note: edited to clarify boundary issue
Suppose $x_i$, $i=1\dots n$, are randomly drawn from a uniform circular distribution between 0 and 1 (using periodic boundaries). Let $d_i$ be the distance between $x_i$ and its nearest neighbour to the right (allowing to wrap at the boundary). Now, the mean for $d_i$ is simply $1/n$. But what is the expectation value for $d_{\min}=\min_i\{d_i\}$? (I reckon it must scale as $E\{d_{\min}\}\propto n^{-2}$ -- correct?)
update The distances between uniformly distributed variables are exponentially distributed with rate $n$, i.e. mean $1/n$. Moreover, the minimum of $k$ independent exponential variables of rate $\lambda_i$ is also exponentially distributed (see Wikipedia, for example) with rate $\lambda_1+\dots+\lambda_k$. So, if the distances were independent, the answer is $E\{d_{\min}\}=1/n^2$. But they aren't independent.