\problemdata      
        {faculty} % latex-friendly label for the prob.
        {Probability-Expectation} % The topic(s) of the problem content.
        {?} % Source (if known)
        {F01, PS11} % Usage (list ones you are aware of).
        {ARM from Christos Kapoutsis S02} % Last revision info (author, date).
\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 the $k$th couple is
sitting together.  Then the number of couples sitting together is $I_1 +
I_2 + \cdots + I_n$.  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*}
\expect{\text{couples sitting together}}
	& = &	\expect{\sum_{k=1}^n I_k} \\
	& = &	\sum_{k=1}^n \expect{I_k} \\
	& = &	\sum_{k=1}^n \pr{I_k = 1} \\
	& = &	\sum_{k=1}^n \frac{2}{2n-1} \\
	& = &	\frac{2n}{2n-1}.
\end{eqnarray*}
\iffalse
Thus, regardless of the size of the party, the expected number of
couples seated together is always close to 1.\fi
Thus, as $n$ increases, the expected number of couples seated together
converges to 1.
}
\end{problem}
