Two points are chosen at random on the surface of a sphere…

Two points are chosen at random on the surface of a sphere of radius $R$. What is the probability that the arc of the great circle through these two points subtends a (central) angle less than $\alpha$ $(\alpha\lt\pi)$?

Step 1: Fix the first point

The sphere looks the same from every point, so the position of the first point does not matter. Put it at the north pole $N$. The second point $M$ is uniformly distributed on the surface: the probability of a region is its area divided by $4\pi R^2$.

Step 2: The favourable region

The central angle $\angle NOM$ is the "polar angle" of $M$. The points with $\angle NOM\lt\alpha$ form a spherical cap around $N$.

Step 3: Area of the cap

In spherical coordinates the area element is $R^2\sin\theta\,d\theta\,d\varphi$. So

\[ S_{\text{cap}} = \int_0^{2\pi}\!\!\int_0^\alpha R^2\sin\theta\,d\theta\,d\varphi = 2\pi R^2(1-\cos\alpha). \]

(Archimedes: the area of a cap equals $2\pi R h$, where $h=R(1-\cos\alpha)$ is its height.)

Step 4: The probability

\[ \boxed{P = \frac{2\pi R^2(1-\cos\alpha)}{4\pi R^2} = \frac{1-\cos\alpha}{2} = \sin^2\frac\alpha2.} \]

Step 5: Checks

Note: the answer is not $\alpha/\pi$. Near the poles the sphere has little area, so small angular distances are rarer than on a circle.