Average number of Heads in a row? If I have a fair coin, how many heads should I get in a row on average? Is it 2 because 1/1-.5 = 2? (This is my intuition, feel free to correct me if I am wrong). I'm guessing this because its .5 to get heads on the first time, it's .25 to get heads on the second time if I get heads on the first time, etc. 
What about when it's an unfair coin and the odds of getting heads is .8? Now is my answer for how many heads on average I should be getting 1/1-.8 = 5? 
 A: If the probability of getting tails is $q = 1-p$, then you expect to get, on average, $q$ tails per throw, because that is exactly what a probability is: the average number of tails per throw.
Expectations are additive, so if you get $q$ tails per throw it requires $\frac nq$ throws in order to expect $n$ tails, and $\frac1q$ throws in order to expect to have one tail.  That means that the number of heads you expect to get just before the first tail is $\frac 1q - 1$.
If the coin is fair, $q=\frac12$, so the expected number of heads in a row is $1$.  If the coin throws heads with probability $0.8$, then $q=0.2$, so the expected number of heads in a row is $4$. In general, if the probability of throwing heads is $p$, the expected number of heads in a row before the first tail is $$\frac1{1-p} - 1 = \frac p{1-p}.$$
Note that although the infinite summation gets the same answer, it is not needed to solve this problem.
A: In order to count occurrence of 'heads in a row' you stop counting just before the first tail.   This is a “count successes before one failure” senario; so the number of heads in a row $X$ given a biased coin, with probability of heads $p$, has a negative binomial probability distribution.
$$\begin{align}
X\mid p &\sim \mathcal{N\!\!B}(p, 1)
\\[2ex] \mathsf P(X=x\mid p) 
 & = p^x(1-p) & : x\in\{0,1,\ldots \}
\\[2ex] \mathsf E[X\mid p]
 & = \sum_{x=0}^\infty x p^x (1-p)
\\[1ex] & = \frac{p}{1-p} 
\\[2ex] \mathsf E[X\mid p=0.5] & = 1
\\[2ex] \mathsf E[X\mid p=0.8] & = 4
\end{align}$$
