[matheist]

> 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.)

[thaumasiotes]

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.

[matheist]

Right, I was taking it as given that the problem of choosing a hemisphere canonically for a point meant "such that the argument works in the same way as for the circle".

Bertrand paradox just doesn't apply here, there's a natural measure on the circle and all higher dimensional spheres. I wouldn't expect an article on this subject to need to make that clarification unless it's dealing with chords or some other situation without a natural measure.

[thaumasiotes]

If I choose my four points as the endpoints of two chords chosen by the "random radial point" method described on the wikipedia page, is it still true that the odds of all four being covered by a semicircle are 50%?

[gowld]

Seems unlikely, because the density of chords is much higher near the center of the circle than the rim. (The density of chords is infinite at exactly the center). This combines with the fact the points are farther apart when their bisector is closer to the center, making it harder for all 4 of them to be on a half-circumference.

[matheist]

I have no idea, but I wouldn't expect so unless it was by coincidence? Not sure what chords have to do with any of this. There's a canonical way to choose 4 points uniformly and independently at random on the circle, and it's got nothing to do with chords.