expected distance in 2D

Question

Suppose there are two jumping children. One randomly chooses positions ($x_1,y_1$) and ($x_2,y_2$). The other randomly chooses the positions ($x_3, y_3$) and ($x_4, y_4$). What is the average distance between the two children (exact solution)? Kindly ask for a hint.

Answer 1

Each of the four pair of points has the same probability $\frac14$ to occur, so the expected distance is

$$ \frac14\left(d_{13}+d_{14}+d_{23}+d_{24}\right)\;, $$

where $d_{jk}=\sqrt{(x_j-x_k)^2+(y_j-y_k)^2}$.