Expected distance between tree nodes
Let $d(i,j)$ denote the distance between $i$ and $j$. Calculation shows that
$$
\sum_{i&lt;j} d(i,j) = 87.
$$
Hence the average distance is
$$
\frac{1}{6^2} \sum_{i,j=1}^6 d(i,j) = \frac{1}{36} \left(\sum_{i&lt;j} d(i,j) + \sum_{i&gt;j} d(i,j) + \sum_{i=j} d(i,j)\right) = \frac{2\cdot 87}{36} = \frac{29}{6}.
$$

A simple way to do the calculation is as follows. Suppose that there are $n$ nodes, and that edge $e$ of weight $w_e$ cuts the tree into parts of size $k_e,n-k_e$. Then the average distance is
$$
\frac{2}{n^2} \sum_e w_e k_e (n-k_e).
$$
For example, in our case we get
$$
\frac{2}{36} (1 \cdot 1 \cdot 5 + 2 \cdot 1 \cdot 5 + 3 \cdot 3 \cdot 3 + 4 \cdot 1 \cdot 5 + 5 \cdot 1 \cdot 5) = \frac{29}{6}.
$$