Average minimum distance between $n$ points generate i.i.d. with uniform dist. Let $U$ be distributed uniformly on $[0,a]$. Now suppose we generate $n$ independent points according to $U$. 
What is the average minimum distance between these $n$ points? That is 
\begin{align}
E\left[ \min_{i,j\in [1,n]: i\neq j}  |U_i-U_j|\right]
\end{align}
Is this a correct formulation of average minimum distance between?
 A: Assume $a=1$ for simplicity. Let $M$ be the minimum distance among these $n$ points. Note $M$ is always at most $\frac1{n-1}$, which occurs when the points are evenly spaced. To calculate $EM$, we first calculate $P(M>m)$, for $0\le m \le 1/(n-1)$.
By symmetry, $P(M>m)=n!\times P(M>m\text{ and } U_1<U_2<\dots<U_n).$ And the latter is equal to the volume of the below subset $S$ of the unit hypercube $Q=[0,1]^n$:
$$
S = \{(x_1,\dots,x_n)\in Q:x_i+m\le x_{i+1}\text{ for }i=1,\dots,n-1\}
$$
Now, consider the transformation $f:S\to [0,1]^n$, given by
$$
(x_1,\dots,x_n)\mapsto (x_1,x_2-m,x_3-2m,\dots,x_n-(n-1)m)
$$
A little thought shows that $f$ is a volume-preserving bijection from $S$ to the region $S'$ of the smaller hypercube $Q'=[0,1-(n-1)m]^n$ given by
$$
S'=\{(y_1,\dots,y_n)\in Q':y_i\le y_{i+1}\text{ for }i=1,\dots,n-1\}
$$
By symmetry of permuting the $y_i$, we have $\text{Vol}(S')=1/n! \ \times\text{Vol}(Q')$. But $\text{Vol}(Q')=(1-(n-1)m)^n$. Putting this all together,
$$
\begin{align}
P(M>m)&=n!\cdot \text{Vol}(S)\\
&=n!\cdot \frac1{n!}(1-(n-1)m)^n\\
&=(1-(n-1)m)^n
\end{align}
$$
and therefore
$$
\begin{align}
EM
&=\int_0^{1/(n-1)}P(M>m)\,dm\\
&=\int_0^{1/(n-1)}(1-(n-1)m)^n\,dm\\
&=\left.\frac{1}{n-1}\frac{(1-(n-1)m)^{n+1}}{n+1}\right|_{0}^{1/(n-1)}\\
&=\boxed{\frac{1}{n^2-1}}
\end{align}
$$
To get the answer for the interval $[0,a]$, simply multiply this result by $a$.
