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?
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 $1$.