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 $\measuredangle AOB < 90°$. 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.
 A: Ok I think I have the answer:
The first step is to calculate the law of the angle $\alpha$
$$\mathbb{P}(\alpha < t) = \frac{( 1 - \cos(t)) 2 \pi}{4\pi}$$
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:


*

*$\alpha < \frac{\pi}{2}$, 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:
$$\mathbb{E}\left[\mathbb{1}_{\alpha < \frac{\pi}{2}}\frac{\alpha + \pi}{2\pi} + \mathbb{1}_{\alpha \geq \frac{\pi}{2}}\frac{\pi - \alpha}{2\pi}\right] $$
$$=\int_0^\frac{\pi}{2} \frac{\sin(t)}{2}\frac{t + \pi}{2\pi} dt + \int_\frac{\pi}{2}^\pi \frac{\sin(t)}{2}\frac{\pi - t}{2\pi} dt$$
$$=\frac{\pi + 2}{4\pi}$$
$\square$
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.
A: Another way to see this is by the inclusion-exclusion formula:
$$\mathbb{P}(E_A\cup E_B \cup E_C) = \mathbb{P}(E_A) + \mathbb{P}(E_B) + \mathbb{P}(E_C) - \mathbb{P}(E_A\cap E_B) - \mathbb{P}(E_A\cap E_C) - \mathbb{P}(E_B\cap E_C) + \mathbb{P}(E_A\cap E_B\cap E_C).$$
Let $E_A = \{(A,B,C) \in (S^2)^3 \mid \measuredangle AOB < \frac{\pi}{2}, \measuredangle AOC < \frac{\pi}{2}  \}$, and similarly define $E_B = \{(A,B,C) \in (S^2)^3 \mid \measuredangle BOA < \frac{\pi}{2}, \measuredangle BOC < \frac{\pi}{2}  \}$ and $E_C = \{(A,B,C) \in (S^2)^3 \mid \measuredangle COB < \frac{\pi}{2}, \measuredangle COA < \frac{\pi}{2}  \}$.
The set we want to measure is  $E_A\cup E_B \cup E_C.$ Note that we have $E_A\cap E_B=E_B\cap E_C = E_A\cap E_C = E_A \cap E_B \cap E_C$, and by symmetry $\mathbb{P}(E_A) = \mathbb{P}(E_B) = \mathbb{P}(E_C) = \frac{1}{4},$ since we can reflect either of the points $B$ or $C$ in $E_A$ (say) through $O$ so that either it lies inside or outside $E_A$ (away from the boundary great circle, which has measure $0$). Applying the inclusion-exclusion formula,
$$\mathbb{P}(E_A\cup E_B \cup E_C) = \frac{3}{4} - 2\mathbb{P}(E_A\cap E_B \cap E_C),$$
and so it remains to find the measure of the last set on the right side of the equation above. Let $\phi = \measuredangle AOB.$ The area of the intersection of the hemispheres $H_A = \{ C \in S^2\mid \measuredangle AOC < \frac{\pi}{2}\}$ and $H_B = \{ C \in S^2\mid \measuredangle BOC < \frac{\pi}{2}\}$ is $2\pi - 2 \phi,$ since its complement in $H_A$ is a slice of $S^2$ by angle $\phi$. Hence, $\mathbb{P}(C \in H_A \cap H_B \mid \measuredangle AOB = \phi) =\frac{1}{4\pi}\left( 2\pi - 2\phi \right) = \frac{1}{2\pi}\left(\pi - \phi \right)$ and we can integrate this over $H_A$ to get 
$$\mathbb{P}(E_A \cap E_B \cap E_C) = \frac{1}{8\pi^2}\int_0^{2\pi} \int_0^{\frac{\pi}{2}} \left(\pi-\phi\right)\sin \phi \, d \phi \, d \theta =  \frac{1}{4\pi}\left( \pi-\int_0^\frac{\pi}{2}\phi \sin \phi \, d\phi  \right) = \frac{\pi-1}{4\pi}.$$
Finally, substituting this back into the formula we had for $\mathbb{P}(E_A\cup E_B \cup E_C) $ gives
$$\mathbb{P}(E_A\cup E_B \cup E_C) = \frac{3}{4} - \frac{2\pi-2}{4\pi} = \frac{\pi+2}{4\pi}$$
as the book claimed.
A: 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)
