This month we had to compute some tricky probabilities involving planetary observability in the Pyrknot star system. We are told that the six “coplanets” to Pyrknot have indepedent uniform positions in the sky, and are not visible in daylight (so we also need to treat the star as a seventh celestial body that can’t be in the sky for a planetary parade to occur). So to find α we can reason in the following way:

From each point on Pyrknot’s surface one can see exactly half of the full sphere of the sky, and so each of the celestial bodies defines a great circle dividing Pyrknot into the half that can see that body and the half that cannot. Now, these great circles chop up the surface of Pyrknot into a graph with a number of vertices V, edges E, and faces (or regions) F, that satisfy the Euler characteristic formula for a sphere of V - E + F = 2. We know

• All pairs of great circles intersect each other at two vertices, so V = 2 * (7 * 6 / 2) = 42.
• Each of the seven great circles is divided into 12 edges from these vertices, so E = 7 * 12 = 84.
• Therefore F = 2 - V + E = 2 - 42 + 84 = 44.

So you end up with 44 different regions that each would see a distinct subset of the planets ignoring the fact that when their star is visible no planets can be seen. Either one or zero of those regions can see a planetary parade at a given moment and so conditional on one region seeing one, you are 1/44 to be inside the same region as that someone (the regions are all equal in expected area by symmetry).

Now, if you build a tower that allows you to see any celestial body visible from a point up to r distance away on the surface of Pyrknot, to first order you are pushing outward the borders of the region by r and so improve your probability proportional to the length of the perimeter of the region. But not quite, because 1/7 of the boundary of the region (in expectation) is caused by the day/night boundary on Pyrknot, and if your tower is near this boundary you could actually ruin seeing a planetary parade by the top of your tower being in daylight. Egads!

The expected perimeter of the region is just

(1/44) * [total perimeter of all regions] = 1/44 * [14 * 2πR] = (7/11)πR.

Now 6/7 of that perimeter expands by a distance r and 1/7 of that perimeter contracts by a distance r. To first order the absolute difference in area is

(6/11)πR⋅r - (1/11)πR⋅r = (5/11)πR⋅r.

As a percentage of the full area of Pyrknot, 4πR², this difference is

[(5/11)πR⋅r] / [4πR²] = (5/44)(r/R)

And thus α = 1/44 and β = 5/44.

Congrats to the solvers able to reason out this tricky astronomical setup!