Expected area of a random triangle in a disc

Solution. This is a classical constant (Woolhouse, 1867): the answer is $$\mathbb{E}[\operatorname{Area}] = \frac{35}{48 \pi}.$$ We sketch the derivation.

Write $P_i = (X_i, Y_i)$ with each $P_i$ uniform in the unit disc. Twice the signed area is $$D = (X_2 - X_1)(Y_3 - Y_1) - (X_3 - X_1)(Y_2 - Y_1).$$ We want $\mathbb{E}[|D|] / 2$.

Reduction. By rotational symmetry, condition on $P_1$ at distance $r_1 \in [0, 1]$ from the origin (WLOG $P_1 = (r_1, 0)$). The marginal density of $r_1$ is $2 r_1$. The remaining integration over $P_2, P_3$ is, for fixed $P_1$, an expectation $\mathbb{E}[|\text{area of } \triangle P_1 P_2 P_3| \mid P_1]$, which depends on $r_1$.

Evaluation. This inner expectation can be computed in closed form using the joint density $(2r_2)(2r_3)/(2\pi)^2$ of $(r_2, \theta_2, r_3, \theta_3)$ and the trigonometric identity $|D| = |r_2 r_3 \sin(\theta_3 - \theta_2) + r_1 r_2 \sin \theta_2 - r_1 r_3 \sin \theta_3|$. After careful integration (spelled out in Klain & Rota, Introduction to Geometric Probability, 1997), the result is $$\boxed{\mathbb{E}[\operatorname{Area}] = \frac{35}{48\pi} \approx 0.2321.} \qedhere$$

import numpy as np
r = np.sqrt(np.random.rand(3, 10**7))
t = np.random.rand(3, 10**7) * 2*np.pi
x, y = r*np.cos(t), r*np.sin(t)
area = 0.5*np.abs((x[1]-x[0])*(y[2]-y[0]) - (x[2]-x[0])*(y[1]-y[0]))
print(f"sim: {area.mean():.5f}   exact: {35/(48*np.pi):.5f}")
# sim: 0.23210   exact: 0.23210