> The same decomposition works in higher dimensions.

I don't think the same argument works in higher dimensions. On a circle, we can canonically pick a semicircle corresponding to each point (we have two choices, let's say we pick the clockwise one).

In higher dimensions there's no canonical choice of half-sphere. In odd dimensions one could pick a canonical half-sphere per point but it might turn out that some other non-chosen half-sphere for that point contains all the other points. In even dimensions there isn't even a way to canonically pick a half-sphere for each point (this is a consequence of the Hairy Ball Theorem).

(For all I know the actual numbers might turn out to be the same, I don't know. I'm just saying that the argument doesn't work.)

Well, it's easy to pick a half-n-sphere corresponding to a point on the surface. You just take the half-sphere centered at that point. For our two-dimensional circle, it would be the arc defined by the diameter that is perpendicular to the radius running between "the point" and the center of the circle.

At that point you've lost the ability to say that only one such half-sphere defined by a dropped point can be a valid solution, and you've also lost the ability to say that if a valid solution exists then there must be a valid solution defined by one of the points you want to include in the half-sphere, but you can define a canonical half-sphere for any point.

I was uncomfortable with the idea of picking "random points on a circle" to begin with, because of https://en.wikipedia.org/wiki/Bertrand_paradox_(probability) , but the article doesn't even address whether the concept is well-defined. We can always choose a point on the perimeter deterministically from any chord (...that isn't a diameter), so the ill-definedness of the problem of choosing a random chord seems like it would infect the problem of choosing a random point on the perimeter.