Expected number of person not to get shot - reloaded

Question

Inspired from this question I came up with a seemingly simpler problem that I could not solve either.

There are $n$ people sit on a round table. At noon, each person shots and kills one of their neighbours, chosen randomly (so, everyone shoots their left neighbour with probability $1/2$ and their right neighbour with probability $1/2$). Which is the expected number of people surviving?

It's easy to compute the probability that nobody survives: if $n$ is odd it is $1/2^{n-1}$ (the only possible cases are that everyone shoots their right neighbour or that everyone shoots their left neighbour) and if $n$ is even it is $1/2^{n-2}$ (there are also the cases of reciprocal killings $(1,2), (3,4), \dots, (n-1,n)$ and $(2,3), (4,5), \dots, (n,1)$.) But then I am stuck.

Answer 1

Every person has probability $\frac14$ of surviving, since they can be shot by either their left or their right neighbour, each independently with probability $\frac12$. Thus, by linearity of expectation, the expected number of survivors is $\frac n4$.