Expected weight of euclidean minimum spanning tree on a unit square
The following is based on J. Michael Steele's paper proving the rigorous bound.

The pigeonhole principle shows that out of any $n$ points in $[0,1]^d$, there are two at distance at most $O(n^{1/d})$, say at most $c_d n^{1/d}$; the idea is to divide $[0,1]^d$ into small enough boxes, and find a box containing two points. Construct a spanning tree by repeatedly adding an edge corresponding to the two closest points and throwing out one of them. This spanning tree has cost at most
$$
c_d (n^{1/d} + (n-1)^{1/d} + \cdots + 2^{1/d}) \leq c_d \int_2^{n+1} x^{1/d} \, dx = O(n^{(d-1)/d}).
$$

For the lower bound, a similar argument dividing $[0,1]^d$ into small boxes shows that for any $\epsilon &gt; 0$, with probability $1-\epsilon$ there are only $\epsilon n$ many edges whose weight is $O_\epsilon(n^{1/d})$, and so with constant probability any spanning tree has cost $\Omega_\epsilon(n^{(d-1)/d})$. With more work, it follows that the expected weight of a minimum spanning tree is $\Omega(n^{(d-1)/d})$.