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For three independent uniform points on a sphere, condition on the central angle $\theta$ between the first two. The third point makes acute angles with both exactly when it lies in the intersection of their positive hemispheres. That intersection is a spherical lune of opening $\pi -\theta$, hence has probability $(\pi -\theta )/(2\pi )$. The first two points also need $\theta <\pi /2$. Integrating against the central angle between independent uniform sphere points density gives $(4\pi {)}^{-1}{\int }_0^{\pi /2}(\pi -\theta )\sin \theta d\theta =(\pi -1)/(4\pi )$.

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 densityand 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 $2R^2(\pi -\gamma )$. ThereforeThe union probability follows from inclusion-exclusion for two hemispheres:Thus the requested conditional answers are“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 isUsing ${\int }_0^{\pi /2}\gamma \sin \gamma d\gamma =1$ and ${\int }_{\pi /2}^{\pi }\gamma \sin \gamma d\gamma =\pi -1$ givesAngles exactly equal to the threshold have probability zero. Connectivity allows relaying and is weaker than requiring direct links between every pair.

Let $\Theta =\angle AOB$ and let the sphere have radius $R$. By rotational invariance, condition on $A$ and place it at the north pole. A band at polar angle $\theta$ has surface area $2\pi R^2\sin \theta d\theta$, while the sphere has area $4\pi R^2$. Thus the central angle between independent uniform sphere points has densitywith density zero outside that interval.

Condition now on $A,B$ with their angle fixed at $\theta$. For $\angle AOC$ and $\angle BOC$ to be acute, the unit position vector of $C$ must have positive dot products with both unit vectors of $A$ and $B$. Hence $C$ must lie in the intersection of the two corresponding open hemispheres. The great circles bounding them meet at the two antipodes perpendicular to the plane of $A,B$. Their common hemisphere region is a spherical lune of opening $\pi -\theta$: when $A=B$ it is a hemisphere of opening $\pi$, and as the two normals separate, that opening decreases by $\theta$. More explicitly, take the common boundary antipodes as poles. The longitude conditions each allow an interval of length $\pi$, with their centres separated by $\theta$, so the intersection has longitude width $\pi -\theta$.

Integrating the spherical area element $R^2\sin \phi d\phi d\lambda$ over that lune gives area $2R^2(\pi -\theta )$. Thus the conditional probability that the two angles involving $C$ are acute is $(\pi -\theta )/(2\pi )$. The remaining angle $\Theta$ must itself be acute, so the desired probability isIntegration by parts gives ${\int }_0^{\pi /2}\theta \sin \theta d\theta =1$, while ${\int }_0^{\pi /2}\sin \theta d\theta =1$. ThereforeThis calculates pairwise acute central angles of three uniform sphere points by conditional area, rather than incorrectly treating the three angle events as independent. Boundary right angles have probability zero, so choosing open rather than closed hemispheres does not change the result.