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".
replyBertrand 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.
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%?
replyI 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.
reply