<s>[INST] Expected number of neighbors Given a row of 16 houses where 10 are red and 6 are blue, what is the expected number of neigbors of a different color?
 [/INST] The chances of a particular neighbour pair being the same colour is
$$ \frac{{14 \choose 8} + {14 \choose 10}}{{16 \choose 6}} = \frac{4004}{8008} = \frac{1}{2}$$
Hence the answer is $\displaystyle 7.5$
This is happening because $\displaystyle {14 \choose 8}, {14 \choose 9}, {14 \choose 10}$ are in arithmetic progession: The number of ways of being same colour is $\displaystyle {14 \choose 8} + {14 \choose 10}$ and the number of ways of being different is $\displaystyle 2{14 \choose 9}$. The probability is $\displaystyle \frac{1}{2}$ if these two are equal.
Interestingly, $\displaystyle {n \choose r}, {n \choose r+1}, {n \choose r+2}$ are in arithmetic progression if and only if $\displaystyle n+2$ is a perfect square and $\displaystyle r$ is given by $\displaystyle r = \frac{n-2 \pm \sqrt{n+2}}{2}$ (see the end of the answer for a proof).
So for instance, the whole bunch of problems:

15 red, 10 blue
21 red, 15 blue

etc give rise to this neat probability of being $\displaystyle \frac{1}{2}$.

Proof that n+2 is a perfect square
$\displaystyle {n \choose r}, {n \choose r+1}, {n \choose r+2}$ are in arithmetic progression iff
$\displaystyle 2{n \choose r+1} = {n \choose r} + {n \choose r+2}$
i.e
$\displaystyle 2 = \frac{r+1}{n-r} + \frac{n-r-1}{r+2}$
Doing some manipulations gives us
$\displaystyle (n-2r-2)^2 = n+2$
Hence
$\displaystyle r = \frac{n-2 \pm \sqrt{n+2}}{2}$
Which has an integer solution iff $\displaystyle n+2$ is a perfect square.
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