We call an arrangement each way of seating $n$ couples at a round table with $2n$ seats. We say that a couple is together if its two members are sitting next to each other. What is the average number of couples together per arrangement.
We count the number of couples together in all arrangements. For all $i$, we denote $N_{i}$ as the number of times where the couple $i$ is together. By symmetry, for all $i, N_{i}=N_{1}$.

The number of arrangements where couple 1 is together is $4 n \cdot(2 n-2)!, 4 n$ choices for the places of couple 1 and $(2 n-2)!$ for seating the other couples. Finally, the number of couples together in all cumulative arrangements is $N_{1}+\cdots+N_{n}$, which is $n N_{1}$. The answer is therefore $\frac{4 n(2 n-2)!}{(2 n)!}=\frac{2 n}{2 n-1}$.