Yup, there's an exact solution. It turns out that the volume of the n-dimensional unit ball is pi^(n/2) / (n/2)!. (The factorial function generalizes to numbers

Yup, there's an exact solution. It turns out that the volume of the n-dimensional unit ball is pi^(n/2) / (n/2)!. (The factorial function generalizes to numbers that aren't integers. For half-integers you get a rational number times sqrt(pi). E.g., if n=3 then we need (3/2)! which happens to equal 3/4.sqrt(pi), so the volume is 4/3.pi.) So for large n we can use Stirling's approximation. To make the formulas a little shorter I'll write m=n/2. Then the volume of the unit n-ball is V(n) = pi^m/m! ~ pi^m (e/m)^m 1/sqrt(2.pi.m) whereas the volume of the smallest n-cube containing that ball is 2^n, so the ratio is R(n) ~ (pi.e/2m)^m / sqrt(2.pi.m) where, remember, m=n/2. Actually, at this point maybe it's nicer to put things back in terms of n: R(n) ~ sqrt[(pi.e/n)^n/pi.n] When n is large, you should see this as R(n) ~ sqrt[(..../n)^n.....] where the "...."s are the less important bits. So for large n this is going to get small. Concrete not-so-large-n example: let's take n=10. Then R(10) ~= sqrt((pi.e/10)^10/10pi) which happens to be about 0.081. That's pretty accurate -- the exact answer is pi^5/3840, which is about 0.0797. So if you fit a 10-ball snugly inside a 10-cube it'll take up about 1/12 of the space. What if we crank the dimension up? With n=100 the ratio is about 2x10^-55. (A naive Monte Carlo simulation will be unable to distinguish this from zero.) So if you do the ball-inside-cube thing in 100 dimensions, the ball occupies less than one billion-billion-billion-billion-billion-billionth of the volume of the cube. High-dimensional cubes are really, really spiky.