Expected number of people seated between A and B $N (\ge 3)$ people sit at a round table. Every seating is equiprobable. Among those people are $3$ called $A$, $B$ and $C$. Let $X$ be a number of people siting between $A$ and $B$ (on the arc which does not include $C$). What is the expected value and variance of $X$? I've tried some combinatorics, but I can't seem to stop overcounting.
 A: Let's create the sequence, that starts with C:
$$CXXXXXXXXXXXX$$
Amongs these Xs there are two guests A and B. Lets mark them with Gs:
$$CXXXXXGXXXGXX$$
Now, as we have $N-3$ Xs, we have:


*

*$1$ way of placing Gs with $N-3$ Xs between

*$2$ ways of placing Gs with $N-4$ Xs between


...


*

*$N-4$ ways of placing Gs with $2$ Xs between

*$N-3$ ways of placing Gs with $1$ X between

*$N-2$ ways of placing Gs with no X between 


Therefore the expected number of Xs between Gs is:
$$E=\frac{\sum\limits_{k=1}^{N-2} k (N-2-k)}{\sum\limits_{k=1}^{N-2}k} =\frac{\frac{1}{6}(N-3)(N-2)(N-1)}{\frac{1}{2}(N-1)(N-2)}= \frac{N-3}{3}$$
A: Let $Y$ be the number of people siting between $A$ and $C$ (on the arc which does not include $B$). Let $Z$ be the number of people siting between $B$ and $C$ (on the arc which does not include $A$). By symmetry, we have:
$$E(X)=E(Y)=E(Z)=\frac{N-3}{3}$$
A: To get both the expected value and the variance, you need to use indicator random variables. I will get you started. 
Number the people who are neither $A,B$ nor $C$ from $1$ to $N-3$. For each $i\in \{1,\dots,N-3\}$, let $X_i$ be a random variable which is $1$ if person $i$ is in the arc between $A$ and $B$ not containing $C$, and $0$ otherwise. Then letting $X=\sum_{i=1}^{N-3}X_i$, it is clear that $X$ is the random variable you want. Therefore,
$$
E[X]=\sum_{i=1}^{N-3}E[X_i]=(N-3)\cdot P(X_1).
$$
Finally, it is obvious that $P(X_1)=\frac13$, since person number $i$ is equally likely to be in each of the three arcs that $A,B,C$ split the circle into.
Variance is a little harder, but not bad. We have
$$
\text{Var}(X)=\text{Cov}(X,X)=\sum_{i=1}^{N-3}\sum_{j=1}^{N-3}\text{Cov}(X_i,X_j)
$$
There are then two cases:


*

*If $i=j$, then $\text{Cov}(X_i,X_j)=E[X_i^2]-E[X_i]^2=P(X_i=1)-P(X_i=1)^2=\frac13-\frac19=\frac29.$

*If $i\neq j$, then $$\text{Cov}(X_i,X_j)=E[X_iX_j]-E[X_i]E[X_j]=P(X_i=1\cap X_j=1)-P(X_i=1)^2$$
All you have to do now is figure out what $P(X_i\cap X_j)$ is. This is the probability that two distinct people, call them $D$ and $E$, are both in the arc between $A$ and $B$ which excludes $C$. To do this, it suffices to consider the circular order of these five people only. 
