= Solution
By rotational invariance, fix the first landing point at the north pole of a <sphere> of radius $R$. A band of central angle $\theta$ and width $d\theta$ has area $2\pi R^2\sin\theta\,d\theta$. Dividing by total area gives the <central angle between independent uniform sphere points> density
$$
\boxed{f_\Theta(\theta)=\tfrac12\sin\theta,\quad0<\theta<\pi,}
$$
and zero outside this range. The radius cancels.
Given the first pair's central angle $\gamma$, the third point can communicate with each of them exactly when it lies in that point's positive <hemisphere>. Choose their intersection line as a polar axis. Each <hemisphere> permits an azimuth interval of length $\pi$, and the intervals' centres are separated by $\gamma$. Their overlap is a <spherical lune> of opening $\pi-\gamma$, with area $2R^2(\pi-\gamma)$. Therefore
$$
\mathbb P(C\text{ linked to both}\mid\gamma)=\frac{\pi-\gamma}{2\pi}.
$$
The union <probability> follows from inclusion-exclusion for two <hemispheres>:
$$
\mathbb P(C\text{ linked to at least one}\mid\gamma)
=\frac12+\frac12-\frac{\pi-\gamma}{2\pi}
=\frac12+\frac{\gamma}{2\pi}.
$$
Thus the requested conditional answers are
$$
\boxed{\mathbb P(C\text{ linked to either}\mid\gamma<\pi/2,\gamma)
=\frac12+\frac{\gamma}{2\pi},}
$$
$$
\boxed{\mathbb P(C\text{ linked to both}\mid\gamma>\pi/2,\gamma)
=\frac12-\frac{\gamma}{2\pi}.}
$$
“Either” here is inclusive: a third ship linked to both is also linked to at least one.
For network connectivity, if $\gamma<\pi/2$ the first pair is already linked, and the third must link to at least one of them. If $\gamma>\pi/2$, the first pair is not linked, and both must link through the third. Consequently the <connectivity of three acute-angle sphere points> is
$$
P_{\mathrm{conn}}=
\int_0^{\pi/2}\left(\frac12+\frac{\gamma}{2\pi}\right)\frac{\sin\gamma}{2}\,d\gamma
+\int_{\pi/2}^{\pi}\left(\frac12-\frac{\gamma}{2\pi}\right)\frac{\sin\gamma}{2}\,d\gamma.
$$
Using $\int_0^{\pi/2}\gamma\sin\gamma\,d\gamma=1$ and $\int_{\pi/2}^{\pi}\gamma\sin\gamma\,d\gamma=\pi-1$ gives
$$
\boxed{P_{\mathrm{conn}}=\frac12+\frac{2-\pi}{4\pi}=\frac{\pi+2}{4\pi}.}
$$
Angles exactly equal to the threshold have <probability> zero. Connectivity allows relaying and is weaker than requiring direct links between every pair.
Back to article page
