We are given a large, round table with $n$ seats. It is easy to see that whenever $p\geq \text{int}(\frac{n}{2}) + 1$ people are seated, at least $2$ people will sit next to each other (here $\text{int}(x)$ denotes the largest integer less or equal than $x$).


Given $n$ seats at a round table, let $s(n)$ be the smallest number so that the probability that at least $2$ people sit next to each other is $\geq 0.5$ when $s(n)$ people choose their places randomly. (That is: first, person number $1$ picks a seat with uniform probability, then number $2$ takes one of the remaining seats with uniform probability, etc.)


Does $\lim\_{n\to \infty}\frac{s(n)}{n}$ exist, and if yes, what is its value?


(Transferred from <https://stats.stackexchange.com/questions/162995/minimum-number-of-people-such-that-2-can-be-expected-to-sit-next-to-each-other> as I didn't get an answer there.)



Minimum number of people such that 2 can be expected to sit next to each other
Some partial answer. Once you fix the first guy then you can see if $k$ persons are seated in the table with distance greater and $1$ then this give by counting the spaces between each one this gives you a partition of $n-k$ in $k$ parts. So the cases that you don't want to count are in correspondence with the ways of summing $n-k$ with $k$ positive integers. This are $n-k-1\choose k-1$, and in total there are $n-1\choose k-1$ cases. At the end you are asking when does $2{n-k-1\choose{ k-1} }< {n-1\choose k-1}$.


