This is a problem from one of the former exams from Probability calculus 1.
There are n people sitting by the table of the form of a regular polygon with n sides (each person by one side). Each person is flipping a coin with a probability 1/2 to eagle and 1/2 to tails. A person is satisfied only when he got an eagle and both of his neighbors did. Calculate the variance of X, which is the number of satisfied people sitting by the table. This is quite difficult for me since I cant figure out how to translate this exercise to probability language. I wanted to build up a variable Xi which is 1 when a person got an eagle and 0 when not. We are interested in the sum of series of Xi which are longer than 3, have only eagles and end up with tails, and we want to calculate their sum minus the number of series. Can someone help, since I havent got really far...
Hint:
Use the setup that you proposed and apply: $$\mathsf{Var}(X_1+\cdots+X_n)=\sum_{i=1}^n\sum_{j=1}^n\mathsf{Cov}(X_i,X_j)=\sum_{i=1}^n\sum_{j=1}^n[\mathsf{E}X_iX_j-\mathsf EX_i\mathsf EX_j]=$$$$\sum_{i=1}^n\sum_{j=1}^n[\mathsf P(X_i=X_j=1)-\mathsf P(X_i=1)\mathsf P(X_j=1)]=$$$$\sum_{i=1}^n\sum_{j=1}^n[\mathsf P(X_i=X_j=1)-\frac18\frac18]=$$$$\sum_{i=1}^n\sum_{j=1}^n\mathsf P(X_i=X_j=1)-\frac{n^2}{64}$$
To find $\mathsf P(X_i=X_j=1)$ discern several cases. Here e.g. $3$ has $2$ and $4$ as neighbors, and $1$ has $n$ and $2$ as neighbors.
It might be handsome to solve it for small $n$ like $n=3,4,5,6$ first.