# What fraction of n-point sets in the unit ball have diameter smaller than 1?

This question is inspired by a recent talk by Matt Kahle on random geometric complexes.

Some simple notation: let $\mathcal{B} \subset \mathbb{R}^d$ be the unit ball in $d$-dimensional Euclidean space with the usual norm $\|\cdot\|$ and let $\mathcal{B}^k$ denote the $k$-fold product of this ball with itself for any positive integer $k$. For each $n \in \mathbb{N}$, define $K_n \subset \mathcal{B}^n$ as follows:

$$K_n = \lbrace(p_1,\ldots,p_n) \in \mathcal{B}^n \text{ so that } \|p_i - p_j\| < 1 \text{ for all } 1 \leq i,j \leq n\rbrace.$$

Thus, $K_n$ consists of those $n$-tuples of points in the unit ball whose diameter is bounded above by $1$. I would like to know what fraction of $\mathcal{B}^n$ lies in $K_n$ asymptotically as $n$ increases. More precisely, let $\mu$ be the usual $d$-dimensional Lebesgue measure, and define $$\chi(n) = \frac{\mu(K_n)}{\mu(\mathcal{B})^n}.$$

How does $\chi(n)$ behave for fixed dimension $d$ as $n \to \infty$?

While an answer to this specific question would be great, I would be much more interested in some insight regarding how one should tackle such problems. The Calculus approach of setting up some integral fails horribly -- at least, I should confess that I tried $n = 2 = d$ and got hopelessly stuck with what appears to be an elliptic integral of the murderous kind -- so I expect that there is no closed formula for $\chi(n)$. But surely someone has worked out asymptotic envelopes for such a basic quantity!

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@Vidit: Thanks for accepting my answer. I would also like to direct everyone's attention to Will Sawin's answer below which gives very sharp asymptotic bounds for $\chi(n)$. – Ricardo Andrade Jun 19 '13 at 11:52
