The spread of five uniforms

Idea. Compute the probability that the range is at most $r$: all numbers lie in a window of length $r$ starting at the minimum. Condition on which draw is the minimum and where it is.

Step 1: the range is at most $r$. Each of the 5 draws is the minimum with the same probability. If the minimum is $m$, the other four must lie in $[m,\min (m+r,1)]$. So $$\begin{array}{rl}\mathbb{P}(R\le r)& =5{\int }_0^{1-r}{r}^{4}dm\\ & +5{\int }_{1-r}^1(1-m{)}^{4}dm\\ & =5r^4(1-r)+r^5=5r^4-4r^5.\end{array}$$

Step 2: $r=\frac{1}{2}$. $$\mathbb{P}\left(R\le \frac{1}{2}\right)=\frac{5}{16}-\frac{4}{32}=\frac{3}{16}.$$

Step 3. $$\mathbb{P}\left(R>\frac{1}{2}\right)=1-\frac{3}{16}=\frac{13}{16}.$$

In general. For $n$ uniforms, $\mathbb{P}(R\le r)=n{r}^{n-1}-(n-1)r^n$, and the expected range is $\frac{n-1}{n+1}$, here $\frac{2}{3}$.