But it's really OK, because we're comparing something like an epsilon neighborhood of the North pole to an epsilon neighborhood of the equator. Even on a 2-sphere the equatorial neighborhood is much bigger. But to sketch a proof of the answer, I used the fact that the average distance can be rigorously calculated as D(n) = 2^n gamma((n+1)/2)^2 / (sqrt(pi) gamma(n+1/2)) and then Stirling's approximation gives lim D(n) = sqrt(2). n—>oo However, this is like the ultimate brute force proof! Is there a far more elegant way to prove this rigorously? —Dan From: Allan Wechsler ----- That is disturbing, isn't it? I still think what I said is true. If you pick any unit vector in N-space as a reference, my intuition is that the dot product of a second, randomly chosen unit vector with the reference vector will follow a distribution more and more tightly centered on 0 as N increases. If this is true, then as Dan points out, for a large enough N, even if you pick a bunch of k different reference vectors, the random vector will tend to be near 90 degrees away from all of them. It may be that N has to increase dramatically to accommodate small increases in k, though. I don't think there is a logical contradiction here. Or my intuition may be completely misguided. Trouble is, there are so many equators ... —Dan ----- ... almost all the hypervolume is near the equator. ----- -----