Random Chord Division

Relevant wiki: Geometric Probability - Problem Solving

Let the angle that one of the points on the circumference of the circle make at the origin, be $\theta_1$ . Let $\theta$ be the angle subtended by the chord joining this point to another independently and uniformly chosen point on the circumference. Given $\theta_1$ , the angle $\theta$ , is just the measurement of the angle of the second point on the circumference with respect to $\theta_1$ . But since the second point is chosen independently of the first point, and is chosen uniformly, $\theta$ is distributed as $\mathcal{U}[0,2\pi)$ , independent of $\theta_1$ . The chord subtending angle $\theta$ at the center of the circle divides the circle into two regions of areas $A_1=\theta-\sin\theta,\ A_2=2\pi-A_1$ . It is easy to check that $A_2\ge A_1$ if $\theta\in[0,\pi]$ , and $A_1>A_2$ , if $\theta\in (\pi,2\pi)$ . Thus the expected area, conditioned on $\theta_1$ is $\frac{1}{2\pi}\left[\int_0^{\pi}(2\pi-\theta+\sin\theta)d\theta+\int_{\pi}^{2\pi}(\theta-\sin\theta)d\theta\right]\\=\frac{1}{2\pi}\left[2\int_0^{\pi}(2\pi-\theta+\sin\theta)d\theta\right]\\=\frac{1}{\pi}\left[2\pi\cdot \pi-\frac{\pi^2}{2}+2\right]=\frac{3\pi}{4}+\frac{1}{\pi}$ Since the result is independent of $\theta_1$ , this is also the unconditional expectation, which gives the answer $\boxed{1.75}$ .