# Theater Row

## Problem

Eight eligible bachelors and seven beautiful models happen randomly to have purchased single seats in the same 15-seat row of a theater. On the average, how many pairs of adjacent seats are ticketed for marriageable couples1?

## Solution

Let $$X$$ be a random variable denoting the number of pairs that are marriageable. We can rewrite $$X$$ as $$X = X_{1} + X_{2} + \ldots + X_{14}$$, where each $$X_{i}$$ is a random variable representing the number of marriageable couples in the $$i^{th}$$ pair of adjacent seats. Note that each $$X_{i}$$ can take on either $$0$$ or $$1$$ only.

We can easily calculate $$E[X_{i}]$$ by considering the two arrangements that result in a marriageable pair, which are Male-Female and Female-Male:

$E[X_{i}] = \left(\frac{8}{15}\right) \left(\frac{7}{14}\right) + \left(\frac{7}{15}\right) \left(\frac{8}{14}\right) = \frac{8}{15}$

Then, applying the Linearity of Expectation property, we can calculate $$E[X]$$ as follows:

$\begin{split} E[X] &= E[X_{1} + X_{2} + \ldots + X_{14}] \\ &= E[X_{1}] + E[X_{2}] + \ldots + E[X_{14}] \\ &= 14\left(\frac{8}{15}\right) \end{split}$

Thus, the expected number of marriageable pairs in the row of $$15$$ seats is $$E[X] \approx 7.46$$.

1. According to outdated cultural norms from the time of this problem’s publication.↩︎