A tower lets you peek past your horizon, so each of the six planets gets a little more likely. But it also catches the sunrise early, and daylight kills the parade. Six boosts, one penalty: beta = 5/44. Seven circles on a ball. That's the whole puzzle.
From the comments: why would a tower help at all?
Because the ground curves. Standing on the surface, your horizon is exactly flat, so each planet is above it for exactly half of all spots on the globe. That is the ½ coin flip from part 3.
In the puzzle, the tower means a body counts as visible if it is visible from any spot within a small distance r of you. Think of it as seeing a little way past your flat horizon, over the curve. A planet just below your own horizon now counts as up. On the globe, the half that can see each planet grows by a thin band of width r along its edge, so each planet's chance goes from ½ to ½(1 + r/R), where R is the planet's radius.
The catch is the star. The tower lets you see past the horizon for the sun too, so dawn comes a little earlier and night gets a little shorter: ½(1 − r/R).
Six planets get a small boost, the star gets the same small penalty: 6 − 1 = 5. That is where β = 5/44 comes from. (And yes, the comment with α = 1/44, β = 5/44 is right.)
Source: Jane Street puzzle, March 2026, "Planetary Parade", and its published solution.