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? 

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.
$$\begin{align}
X\mid p &\sim \mathcal{N\!\!B}(p, 1)
\\[2ex] \mathsf P(X=x\mid p) 
 & = p^x(1-p) & : x\in\{0,1,\ldots \}
\\[2ex] \mathsf E[X\mid p]
 & = \sum_{x=0}^\infty x p^x (1-p)
\\[1ex] & = \frac{p}{1-p} 
\\[2ex] \mathsf E[X\mid p=0.5] & = 1
\\[2ex] \mathsf E[X\mid p=0.8] & = 4
\end{align}$$
