Sample $2n$ points $X_1,\ldots,X_{2n}$ from the uniform distribution on $[0,1]$ and order them, to get $0\le X_{(1)} \le \ldots \le X_{(2n)} \le 1$. Turn this collection of $2n$ points into $n$ possibly overlapping subintervals of $[0,1]$ by repeatedly sampling (uniformly without replacement) two points to form an interval.
For example, for $n=2$ there are $3$ ways to form such subintervals that occur with equal probability: $[X_{(1)},X_{(2)}],[X_{(3)},X_{(4)}]$ and $[X_{(1)},X_{(3)}],[X_{(2)},X_{(4)}]$ and $[X_{(1)},X_{(4)}],[X_{(2)},X_{(3)}]$.
If $L_i$ denotes the length of the $i$-th longest interval, can we say something about the distribution of $L_i$? More generally, can we say something about the joint distribution $(L_1,\ldots, L_n)$?
I found this question in an old Russian book preparing for math competitions but I'm completely stuck.
The ordering of the points is just a distraction, since you then uniformly randomly sample from them, ignoring the order. What you end up with is simply $n$ independently sampled intervals, each generated by uniformly sampling two points from $[0,1]$. The length $L$ occurs in a range of positions of length $1-L$ and thus has density proportional to $1-L$; then normalization yields the density $f(L)=2(1-L)$, and thus the cumulative distribution function $F(L)=2L-L^2$. Then you can apply the standard treatment of order statistics to find the distributions you're looking for.