\problemdata
{expected value}
{Fall 1999}
{Eric Ho, S02}
\begin{problem} Eight men and seven women, all single, happen randomly to have
purchased single seats in the same 15-seat row of a theater.

\begin{problemparts}

\problempart
What is the probability that the first two seats contain a marriageable
couple, i.e., a single man next to a single woman?

\solution{Here's one way to do it.  There are $\binom{15}{2} = 105$
possible pairs that could occupy the first two seats, but only
$8\cdot7 = 56$ of them are marriageable couples.  So the probability
is $56/105 = 8/15$.
}

\problempart What is the expected number of pairs of adjacent seats which
contain marriageable couples?  (For example, if the sequence of men and
women in the seats is
\begin{center}
MMMWMMMMWWWWWWM,
\end{center}
then there are four possible couples sitting in the pairs of adjacent
seats that start at the 3rd, 4th, 7th and 14th seats, respectively.)

\solution{The approach that we are going to use is to assign an
indicator random variable (i.e. value 0 or 1) to each event that we
want to count, and then use the expectation of sum formula to compute
the expected number of events that occur.

Let the random variable S be the number of pairs of seats
containing a marriageble couple; we want to find $\expect{S}$.  We can
solve this problem by introducing an indicator random variable $S_i$,
where $S_i = 1$ if there is a marriageable couple in seats $i$ and
$i+1$, and $S_i = 0$ otherwise.  So we have $S = \sum_{i=1}^{14} S_i$.

Since each seat is like any other, the probability that seats $i$ and
$i+1$ contain a marriageable couple is the same as in the first
part, $8/15$.  Thus
\[
\expect{S_i} = 1\cdot\Pr(S_i=1) + 0\cdot\Pr(S_i=0) = \frac{8}{15}.
\]
Finally, because expectation is linear,
\[
\expect{S} =  \sum_{i=1}^{14} \expect{S_i}  =  14\cdot \frac{8}{15}.
\]
Note that this approach works despite the fact that the indicator random
variables $S_i$ are not independent.  That's the beauty of linearity of
expectation!

}

\iffalse
\problempart Use a similar approach to calculate the expected number
of ``triples'' of people having the same birthdate\footnote{By
birthdate we mean born on the same day of the same month. Assume that
all birthdates are equally likely and independently chosen (by a
higher force).} in a classroom with $m$ people on a planet with $N$
days in a year. How many such triples would you expect to see in this
class ($m=100,N=365$)?

\solution{
The probability that a triple of students has the same birthday is
$1/N^2$. The number of possible triples is $\binom{m}{3}$ which is
$\frac{m(m-1)(m-2)}{6}$. Problem is that these triples are not
independent of each other, so computing the probability of $i$ triples
is very hard. However if we use the linearity of expectations, then we
get to ignore this thorny issue. As in the part(b), each triple gets
assigned an indicator random variable and the expected value of that
variable is $1/N^2$ (feel free to verify rigorously). Let $T$ be the
random variable representing the number of triples with the same
birthdate. Then,

$$\expect{T} = \frac{m(m-1)(m-2)}{6N^2}$$

For this class that is $100.99.98/6.365.365 = 1.2 (approx)$
}
\fi

\end{problemparts}
\end{problem}
