Probability/Combinatorics question: boys and girls around a table

Question

There are 9 boys and 7 girls sitting around a table. If a boy sits next to a girl we call it a switch. Example: In BBBBBGGGBBGBGGG there are 6 switches.(If a boy sits next to 2 girls, there are 2 switches counted.) What is the expected number of switches?

Answer 1

Give the boys a number. For boy $i$ let $G_i$ denote the number of girls that sit next to him.

Then you are searching for: $$\mathbb E(G_1+\cdots+G_9)=\mathbb EG_1+\cdots+\mathbb EG_9=9\mathbb EG_1$$

Let Alex (the boy having number $1$) take place.

The probability that Alex will get $2$ boys as neighbor is $\frac{8}{15}\frac{7}{14}=\frac{4}{15}$.

The probability that Alex will get $2$ girls as neighbor is $\frac{7}{15}\frac{6}{14}=\frac{3}{15}$.

The probability that Alex will get a girl and a boy as neighbor is $2\frac{8}{15}\frac{7}{14}=\frac{8}{15}$.

So $\mathbb EG_1=0\times\frac{4}{15}+2\times\frac{3}{15}+1\times\frac{8}{15}=\frac{14}{15}$

We come to an expectation of $9\times\frac{14}{15}=\frac{42}{5}$ switches.

Answer 2

Let $X_i\in\{0\,,1\,,2\}$ be the count of switches next to boy $i$.  

Then the expected number of switches is : $\;\mathsf E\left(\sum\limits_{i=1}^9{X_i}\right)$