TITLE: Why is volume of a high-D ball concentrated near its surface?
QUESTION [6 upvotes]: I came across the following sentence while reading a book on applied math:

Volume of a high dimensional unit ball is concentrated near its
  surface and is also concentrated at its equator.

This is from book's introduction, and I believe the sentence will be explained at some later point in the book. However, it is hard to me to accept it on intuition level. Can some of you explain this sentence to me, in a sort of laymen-style, if possible?

REPLY [6 votes]: The book probably means that the volume of the unit ball in $\mathbb{R}^n$ has an unexpected behaviour as a function of $n$: it is given by
$$ V_n = \frac{\pi^{n/2}}{\Gamma\left(1+\frac{n}{2}\right)} $$
and the surface area is given by
$$ A_n = n V_n = \frac{n\pi^{n/2}}{\Gamma\left(1+\frac{n}{2}\right)}. $$
In particular, $\frac{V_n}{V_{n-1}}$ (that is a measure of "concentration around the equator") behaves like $\sqrt{\frac{\pi}{n}}$ and $\frac{V_n}{A_n}$ (that is a measure of "concentration around the boundary") is exactly $\frac{1}{n}$. Additionally, by the central limit theorem the behaviour of
$$ A(\tau)=\mu\{(x_1,\ldots,x_n)\in\mathbb{R}^n: x_1+\ldots+x_n = \tau, x_1^2+\ldots+x_n^2\leq 1\} $$
is approximately gaussian and more and more concentrated around $\tau=0$ as $n\to +\infty$.