EXAMPLE 5.13

Five friends have purchased tickets to a certain concert. If the tickets are for seats 1-5 in a particular row and the tickets are randomly distributed among the five, what is the expected number of seats separating any particular two of the five? Let $X$ and $Y$ denote the seat numbers of the first and second individuals, respectively. Possible $\left(X,Y\right)$ pairs are $\{\left(1,2\right),\left(1,3\right),\dots ,\left(5,4\right)\}$ , and the joint pmf of $\left(X,Y\right)$ is

$$p\left(x,y\right)=\{\begin{array}{ll}\frac{1}{20}& x=1,\dots ,5;y=1,\dots ,5;x\ne y\\ 0& \text{ otherwise }\end{array}$$

The number of seats separating the two individuals is $h\left(X,Y\right)=\left|X-Y\right|-1$ . The accompanying table gives $h\left(x,y\right)$ for each possible $\left(x,y\right)$ pair.

$h(x,y)$	1	2	3	4	5
1	$-$	0	1	2	3
2	0	$-$	0	1	2
3	1	0	$-$	0	1
4	2	1	0	$-$	0
5	3	2	1	0	$-$

Thus

$$E\left[h\left(X,Y\right)\right]=\sum _{\left(x,y\right)}h\left(x,y\right)\cdot p\left(x,y\right)=\sum _{\begin{array}{c}x=1\\ x\ne y\end{array}}^5\sum _{\begin{array}{c}y=1\\ x\ne y\end{array}}^5\left(\left|x-y\right|-1\right)\cdot \frac{1}{20}=1$$