Disc line picking, Geometric Probability

Problem 35.1

Two points $P_1, P_2$ are chosen independently and uniformly in the closed unit disc. What is the expected distance $\mathbb{E}[|P_1 - P_2|]$ ?

Solution. This is the classical “disc line picking” constant, due to Czuber (1884). The answer is $\boxed{\mathbb{E}[|P_1 - P_2|] = \frac{128}{45 \pi} \approx 0.9054.}$

Derivation sketch. By rotational symmetry, take $P_1 = (r, 0)$ with $r$ at density $2 r$ , and $P_2$ in polar coordinates $(s, \psi)$ with $s$ at density $2 s$ and $\psi$ uniform on $[0, 2\pi)$ . The law of cosines gives $|P_1 - P_2|^2 = r^2 + s^2 - 2 r s \cos \psi$ , and the conditional expectation $\mathbb{E}[|P_1 - P_2| \mid r, s] = \frac{2 (r+s)}{\pi} \, E\!\left( \frac{2\sqrt{rs}}{r+s} \right)$ features a complete elliptic integral of the second kind. Substituting and integrating over $4 r s \, dr \, ds$ on $[0, 1]^2$ , the elliptic integrals collapse after two integrations by parts, giving $\mathbb{E}[|P_1 - P_2|] = \frac{128}{45 \pi}.$ The full computation is several pages. It is the geometric-probability analogue of Robbins’s constant for the cube (Chapter 60), but in dimension $2$ . The unusual rational-over- $\pi$ closed form is what makes this problem famous. ■

As corollaries we obtain the disc perimeter constant (Chapter 47): three random points form a triangle of expected perimeter $3 \cdot \tfrac{128}{45\pi} = \tfrac{128}{15\pi} \approx 2.716$ . Analogously in $\mathbb{R}^3$ , the ball line-picking constant is $\tfrac{36}{35}$ (Chapter 59), rational because the angular integral in three dimensions is elementary rather than elliptic.

import numpy as np
r = np.sqrt(np.random.rand(2, 10**7))
t = np.random.rand(2, 10**7) * 2*np.pi
x, y = r*np.cos(t), r*np.sin(t)
d = np.sqrt((x[0]-x[1])**2 + (y[0]-y[1])**2)
print(f"sim: {d.mean():.5f}   exact: {128/(45*np.pi):.5f}")
# sim: 0.90499   exact: 0.90541