expected value of getting couples on adjoining seats for b bachelors and m models

Question

Ref: The Theater Row Puzzle from the book Fifty Challenging Problems in Probability by Mosteller (1965) reads:

"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 couples?"

I understand the solution of the book (explicited there, it is v. helpful to have a look to understand the situation I think) and why and how we use linearity of expectations. My issue is that I tried to answer it looking at chunks of 3 seats (rather considering pairs of seats) and get a different result. What is wrong with my logic?

Consider a chunk of 3 seats: XYZ. Denote male by M, female by F. Consider only the possibilities where a non-zero number of couple arises.

You can have the case of 2 males: MFM, MMF, FMM (each with probability: $8/15*1/2*7/13 = .1436$) or 2 women: FMF, FFM, MFF (each with probability: $7/15*6/14*8/13 = .123$).

Computing expected value of number of couples for a 3-seat chunk:

$$.1436*2+.1436*1+.1436*1 + .123*2 + .123*1 + .123*1 = 1.0667$$

Now multiply this by 13 (the number of 3 seat chunks on a 15 seats row) and you get about twice as many couples as in the actual solution.

What has gone wrong in my reasoning?

Answer 1

You're double-counting couples. For example say that the row begins MMFF... ; you'll count the couple at positions 2 and 3 twice, one as part of the chunk MMF coming from positions 1-3 and one as part of the chunk MFF coming from positions 2-4.

Your answer comes out to be somewhat less than twice the correct answer because you don't double-count every couple. In particular a couple at positions 1 and 2 or at positions 14 and 15 (i. e. at either end of the row) would only be counted once by your method.