Two People, One Hour, and Seven Chances in Sixteen

Two people agree to meet somewhere between one and two in the afternoon, without fixing a time. Each arrives at a moment chosen uniformly at random inside that hour, independently of the other. Each is willing to wait fifteen minutes and then leaves. How often do they actually meet?

Fifteen minutes is a quarter of an hour, so the reflex answer is one in four. That reflex has a real thought behind it, which is why it is worth answering carefully rather than waving away. The true figure is $7/16$, which is $0.4375$, just under a coin flip and nearly twice the guess.

Turning two clocks into one point

Let $X$ and $Y$be the two arrival times in minutes after one o'clock, so each is uniform on $[0, 60]$ and the two are independent. Then the pair $(X, Y)$ is uniform on the square $[0,60]^2$, and every probability in this problem becomes an area divided by $3600$. That single move is what makes the question tractable, and it deserves a name.

The next step is the one people skip, and skipping it is how the answer goes wrong. What exactly is the meeting event, written in terms of $X$ and $Y$?

Suppose the first person arrives at time $X$ and waits until $X + 15$. The second arrives at $Y$. They overlap if $Y \le X + 15$. If instead $Y$ came first, the same reasoning gives $X \le Y + 15$. Both conditions together are one condition:

The condition is symmetric in the two people, which it has to be, since neither of them is distinguished. Nothing about who arrives first survives into the final inequality, and that is the sign that the modelling step is right.

Two triangles, and the mistake that halves them

Inequality (1) carves a diagonal band out of the square, running parallel to the line $x = y$ with a vertical extent of fifteen minutes on either side. What it leaves out is a pair of corner regions, and getting their combined area right is the whole computation.

The region where $x - y > 15$ is the triangle with vertices $(15, 0)$, $(60, 0)$ and $(60, 45)$. Its two legs both measure $60 - 15 = 45$. The region where $y - x > 15$ is its mirror image across the diagonal, with the same two legs. So there are two congruent right triangles of leg 45, and two congruent right triangles with equal legs fit together into the full square on that leg:

That factor of two is the single place this problem punishes carelessness. Compute the area of one triangle, $45^2/2 = 1012.5$, forget that there are two of them, and you get a miss probability of $1012.5/3600 = 9/32$ and an answer of $23/32$. Or take the band's half width of 15 as the leg of the omitted square rather than the $60 - 15$ that it is, giving a miss of $(15/60)^2 = 1/16$ and an answer of $15/16$, which is confidently wrong in the other direction. The safeguard is the picture: the two triangles visibly assemble into a square, so the omitted area is a square of side 45 and nothing else.

With the omitted area in hand the arithmetic is short. The miss probability is the square of side 45 over the square of side 60:

and the answer is the complement:

The general wait

Nothing above used the particular numbers, so replace them. Let the window have length $T$ and let each person wait $n$, with $0 \le n \le T$. The omitted triangles now have legs $T - n$, so

At $T = 60$ this is $n(120-n)/3600$. Substituting $n = 15$ returns $7/16$, and a half hour of patience gives $30 \cdot 90 / 3600 = 3/4$. The condition $n \le T$ matters: past $n = T$ the formula keeps climbing above 1 and stops describing anything, because the triangles have already vanished and there is no more area to reclaim.

The shape of that curve explains the gap between $1/4$ and $7/16$. Differentiating (5) gives $2(T-n)/T^2$, so the first minutes of patience are the most valuable and the return falls off linearly. Waiting is front loaded. The first quarter of an hour buys nearly half the possible meetings, and the last quarter hardly buys anything, because by then only near-opposite arrival times are still failing.

Why one in four is a reasonable wrong answer

The guess is not pulled from nowhere. Fifteen minutes really is a quarter of the window, and if you fix one person's arrival at, say, one thirty, the other person meets them exactly when they land in a thirty-minute stretch, which is half the hour. The reflex takes a length of time and reads it directly as a probability. What it misses is that the stretch available to the second person depends on when the first arrived, and it depends on it asymmetrically: someone arriving at one o'clock has only fifteen minutes of cover ahead of them, while someone arriving in the middle of the hour has thirty. Averaging that varying window over the hour is precisely what equation (3) does.

Two readings that do not matter, and one that does

Two boundary questions come up every time this problem is posed, and both are harmless.

The first is whether "meet" needs a strict inequality. It makes no difference: $|X - Y| = 15$ is a line segment in the square, an area of zero, so $P(|X-Y| \le 15)$ and $P(|X-Y| < 15)$ are the same number. The same goes for the event that the two arrive at exactly the same instant.

The second is whether somebody who arrives at 1:55 is allowed to wait past two o'clock. The problem as stated lets them, and the square in Fig. 1 carries no truncation along its top and right edges for that reason. This one is not cosmetic. If waiting is cut off at two o'clock, late arrivals get less than their fifteen minutes, the band gets clipped near the far corner, and the answer changes. So the reading has to be declared rather than assumed, and the version above is the one where the fifteen minutes is always available in full.

Sources and further reading

• The distribution both arrival times follow — Continuous uniform distribution
• The area-for-probability trade in the second section — Geometric probability
• Why the complement in equation (4) is the easier side to compute — Complementary event
• The distribution of the gap $X - Y$ itself — Triangular distribution

The figure $7/16$ was checked four ways before publication. Symbolic algebra confirms equations (2) through (5). An exact lattice count on an $m \times m$ grid of cell centres returns $7/16 + 3/(4m)$ identically, which holds at $m = 240$ as $141/320$ and at $m = 1200$ as $701/1600$, so the limit is $7/16$ exactly rather than merely close to it. Four million simulated afternoons give $0.4374935$. Finally, a separate simulation that never uses inequality (1) at all, and instead builds each person's fifteen-minute occupancy interval and tests whether the two intervals intersect, gives $0.43784$ on 200000 trials, which puts the modelling step under test rather than the arithmetic alone.