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  1. 051You and a friend agree to meet at a spot some time between 5 and 6 pm. Each of you arrives at an independent, uniformly random time in that hour and waits 20 minutes for the other before leaving (or until 6 pm, whichever comes first). What is the probability you meet?Continuous and geometric probabilityCoreJane StreetNew York · 2026

    Try it first

    Before you draw anything: what is the chance you meet?

    Show the worked solution

    5/9, about 55.6%. Put your arrival time on one axis and your friend's on the other, so every outcome is a point in a unit square. You meet when the two times differ by at most a third of an hour, a band along the diagonal. The two corner triangles outside the band each have legs of 2/3, area 2/9, so the band is 1 - 4/9 = 5/9.

    Why turn two arrival times into a square?

    Think of two people trying to catch each other at a tea stall with no phones. Nothing about the answer depends on who is you and who is the friend; it depends only on the pair of times. With two independent uniform times, every pair is equally likely, so the pair is a point spread evenly over a square and any probability is simply an area. The event you meet becomes a region: the set of points where the two times are within 20 minutes of each other. That region is a diagonal band, because the line x = y is where you arrive together.

    Every pair of arrival times is a point; you meet inside the bandMiss2/9Miss2/9Meet: 5/900202040406060Your arrival, minutes past the hourFriend's arrivalCount the white, not the greenWhole square1One miss triangle: legs of 40 minutes(2/3) x (2/3) / 2 = 2/9Both miss triangles4/9Meeting band1 - 4/9 = 5/9P(meet) = 5/9 = 55.6%
    Plotting your arrival against your friend's, the meeting region is the diagonal band where the times differ by 20 minutes or less; the two white corner triangles, each 2/9 of the square, are the misses, so you meet with probability 5/9.

    How do you get the area without any integration?

    Count the region you do not want. The two corners where one person arrives more than 20 minutes after the other are right triangles with both legs 40 minutes long, which is 2/3 of the side. Each has area (2/3) x (2/3) / 2 = 2/9, together 4/9, so the band is 5/9. Complements are the fastest route here, as they are for most geometric probability questions, because the leftover pieces are usually triangles.

    The relationship
    P(∣X−Y∣≤w)=1−(1−w)2=2w−w2w=13: 1−(23)2=59P(|X-Y|\le w) = 1-(1-w)^2 = 2w - w^2 \qquad w=\tfrac13:\ 1-\left(\tfrac23\right)^2 = \tfrac59
    X, Ythe two arrival times as fractions of the hour, independent and uniform on 0 to 1
    wthe waiting time as a fraction of the hour, here 20 of 60 minutes
    What it says in wordsThe chance of meeting is one minus the two corner triangles, whose legs are each one minus the waiting time.

    What does the general formula tell you that the number does not?

    Read 2w - w squared term by term. The 2w is the naive answer of either person waiting, and the minus w squared removes the double count and the clipping at the edges of the hour. It also tells you how waiting time buys certainty: to meet half the time each person must wait about 17.6 minutes, and to be sure each must wait the whole hour. The same picture prices any tolerance between two random arrivals, such as two orders landing in the same matching window of an auction.

    Where candidates lose it

    The common answer is 1/3, from reading 20 minutes as a third of the hour. It ignores that either person can be the one who waits, and it has no way to handle the edges of the hour, where a person arriving at 5:55 can only wait five minutes.

    The second trap is trying to integrate over one person's arrival time case by case near the edges. It works but wastes three minutes. Draw the square first and subtract the two triangles out loud.

    What the interviewer asks next

    • How long would each person need to wait for a 50% chance of meeting?
    • You wait 10 minutes and your friend waits 30. What is the chance now?
    • Three people arrive at random in the hour and each waits 20 minutes. What is the chance all three are together at some moment?

    Asked at Jane Street, Technology, New York, 2026 (Wall Street Oasis): 1v1 math problems. bus stop. two people meeting probelm

  2. 074A point is dropped uniformly at random in a unit square. What is the expected distance from the point to the nearest edge of the square?Continuous and geometric probabilityCoreHudson River TradingNew York · 2024

    Try it first

    What is the expected distance to the nearest edge?

    Show the worked solution

    1/6. Let D be the distance to the nearest edge. D exceeds d exactly when the point lies in the inner square of side 1 - 2d, so P(D > d) = (1 - 2d)^2 for d up to 1/2. The expected value of a non-negative variable is the integral of its tail, and the integral of (1 - 2d)^2 from 0 to 1/2 is 1/6.

    Why work with the chance of being far rather than the distance itself?

    Think of a sandpit where a child stands at a random spot and the question is how far they are from the nearest edge. Writing the distance as min(x, 1 - x, y, 1 - y) and integrating a minimum of four things means splitting the square into four triangles. Asking instead when the point is farther than d from every edge has a one-picture answer: the point must lie in a smaller square, shrunk by d on every side. That square has side 1 - 2d, so its area, (1 - 2d)^2, is the tail probability. One formula replaces four cases.

    Farther than d from every edge means inside a square of side 1 - 2d0.640.360.160.04nearest edgenumbers: P(distance > d) for d = 0.1 to 0.400.10.20.30.40.50.51P(D > d) = (1 - 2d)^2area = 1/6d, distance to the nearest edgesimulated: 0.1669exact: 1/6 = 0.1667
    A point is more than d from every edge only inside the inner square of side 1 - 2d, so the chance of being farther than 0.1, 0.2, 0.3 and 0.4 is 0.64, 0.36, 0.16 and 0.04, and the area under that tail curve is the expected distance, 1/6.

    How does the tail give the expectation?

