Random chord longer than the radius, Geometric Probability

Problem 15.1

Let $P, Q$ be two points chosen independently and uniformly on a circle of radius $r$ . What is the probability that the chord $PQ$ is longer than $r$ ?

Solution. As in Chapter 13, parametrise $P, Q$ by angles $\alpha, \beta$ iid uniform on $[0, 2\pi)$ . The chord length is $|PQ| = 2 r \sin\tfrac{\phi}{2}$ , where $\phi = \alpha - \beta \pmod{2\pi} \in [0, 2\pi)$ is uniform.

On $[0, 2\pi)$ , the chord exceeds $r$ iff $2 r \sin\tfrac{\phi}{2} > r$ , i.e., $\sin\tfrac{\phi}{2} > \tfrac12$ . Since $\phi/2 \in [0, \pi)$ , this is equivalent to $\phi/2 \in (\pi/6, 5\pi/6)$ , i.e., $\phi \in (\pi/3, \; 5\pi/3).$

The length of this set is $\tfrac{5\pi}{3} - \tfrac{\pi}{3} = \tfrac{4\pi}{3}$ . The sample space $[0, 2\pi)$ has length $2\pi$ , so $\boxed{\mathbb{P}(|PQ| > r) = \frac{4\pi/3}{2\pi} = \frac{2}{3}.} \qedhere$

A classical note. The side of an equilateral triangle inscribed in a circle of radius $r$ is $r \sqrt 3$ . Bertrand’s paradox asks for the probability that a random chord is longer than this side, under three different sampling rules; with uniform endpoints one gets $\tfrac13$ . The problem in this chapter asks a simpler question — chord longer than the radius — and the answer $\tfrac23$ does not suffer any such paradox, because the “random chord” is fully specified by uniform endpoints.

import numpy as np
theta = np.random.rand(2, 10**7) * 2*np.pi
chord = 2 * np.abs(np.sin((theta[0] - theta[1]) / 2))
print(f"sim: {(chord > 1).mean():.5f}   exact: {2/3:.5f}")
# sim: 0.66660   exact: 0.66667