The meeting problem, Geometric Probability

Problem 5.1

Alice and Bob each arrive at a café at an independent uniform random time between noon and 1:00 PM. Each waits $15$ minutes before leaving. What is the probability that they meet?

Solution. Let $x, y \in [0, 1]$ be the arrival times of Alice and Bob (in units of an hour, with $0$ = noon). The parameter space is the unit square $[0,1]^2$ with total area $1$ . Alice and Bob meet iff their arrival times are within a quarter hour of each other: $|x - y| \leq \tfrac14.$

The region $\{|x - y| > \tfrac14\}$ is the union of two corner triangles: one above the line $y = x + \tfrac14$ , and one below the line $y = x - \tfrac14$ . Each is a right triangle with legs of length $\tfrac34$ , so each has area $\tfrac12 \cdot (\tfrac34)^2 = \tfrac{9}{32}$ . Their combined area is $\tfrac{9}{16}$ .

Hence the meeting region has area $1 - \tfrac{9}{16} = \tfrac{7}{16}$ , and $\boxed{\mathbb{P}(\text{meet}) = \frac{7}{16}.} \qedhere$

The meeting problem. The unit square of arrival times (x, y) ; Alice and Bob meet iff |x - y| \leq \tfrac14 (sage band). The two corner triangles (copper) with total area \tfrac{9}{16} are the “miss” region. Meeting probability = \tfrac{7}{16} . — The meeting problem. The unit square of arrival times $(x, y)$ ; Alice and Bob meet iff $|x - y| \leq \tfrac14$ (sage band). The two corner triangles (copper) with total area $\tfrac{9}{16}$ are the “miss” region. Meeting probability $= \tfrac{7}{16}$ .

The meeting problem generalises: if Alice waits $\alpha$ hours and Bob waits $\beta$ hours (both in a 1-hour window), the meeting region is the unit square with two opposite corner triangles removed, and the probability is $1 - \tfrac{(1 - \alpha)^2}{2} - \tfrac{(1 - \beta)^2}{2}$ for $\alpha, \beta \leq 1$ . For $\alpha = \beta = \tfrac14$ , this gives $1 - (3/4)^2 = 7/16$ .

import numpy as np
x, y = np.random.rand(2, 10**7)
p = (np.abs(x - y) <= 0.25).mean()
print(f"sim: {p:.5f}   exact: {7/16:.5f}")
# sim: 0.43746   exact: 0.43750