Two random arcs on a circle, Geometric Probability

Problem 16.1

Two arcs, each of length $a$ (with $a \leq \tfrac12$ ), are placed independently at uniformly random positions on a circle of unit circumference. What is the probability that they overlap?

Solution. Parametrise each arc by the position of its centre on the circle; measure positions as fractions of the total circumference, so each centre is uniform on $[0, 1)$ , and positions are taken modulo $1$ .

By rotational symmetry, fix the first arc’s centre at $0$ : it then covers the set $[-a/2, a/2]$ (mod $1$ ). Let $U \in [0, 1)$ be the uniform position of the second arc’s centre; the second arc covers $[U - a/2, U + a/2]$ (mod $1$ ).

The two arcs overlap iff their centres are within $a$ of each other on the circle, i.e., iff $U \in [-a, a]$ (mod $1$ ), which is the set $[0, a] \cup [1 - a, 1)$ of total length $2a$ . (Here we use $a \leq \tfrac12$ so the two pieces do not overlap.)

Hence $\boxed{\mathbb{P}(\text{overlap}) = 2a.} \qedhere$

Two arcs of length a = 0.25 (proportion of circumference). Arc 1 is fixed, arc 2 is placed with centre at u = 0.15 . Since |0.15| a as a circular distance, the arcs overlap in the rose region. — Two arcs of length $a = 0.25$ (proportion of circumference). Arc 1 is fixed, arc 2 is placed with centre at $u = 0.15$ . Since $|0.15| < a$ as a circular distance, the arcs overlap in the rose region.

Variation. If there are $n$ independent arcs of length $a$ , the probability that no two overlap is $1 - n(n-1) a + O(a^2)$ for small $a$ (union bound). The exact formula is more involved. A classical generalisation is Stevens’s 1939 theorem: $n$ iid arcs of length $a$ on a unit circle cover the entire circle with probability $\sum_{k = 0}^{\lfloor 1/a \rfloor} (-1)^k \binom{n}{k} (1 - k a)^{n-1},$ a remarkable closed form.

import numpy as np
a, N = 0.2, 10**7
# First arc centred at 0; second arc centred at uniform u on unit circle
u = np.random.rand(N)  # fraction of circumference
# Circular distance between centres
d = np.minimum(u, 1 - u)
p = (d <= a).mean()
print(f"sim: {p:.5f}   exact: {2*a:.5f}")
# sim: 0.39998   exact: 0.40000