    For any non-negative random variable, the expected value equals the integral of the chance that it exceeds each level, E[D] = integral of P(D > d). Here that is the integral of (1 - 2d)^2 from 0 to 1/2. Substitute u = 1 - 2d and it becomes half the integral of u^2 from 0 to 1, which is 1/6. A simulation with 200,000 random points gives 0.1669, against the exact 0.1667.

    The relationship
    E[D]=∫01/2P(D>d) dd=∫01/2(1−2d)2 dd=[−(1−2d)36]01/2=16E[D] = \int_0^{1/2} P(D>d)\,dd = \int_0^{1/2} (1-2d)^2\,dd = \Big[-\tfrac{(1-2d)^3}{6}\Big]_0^{1/2} = \tfrac16
    Dthe distance from the random point to the nearest edge
    P(D > d)the area of the inner square of side 1 - 2d
    What it says in wordsAdd up the chance of being farther than each distance, and the total is the expected distance.

    How do you sanity-check 1/6 against simpler cases?

    Build up the number of edges. The distance to one fixed edge averages 1/2, to the nearer of two opposite edges averages 1/4, and to the nearest of all four it falls to 1/6, so each added constraint pulls the minimum closer. The density of D is the slope of the tail, 4(1 - 2d), largest at the edge, which says most random points are near the boundary. That is the same reason most of the volume of a high-dimensional cube sits near its surface, a fact that matters when sampling scenarios in many risk factors at once.

    Where candidates lose it

    The trap answers are 1/2 and 1/4, from handling one edge or one axis and forgetting that the nearest of four edges is a minimum. A candidate who integrates min(x, 1 - x, y, 1 - y) directly often splits the square wrongly and lands on a different number.

    Draw the inner square and say tail integral; the whole calculation is then one line.

    What the interviewer asks next

    • What is the expected distance to the nearest edge in a unit cube?
    • What is the expected distance to the nearest corner of the square?
    • What is the density of the distance to the nearest edge, and where is it highest?

    Asked at Hudson River Trading, Campus Algo Dev Interview, New York, 2024 (Wall Street Oasis): I was asked a expected value question involving the expected value among distance to an edge, with a randomly placed object.

  3. 082Two independent waiting times are each exponentially distributed with a mean of one minute. What is the probability that their total is less than one minute?Continuous and geometric probabilityCoreCitadelChicago · 2025

    Try it first

    Pick the closest value.

    Show the worked solution

    1 - 2/e, about 26.4%. Convolving two exponential densities gives the total the density x e^-x, which starts at zero and peaks at one minute. Its area below one minute is 1 - 2/e. The same number drops out of the Poisson view: the total is under a minute exactly when at least two arrivals land in the first minute of a rate-one Poisson process.

    Why does adding two waits change the shape so much?

    Suppose you need two buses, one after the other, and each arrives on average a minute after you reach its stop. Catching the first bus quickly is common; catching both quickly is rare, because both have to cooperate. A single exponential wait is most likely near zero, but a sum of two is almost never near zero, so its density starts at zero and rises into a hump. That shift of mass away from zero is why the answer is much smaller than the 63.2% chance that one wait is under a minute.

    Two memoryless waits add up to a hump: little mass near zero012345minutes0.51.0one wait: e^-xsum of two waits: x e^-x26.4%One wait under 11 - 1/e = 63.2%both: 40.0%Total under 11 - 2/e = 26.4%the shaded area
    The single exponential wait puts most of its mass near zero, but the total of two waits has density x e^-x, which starts at zero and peaks at one minute, so only 26.4% of its area, shaded, falls below one minute.
    The relationship
    fS(s)=∫0se−xe−(s−x) dx=s e−s,P(S<1)=∫01s e−s ds=1−2e≈0.264f_{S}(s) = \int_0^s e^{-x}e^{-(s-x)}\,dx = s\,e^{-s}, \qquad P(S<1) = \int_0^1 s\,e^{-s}\,ds = 1 - \frac{2}{e} \approx 0.264
    Sthe total of the two waits
    xthe first wait, which can be anything from 0 to s
    e^{-x}the exponential density with mean 1
    What it says in wordsTo land on a total of s, the first wait takes any value x and the second makes up the rest; adding over all x gives s e^-s.

    Is there a way to get 1 - 2/e without integrating?

    Yes, and it is the cleaner answer to give aloud. Exponential waits with mean one are the gaps between arrivals of a Poisson process with rate one per minute. The second arrival comes before one minute exactly when at least two arrivals land in the first minute, and the Poisson chance of zero or one arrival is e^-1 + e^-1 = 2/e. So the answer is 1 - 2/e, about 26.4%, with no calculus at all.

    Sanity-check the size. Both waits being under a minute has probability (1 - 1/e)^2, about 40.0%, and the total being under a minute is a stricter event, so the answer must be smaller: 26.4% is. The limitation is the independence assumption; if the two waits were driven by the same traffic, they would move together and the total would be more spread out.

    Where candidates lose it

    The frequent wrong answer is (1 - 1/e)^2, about 40%, which is the chance that each wait is under a minute. The question asks about the total, and two waits of 0.7 minutes each pass that test while failing this one.

    The second loss is starting a convolution integral and getting lost in the limits. Say the Poisson route first: at least two arrivals in the first minute, one minus the chance of zero or one.

    What the interviewer asks next

    • What is the probability that the sum of three such waits is under one minute?
    • Given the total is exactly 2 minutes, what is the distribution of the first wait?
    • What is the probability that the first wait is shorter than the second?

    Asked at Citadel, Quant Research Interview, Chicago, 2025 (Wall Street Oasis): if i knew this was about convolutions, i would have answered better