March 12th, 2015, 3:15 pm
Yes, I think Vanubis is on to the right idea. For the case of N=D+1 points [$]\vec{p_i}[$] on a hypersphere of dimension D, the convex hull of the points is the set [$]H(p) = \{\sum_{i=0}^{N-1} w_i \vec{p_i} | \sum_i w_i = 1, w_i \geq 0\}[$].We want to know if the origin is in [$]H(p)[$]. To do this, we just drop the condition [$]w_i \geq 0[$] and solve the [$]N[$] linear equations for the [$]w_i[$] that make [$]\sum_{i=0}^{N-1} w_i \vec{p_i} = 0[$] with [$]\sum_i w_i = 1[$]. This is [$]N[$] linear equations with [$]N[$] unknowns, so there is a unique solution (degenerate cases are measure zero since the points are random on the sphere). The origin is in [$]H(p)[$] if and only if all the [$]w_i[$] in the solution are positive. For any generic set of points [$]p_i[$], we can consider the [$]2^N-1[$] other sets where we have flipped the sign on some subset of the points (i.e. reflected through the origin). For any point [$]p_i[$] with negative coefficient [$]w_i[$], we can just use the flipped point [$]-p_i[$] and make [$]w_i[$] positive. That automatically gives one out of [$]2^N[$] combinations with the origin in [$]H(p)[$]. There is a second combination where we flip the signs on ALL the points - that will also have the origin in [$]H(p)[$]. The other combinations of flips don't work, so the final answer is:P = probability of containing the origin = [$]1/2^{N-1} = 1 / 2^D[$]