average of pairs sitting next to each other - harder version The Problem
This question comes from "50 Challenging Problems in Probability" By Mosteller. The question is:

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 have a marriageable couple in them?

It's fairly simple if you look at each seat and say "what's the probability that the seat next to me has a person of the opposite sex next to me". For a single seat it's just the sum of probabilities that it's guy&girl + girl&guy: $(\frac{8}{15}\times\frac{7}{14}) + (\frac{7}{15}\times\frac{8}{14})$
We define X to be a random variable indicating probability of a couple in seat $1$ so expectation is: $E[X_1] = 1 \times [(\frac{8}{15}\times\frac{7}{14}) + (\frac{7}{15}\times\frac{8}{14})]$
Due to the linearity of expectation we have: $\sum_{i=1}^{14} E[X_i] = E[X_1] + E[X_2] +\ .... E[X_{14}] = 14 \times E[X_1] = 7\frac{7}{15}$

My Question
This question kind of double counts couples. It says that if the first three seats for example are: Bachelor1-Model1-Bachelor2, that those count as two couples. Which can be right in a sense...
But I would count that as one couple. Because Model1 is either married to Bachelor1 or Bachelor2. Whichever way you pick it, (B1-M1 + B2 or B1 + M1-B2), there is only one wedding that's gonna happen. This is what threw me off when I tried to solve that problem
My question is: under the above assumption, how would I solve this question? Seems like the linearity of expectation here doesn't hold, right? To put it more bluntly: How many weddings are we expecting if every couple that sits next to each other gets married? (Polygamy not allowed)
It's a much harder problem when formulated this way I think...
 A: If you’re going to quote problems that make assumptions about binary gender and heterosexuality, I’d prefer if you make those assumptions explicit.
You don’t explicate how your modified approach should work for more than $3$ people in a row, but I would assume that in the spirit of your modification, you’d want to count a stretch of alternating binary genders of odd length $2n+1$ or of even length $2n$ as $n$ weddings. If so, that would be half as many weddings as the unmodified problem counts couples, plus half a wedding per stretch of even length – so we need the expected number of stretches of even length. (Here a “stretch” is a maximal stretch, i.e. not part of a longer stretch.)
With $15$ seats, there are $15-2n+1$ opportunities for forming a stretch of $2n$ alternating genders. All but the first and last of these form a stretch if a non-matching gender sits at either end, so a stretch of length $2n$ requires $n+1$ specifically placed people of either gender. There are $\binom{15}8$ ways to choose seats for the bachelors, and fixing those $2n+2$ seats leaves $\binom{15-(2n+2)}{8-(n+1)}$ ways to choose the remaining bachelor seats, so the probability for this to happen is
$$
\frac{\binom{15-(2n+2)}{8-(n+1)}}{\binom{15}8}\;.
$$
We need to multiply that by $2$ since the genders can alternate in two different ways.
The first and last potential stretch are a bit more complicated, since there’s no need to terminate them on one end. Thus, only $2n+1$ seats are fixed, and it can be one more of either gender, so the probability for this is
$$
\frac{\binom{15-(2n+1)}{8-(n+1)}+\binom{15-(2n+1)}{8-n}}{\binom{15}8}\;.
$$
In all, we expect
$$
\binom{15}8^{-1}\sum_{n=1}^72\left((15-2n-1)\binom{15-(2n+2)}{8-(n+1)}+2\left(\binom{15-(2n+1)}{8-(n+1)}+\binom{15-(2n+1)}{8-n}\right)\right)\\
=\frac{19022}{6435}\approx2.956
$$
stretches of even length, and thus
$$
\frac12\left(\frac{112}{15}+\frac{19022}{6435}\right)=\frac{6707}{1287}\approx5.211
$$
weddings.
