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?
 A: 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.
A: The answer is 7.5.
Let $X$ be the number of neighbors of a different color.  Let $X_i$ be 1 if houses $i$ and $i+1$ are different colors and 0 otherwise.  Then $X = \sum_{i=1}^{15} X_i.$  
So the expected number of neighbors of a different color is 
$$E\left[\sum_{i=1}^{15} X_i\right] = \sum_{i=1}^{15} E[X_i] = \sum_{i=1}^{15} P(X_i = 1).$$
Now, $$P(X_i = 1) = P(i \text{ is red})P(i+1 \text{ is blue}|i \text{ is red}) + P(i \text{ is blue})P(i+1 \text{ is red}|i \text{ is blue}) $$
$$ = \frac{10}{16} \frac{6}{15} + \frac{6}{16} \frac{10}{15} = \frac{120}{16(15)} = \frac{1}{2}.$$
Thus $$E[X] = \sum_{i=1}^{15} \frac{1}{2} = 7.5.$$
More generally, using indicator variables is often an effective way to calculate expected values.  The linearity of the expectation operator allows you to sidestep all the nasty dependencies you would otherwise be forced to deal with.  For lots of interesting examples using this approach, see Section 7.2 of Sheldon Ross's A First Course in Probability.
Given that the answer comes out so nicely I would not be surprised if there is a cleaner argument than mine.  If someone comes up with one I would love to see it. 

Moron's generalization has inspired me to generalize mine. :)  
Suppose you have $m$ red houses and $n$ blue houses.  Then the expected number of neighbors of a different color is $$\frac{2mn}{m+n}.$$
Similar to my argument above,
$$P(X_i = 1) = P(i \text{ is red})P(i+1 \text{ is blue}|i \text{ is red}) + P(i \text{ is blue})P(i+1 \text{ is red}|i \text{ is blue}) $$
$$ = \frac{m}{m+n} \frac{n}{m+n-1} + \frac{n}{m+n} \frac{m}{m+n-1} = \frac{2mn}{(m+n)(m+n-1)}.$$
Thus $$E[X] = \sum_{i=1}^{m+n-1} \frac{2mn}{(m+n)(m+n-1)} = \frac{2mn}{m+n}.$$
