TITLE: Have I found an error in Williams' "Martingales" exercises?
QUESTION [7 upvotes]: I want to solve Problem EG.2. from Probability with Martingales:

Planet X is a ball with centre 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 $\measuredangle AOB < 90^\circ$. Show that the probability that they can keep in touch (with, for example, A communicating with B via C if necessary) is $(\pi + 2)/(4\pi)$.

I believe I have shown that the probability is in fact $(2 \pi + 1) / (4 \pi)$. To begin, we denote by $AB$ (resp. $AC$, $BC$) the event that $A$ and $B$ are within $90^\circ$ of each other. We then observe that the probability that communication is possible is given by:
$$P(comms) = P(AB \cup (AC \cap BC)),$$
which after some basic manipulation may be re-written as
$$P(comms) = P(AB) + P(AC \cap BC) - P(AB \cap AC \cap BC).$$
We then proceed by computing each term on the right-hand side. (A warning that my argument and notation is somewhat loose, and I would appreciate any ideas about tightening it.) For $P(AB)$, we have
$$P(AB) = \iint_S P(AB | A = x) \frac{1}{|S|} dS(x),$$
where $S$ denotes the planet's surface and $|S|$ its surface area. By symmetry, we see $P(AB|A) = 1/2$, so that $P(AB) = (1/2)(1/|S|)|S|=1/2$.
Next,
$$P(AC \cap BC) = \iint_S P(AC|C = x) P(BC|C = x) \frac{1}{|S|} dS(x).$$
By the same argument as above, $P(AC|C) = P(BC|C) = 1/2$, and hence $P(AC \cap BC) = 1/4$.
Finally, we have
$$P(AB \cap AC \cap BC | A=x) = \iint_{S/x} P(AC \cap BC | A = x, B = y) \frac{1}{2|S|} dS(y),$$
where $S/x$ denotes the hemisphere of $S$ centred on $x$. Now, $P(AC \cap BC|A=x,B=y)$ corresponds to the fraction of $S$ covered by the intersection of $S/x$ and $S/y$. If we define a spherical coordinate system on $S$ with $x$ corresponding to polar angle $\theta = 0$, then this fraction is given by $1/2 - \theta(y)/(2 \pi)$ for $0 \leq \theta(y) \leq \pi/2$ (i.e. $\theta(y)$ in $S/x$), and hence
$$P(AB \cap AC \cap BC | A=x) = \frac{1}{4} - \frac{1}{4\pi}$$
(where I skip performing the double integral above here explicitly for brevity).
All up, we then obtain
$$P(comms) = \frac{2 \pi + 1}{4 \pi},$$
as stated above, which disagrees with Williams' stated result, although no mistake is obvious to me. Moreover, I have written a Monte Carlo simulator in Python that seems to confirm my result (and which I can provide if anyone would find it useful).
In summary, I ask the following: (1) Is Williams correct, or am I?; and (2) How can my argument be made tighter, more rigorous, or more elegant?

REPLY [6 votes]: The problem is ambiguously worded. You sensibly took "they" to refer to spaceships A and B, the subject of the previous sentence, but it's meant to refer to all three spaceships, the subject of the sentence before that.
Your calculations are correct, and for the probability that all three spaceships are in contact they yield
\begin{align}
P(AB\cap BC\cap CA)+3P(AB\cap BC\cap\overline{CA})
&=
3P(AB\cap BC)-2P(AB\cap BC\cap CA)
\\
&=
3\cdot\frac14-2\cdot\left(\frac14-\frac1{4\pi}\right)
\\
&=
\frac{\pi+2}{4\pi}\;.
\end{align}