\problemdata
{f01 ps11}
{random variable}

\begin{problem} The faculty holds a dinner party for $n \geq 2$ couples.  The
hostess chooses a seating arrangement for the $2n$ people around a
circular table uniformly at random. What is the expected number of
couples such that the husband and wife are sitting next to each other?

\solution{
Let $I_k$ be an indicator variable for the event that couple
$k$ is sitting together.  Then the number of couples sitting together
is $I_1 + I_2 + \ldots + I_n$.  The probability that a couple is
seated together is $2 / (2n-1)$, since a wife is equally likely to
occupy each of the $2n-1$ positions relative to her husband and there
are 2 positions right next to him.  Consequently, $\Pr(I_k = 1) = 2 /
(2n-1)$, and we can reason as follows:

\begin{eqnarray*}
\Ex(\text{couples sitting together})
	& = &	\Ex\left(\sum_{k=1}^n I_k\right) \\
	& = &	\sum_{k=1}^n \Ex(I_k) \\
	& = &	\sum_{k=1}^n \Pr(I_k = 1) \\
	& = &	\sum_{k=1}^n \frac{2}{2n-1} \\
	& = &	\frac{2n}{2n-1}
\end{eqnarray*}
%Thus, regardless of the size of the party, the expected number of
%couples seated together is always close to 1.
Thus, as $n$ increases, the expected number of couples seated together converges to 1.
}

\end{problem}