You can get upper and lower bounds on the answer. The problem is: N gas atoms in an 0-centers D-dimensional ball. (D>0.) What is chance there exists an empty hemi-ball? Lower bound: The left hemiball (x<0) is empty with chance 2^(-N). Better lower bound: Pempty >= 2^(1-N) by considering both the x<0 and x>0 hemiballs. Still better lower bound: Pempty >= N * 2^(1-N) by considering all hemiballs arising from semicircles in xy plane (assuming Salamin correct about the exact solution when D=2). And when D=4 this would give the lower bound (2*N)^2 * 2^(-N) by considering xy and zt planes. And when D=3 it would give (2*N+2)*2^(-N) by considering xy plane and the z<0 and z>0 hemiballs. Upper bound: If all 2^D quadrants are occupied, then there is no empty hemiball, because any hyperplane passing thru ball-center must contain at least one entire quadrant on one side. Better upper bound: divide D-space into D+1 equal regions by "coning off" the D+1 faces of a regular D-simplex centered at 0, to the origin. If every region is occupied, then there can be no empty hemiball. (Because any hyperplane passing thru ball-center must contain at least one entire region on one side.) The chance that all D+1 regions are occupied is lower bounded by D^N / (D+1)^N <= 1-Pempty. On-sphere lemma: (As somebody already noted) If all N points lie on the unit sphere, rather than in it, that makes no difference (Pempty unaffected). Special argument when D=2: The question is equivalent to this: is the longest side of the N-gon formed by N random points on the unit circle, on the "wrong" side of 0? (I.e. 0 lies on opposite side from all the other points.) There are N sides. Special argument when D=3: The question is equivalent to this: is the largest-circumradius-triangular-face of the polyhedron that is the convex hull of N random points on the unit sphere, on the "wrong" side of 0? (I.e. 0 lies on opposite side from all the other points.) There are 2*N-4 such faces (generically triangular, i.e. the probability a nontriangular face exists, equals 0). It seems to me that in both the D=2 and D=3 cases, the answer can be written in closed form as a definite integral using (a) the two special observations above, (b) the fact that it is impossible for more than 1 "special face" to exist, it automatically is unique one. So I think when D=2 you are going to get Pempty = N * int(0<u<1) (u/2)^(N-2) * du where u=theta/pi, if you see what I mean. I admit, this is off by a factor N-1 versus Salamin's claimed answer, so I'm probably missing something. Perhaps I was suppose to divide by int(0<u<2) (u/2)^(N-2) * du to normalize probability distribution or something. I think that yields Salamin's claimed answer. When D=3 you similarly (presumably also wrongly + missing something) would get Pempty = C * (2*N+4) * int(0<u<1) (u/2)^(N-3) * (1-(1-u)^2)^(3/2) * du using Archimedes in the direction orthogonal to the face, for some positive constant C I have not bothered to work out. If the same "normalize by dividing" fix worked then Pempty = (2*N+4) * int(0<u<1) (u/2)^(N-3) * (1-(1-u)^2)^(3/2) * du / int(0<u<2) (u/2)^(N-3) * (1-(1-u)^2)^(3/2) * du. So anyhow, I have not worked out the full details, but if this approach works you should be able using my hints, to get exact 2D and 3D answers; plus I have shown how to find upper & lower bounds in any dimension (which are undoubtably improvable).