probability and expectation IMO book question I was trying to solve this problem but didn't understand the solution when I saw it.

Problem: There are $8$ girls and $7$ boys  in a  social party, sitting around a round table. If all  the  girls  sit  together,  there  are  then  only  two  girls adjacent to boys. If girls and boys sit as alternately as possible,  then there are $14$ pairs of seats that are girl and boy  adjacent.  How many pairs of seats are there in average that are girl and boy adjacent

Comments:
My Issue is that when I looked at the solution I did not understand why is it that they took the probability of $1$ pair and multiplied by $15$ (the total no. of seats). I'm not convinced that the event of having a pair at one seat is independent of having a pair at another seat since the amount of remaining boys/girls differ.
Can someone please help me understand what's wrong with my reasoning and why is the probability of seat $i,j$ having a pair independent of seat $j,j+1$ having a pair?
 A: Let $A$ be the cyclic abelian group $\Bbb Z/15$ with $15$ elements. Consider the space $\Omega$ of all $\omega:A\to\{0,1\}\subset \Bbb R$, so that $\sum \omega=8$. Here we identify $\omega$ with a tuple $(\omega_0,\omega_1,\dots,\omega_{14})$ and $\sum\omega$ is the sum of the components of $\omega$. We define the random variables $X_a$ for $a\in A$ defined by $X_a(\omega)=\omega_a$.
(We consider a girl to correspoin to an $1$ entry in $\omega$, a boy to a $0$ entry, and use the cyclic order of the indices to let them sit in the same order cyclically around the round table.)
The function giving the number of adjacent pairs $01$ and/or $10$ is the random variable $Z$...
$$
\begin{aligned}
Z(\omega)&=\sum_{a\in A}|\omega_a-\omega_{a+1}|\ ,\text{ so}\\
Z&=\sum_a|X_a-X_{a+1}|\ .
\\
&\qquad\text{ Then:}
\\
\Bbb E Z
&=\Bbb E\sum_a|X_a-X_{a+1}|\\
&=\sum_a\Bbb E|X_a-X_{a+1}|\\
&=|A|\cdot\Bbb E|X_0-X_1|\ ,
\end{aligned}
$$
the last line using the cyclic symmetry on $\Omega$ induced by the action of $A$.
This argument "desaggregates" the information, and lets us look only at the seats labelled $0$ and $1$.
