What is the expected number of flips of an unfair coin until you have 2 more heads than tails? $p$ is the probability of heads. Note if $p \le 0.5$, the answer is infinity, so assume $p > 0.5$. 
What is the expected number of flips of the coin where we have 2 more heads than tails?
Note you would stop flipping the coin when you first encounter the situation where you have 2 more heads than tails.
 A: We only ever need to keep track of the difference between the number of heads, and the number of tails. Moreover, the time it takes to increase this difference by $2$ is twice the time it takes to increase this difference by $1$.
So let's call $x$ the time it takes to go from a difference of $k$ to a difference of $k+1$: for example, the time it takes from the start until you have flipped heads once more than tails.
After the very first coinflip, we're either done (with probability $p$), or we have made progress in the wrong direction and have a difference of $2$ to make up for (with probablity $1-p$). So we have $$x = 1 + (1-p) \cdot 2x$$ or $x = \frac1{2p-1}$. 
The answer you want is $2x = \frac2{2p-1}$.
A: This corresponds to calculating the expectation of the first-passage time of an asymmetric one-dimensional random walk. I believe that even the asymmetric one-dimensional random walk satisfies the (strong) Markov property, so therefore we can assume that the random walk "starts over" at every step. (See p. 4)
I.e. if $T(1)$ is the random variable indicating the first time we have one more head than tail, then $T(2)$ is the sum of two independent copies of $T(1)$. So by linearity of expectation: $$\mathbb{E}[T(2)]=\mathbb{E}[T(1)]+\mathbb{E}[T(1)]=2\mathbb{E}[T(1)]. $$
Now, $\mathbb{E}[T(1)]$ is a lot easier to calculate.  According to this document (p.3 & p.5), $$\mathbb{E}[T(1)] = \Phi'(1)\,, \quad \text{where} \quad \Phi(t) = \frac{1 - \sqrt{1-4p(1-p)t^2}}{2(1-p)t} $$ i.e. $$\mathbb{E}[T(1)] = \frac{1 }{2p-1}  \quad (p > \frac{1}{2})\,.$$ (In the document, $N= T(1)$.)
