Three friends at the café, Geometric Probability

Problem 11.1

Three friends each arrive at a café at an independent uniform random time in a one-hour window, and each waits exactly $w$ hours before leaving, where $0 \leq w \leq 1$ . What is the probability that all three end up meeting (i.e., the last to arrive reaches the café before the first to leave)?

Solution. Let $T_1, T_2, T_3$ be iid uniform arrivals in $[0, 1]$ (units of one hour). Friend $i$ is present during the interval $[T_i, T_i + w]$ . All three are simultaneously present iff the intersection of their intervals is non-empty, i.e., iff $T_{\max} \leq T_{\min} + w \;\;\iff\;\; \max T_i - \min T_i \leq w,$ where $T_{\max}, T_{\min}$ are the largest and smallest arrival times.

For $n$ iid uniform random variables in $[0, 1]$ , the range $R = T_{(n)} - T_{(1)}$ has the classical density $f_R(r) = n(n-1) r^{n-2} (1 - r)$ on $[0, 1]$ . Integrating, $\mathbb{P}(R \leq w) = \int_0^w n(n-1) r^{n-2}(1 - r) \, dr = n w^{n-1} - (n-1) w^n.$ For $n = 3$ : $\boxed{\mathbb{P}(\text{all meet}) = 3 w^2 - 2 w^3.} \qedhere$

For common waiting times:

$w = \tfrac14$ : $\mathbb{P} = \tfrac{3}{16} - \tfrac{1}{32} = \tfrac{5}{32} \approx 0.156$ .
$w = \tfrac13$ : $\mathbb{P} = \tfrac{1}{3} - \tfrac{2}{27} = \tfrac{7}{27} \approx 0.259$ .
$w = \tfrac12$ : $\mathbb{P} = \tfrac34 - \tfrac14 = \tfrac12$ .
$w = \tfrac23$ : $\mathbb{P} = \tfrac43 - \tfrac{16}{27} = \tfrac{20}{27} \approx 0.741$ .

Compare with the two-friend case (Chapter 5), where the meeting probability is $1 - (1 - w)^2$ . For $w = \tfrac14$ the two-friend case gives $\tfrac{7}{16}$ , far exceeding the three-friend $\tfrac{5}{32}$ . Adding a third friend sharply reduces the chance of simultaneous meeting.

import numpy as np
w = 0.4
arr = np.random.rand(3, 10**7)
meet = (arr.max(axis=0) - arr.min(axis=0)) <= w
print(f"sim: {meet.mean():.5f}   exact: {3*w**2 - 2*w**3:.5f}")
# sim: 0.35272   exact: 0.35200