TITLE: How to solve probability circular table problem with neighbors.
QUESTION [5 upvotes]: 7 people are seated around a round table eating dinner. They then all get
up, get dessert and then sit down at random. What is the probability that
each person has two new neighbors? What is the answer to this question
if there are 8 people? How about for 9 people?

REPLY [2 votes]: Consider the graph $K_5$, there is a bijection between the valid arrangements and the hamiltonian cycles contained by $K_5$ after substracting the edges of a hamiltonian cycle of $K_5$. Let $Q_n$ be that number.
Thanks to the help of mathoverflow (Brendan McKay) I now know $Q_n$ is given by this sequence. On the other hand there are clearly $(n-1)!$ cycles in $K_n$ total, thus what we want is $\frac{Q_n}{(n-1)!}$.Since $Q_7=23$ the probability is $\frac{23}{6!}\approx3.2\%$