sorry re my not-very-good coherence before re the upper bounds. Now let me try to work out the idea, which I explained badly and with two errors before, for exact results in 2 and 3 dimensions. Specifically in D=3 dimensions. [My errors: a wrong factor in integral, and the wrong claim at most one face of the convex hull of N points on sphere, could separate the polyhedron from 0.] There are N points random on unit sphere. They define a convex hull polyhedron with 2N-4 faces, each a triangle. WLOG let the face F with largest circumcircle be parallel to xy plane and facing outward in the negative-z direction. All N points lie in one hemisphere if and only if F lies above the xy plane, at altitude z>0, and all N-3 other points lie above it. Using Archimedes's uniform-z "cylinder-sphere" surface-area lemma with great simplifying effect, we see Prob(empty hemiball exists) = 2* integral(0<z<1) (2-z)^(N-3) * dz / integral(|z|<1) (2-z)^(N-3) * dz = 2*(9/4)*(2^N-4)/(3^N-9). The factor 2 is because of freedom to turn everything upside down. This formula is completely unchecked via Monte carlo. Unless I made another mistake, which is reasonably likely since this is tricky, it ought to be valid if N>=3. It gives the right answer 1 when N=3. When N=4 it gives 3/4. Now let us try to derive the formula in D=2 dimensions. The longest side S of the convex hull N-gon of the N points on unit circle WLOG is parallel to x-axis and at altitude y. All N points lie in one hemicircle if and only if S lies above the xy plane, at altitude z>0, and all N-2 other points lie above it. We see Prob(empty hemiball exists) = 2* integral(0<t<pi/2) (t)^(N-2) * dt / integral(0<t<pi) (y)^(N-2) * dt = 2^(2-N) if N>=2 which disagrees with Salamin's claim the formula is N*2^(1-N). Crap. Off by a factor of N/2. I can re-derive Salamin's formula. The probability that N points on circle, one of them red, rest blue, has property all blue points lie on on the left side of the diameter with one end at the red point, is 2^(1-N). The uncolored version involves a factor of N versus the colored version. QED Hence, my approach with these integrals is wrong. I presume the reason for the wrongness is that the "largest circumball face" is not distributed a priori in the same way as "a D-tuple of points on the sphere." You could try to fix my D=3 formula by saying there ought to be a fudge factor of 2N-4 for the "one red face" out of 2N-4 total faces, but that won't work because in the 3D case it is possible for TWO faces to separate from 0, unlike the D=2 case where only one side of the N-gon can separate. So it looks like this approach is dead for D=3, sorry. ---------------Other remarks: First of all, I don't see this claim by Miller the "N points on sphere, empty circle" problem is "not exactly the same" as the "N points in ball, empty hemisphere problem." It IS exactly the same. Proof: Mapping points in ball radially up to surface leaves emptiness of hemiballs unaltered. The mathpages guy Miller cited, assuming he did what he did right, using dodecahedral finite hemisphere set, should get a valid lower bound on Pempty(N). http://www.mathpages.com/home/kmath327/kmath327.htm Second, I have no idea what Veit E. was talking about with "N axes." Huh? If what he was trying to say was "a linear transformation of D-space preserves the fact points are on opposite sides of a hyperplane" then this is true but seems useless because the probability measure on hyperplanes and points in a ball is distorted by linear transformations... you'd have to use orthogonal transforms only (hence orthogonal axes only) in which case I fail to see what was being accomplished. Third, the inclusion exclusion approach can be made to work in any dimension using an infinite set of (all) hemispheres (actually cardinality-H set of random ones, then take limit H-->infinity if desired)... but it gets nasty. The main term is 2^(-N) is the chance a given hemiball is empty. The 2nd term involves the expected volume of the intersection of 2 random hemiballs, which still is pretty easy to calculate. The Kth term involves expected volume of intersection of K random hemiballs. When K>3 things are going to be ultra-nasty, and the approximation got by only using the first 2 terms, say, seems pretty useless.