0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\frac n4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.