The contraction algorithm for min-cut is: pick an edge $(u,v)$ uniformly at random, and "contract" it by merging $u$ and $v$ into a single vertex, deleting self-loops. Continue until two vertices remain; return this cut.
The probability of finding the min cut can be shown to be at least
$$ \frac{1}{{n \choose 2}} . $$
This feels like it should have some clever combinatorial interpretation or proof. (Something that sounds like e.g. "we place the vertices in a uniformly random order and if the last pair is correct, the algorithm succeeds.") But every proof I have found involves brute-forcing the product $$ \frac{n-2}{n} \frac{n-3}{n-1} \cdots \frac{1}{3} = \frac{2}{n(n-1)} . $$
Is there any direct combinatorial argument that the probability exceeds ${n \choose 2}^{-1}$?
