until they have burned one tankful between them, then one car refills the rest and stops. That happens after 1,000/n miles, then 1,000/(n - 1), down to the last car's 1,000. The sum has no ceiling, but it grows only like the logarithm of n.

    Why does a car drop out after exactly 1,000/n miles?

    Picture a group of hikers sharing water on a long desert walk. Once the group has drunk one person's worth between them, that person can pour what is left into the others' bottles, fill every one back up, and wait by the path. Carrying that hiker any further only costs water. The cars work the same way. After d miles each car has burned d miles of fuel and has room for d more. One car can refill the other n - 1 cars exactly when (n - 1) x d equals what it has left, 1,000 - d, which solves to d = 1,000/n.

    Each car that drops out hands over one tank; each stretch is shorter than the last3 cars333 miles2 cars500 miles1 car, its own full tank1,000 miles03338331,833 miles0 to 333: three cars burn 3 x 333 = 1,000 miles of fuel. Car 3 tops up cars 1 and 2, then stops.333 to 833: two cars burn 2 x 500 = 1,000 miles of fuel. Car 2 tops up car 1, then stops.833 to 1,833: car 1 drives the whole of its own tank, 1,000 miles.Total distance by number of cars: no ceiling, but it grows like 1,000 x ln n1 car1,0002 cars1,5003 cars1,83310 cars2,929100 cars5,187
    Three cars drive 333 miles together, two cars drive the next 500 and the last car drives its own 1,000, for 1,833 miles in all; each stretch burns one full tank, and ten cars reach only 2,929 miles.

    Why do three cars buy 1,833 miles and not 3,000?

    Follow the road left to right. Three cars run 333 miles together and burn 1,000 miles of fuel between them, one full tank. Car 3 has 667 miles of fuel left and hands 333 to each of the other two, which fills them. Two cars then run 500 miles, burning another tankful, and car 2 refills car 1. Car 1 runs its own 1,000. Every stretch burns exactly one tank, but each extra car buys a shorter stretch, because its tank has to move more cars.

    The relationship
    D(n)=1000(1+12+13+⋯+1n)≈1000 (ln⁡n+0.577)D(n) = 1000\left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots + \tfrac{1}{n}\right) \approx 1000\,(\ln n + 0.577)
    D(n)the farthest one car gets with n cars, in miles
    1/kthe stretch driven while k cars are still on the road, in tankfuls
    0.577the Euler-Mascheroni constant, the gap between the harmonic sum and ln n
    What it says in wordsAdd one over k for every car count from n down to one; the total grows like the natural logarithm of n.

    What happens as n goes to infinity?

    The harmonic series never converges, so enough cars can reach any distance at all. But the growth is logarithmic: ten cars reach about 2,929 miles and a hundred cars about 5,187, so every tenfold increase in cars adds only about 2,300 miles. When asked for the asymptotic answer, say both halves: unbounded, and slow. State the assumptions too: fuel moves between cars without loss, and a car that drops out is simply left behind. If the helper cars had to drive home, the answer would shrink sharply.

    Where candidates lose it

    The two fast wrong answers are 1,000 miles, because fuel cannot be created, and 3,000 miles, because three tanks were bought. Both miss that every car on the road burns fuel at the same rate, so the fleet spends most of its fuel carrying itself forward.

    The second loss comes at the asymptotic part. Candidates who say the distance levels off, or that it grows in proportion to n, lose the point. Say unbounded, growing like 1,000 times ln n, and give the ten and hundred car figures to show you can use the result.

    What the interviewer asks next

    • How many cars do you need to cover 3,000 miles? (11)
    • What changes if every helper car must keep enough fuel to drive back to the start?
    • Where else does the harmonic series turn up, in probability or in markets?

    Asked at Millennium Management, Investments, London, 2024 (Wall Street Oasis): What is the maximum distance you can get with the cars if you can transfer petrol in the middle of the journey?