Exercise EG.1.1 from David Williams' “Probability with Martingales” page 224 I have a hard time with the following exercise:

Planet X is a ball with center O. Three spaceships A, B and C land at random on its surface, their positions being independent and each uniformly distributed on the surface. Spaceships A and B can communicate directly by radio if $\measuredangle AOB &lt; 90°$. Show that the probability that they can keep in touch (with, for example, A communicating with B via C if necessary) is $$\frac{\pi + 2}{4 \pi}$$.
I can reach a result if I consider a circle instead of a sphere, but I have a hard time to define the coordinates of point placed randomly on a sphere.
Ok I think I have the answer:

The first step is to calculate the law of the angle $\alpha$

$$\mathbb{P}(\alpha &lt; t) = \frac{( 1 - \cos(t)) 2 \pi}{4\pi}$$

whenever $0 \leq t \leq \pi$.

Explanation: let's consider the unit sphere, and an arbitrary point on it, the surface defined by all the point that makes an angle of $t$ or less with is point is $( 1 - \cos(t)) 2 \pi$

Now that we know the cumulative function distribution of $\alpha$, we can deduce its density function:

$$f_\alpha(x)=\frac{\sin(x)}{2}$$

whenever $0 \leq x \leq \pi$.

Let's consider now the angle $\measuredangle AOB$, let's call it $\alpha$. We have two cases:


$\alpha &lt; \frac{\pi}{2}$, in this case the chances that all spaceships communicate together are the chances that $C$ is on a place where it could reach either $A$ or $B$:
$$\frac{\alpha + \pi}{2\pi}$$
$\alpha \geq \frac{\pi}{2}$, in this case the chances that all spaceships communicate together are the chances that $C$ is on a place where it could reach either $A$ and $B$:
$$\frac{\pi - \alpha}{2\pi}$$


The probability that all spaceships can communicate is given by:

$$\mathbb{E}\left[\mathbb{1}_{\alpha &lt; \frac{\pi}{2}}\frac{\alpha + \pi}{2\pi} + \mathbb{1}_{\alpha \geq \frac{\pi}{2}}\frac{\pi - \alpha}{2\pi}\right] $$
$$=\int_0^\frac{\pi}{2} \frac{\sin(t)}{2}\frac{t + \pi}{2\pi} dt + \int_\frac{\pi}{2}^\pi \frac{\sin(t)}{2}\frac{\pi - t}{2\pi} dt$$
$$=\frac{\pi + 2}{4\pi}$$

$\square$

Thanks to the author of this article:

http://weberprobability.blogspot.fr/2014/02/planet-zog-questions.html

who gave me all the hints to reach a complete solution.
verified that the answer is correct, notice that:

the integral of [0~pi/2] of (sin(t)/2) * (t + pi)/(2 pi) dt is (pi + 1) / (4 pi)

and the integral of [pi/2~pi] of (sin(t)/2) * (pi - t)/(2 pi) dt is 1 / (4 pi)