Hello,
I'd like to find the expected number of Bernoulli trials that I'll need before I will get *exactly* n more heads than tails, given a coin which gets a heads with probability p.

My approach for this problem has been as follows: a) The probability of getting n more heads is equal to getting any permutation of N+r-1 heads and r tails, followed by a single head, thus $\sum_{r=0}^{\infty} p^{n+r}(1-p)^{r}$ ${n+2r-1}\choose{r}$

b) To find the expected number, I'll just have consider $\sum_{r=0}^{\infty} (n+2r) p^{n+r}(1-p)^{r}$ ${n+2r-1}\choose{r}$

c) I need to find the expectation over the trial lengths of $\gamma^{l}$ ($\gamma$ is a constant $<1$), i.e. $\sum_{r=0}^{\infty} \gamma^{n+2r} p^{n+r}(1-p)^{r}$ ${n+2r-1}\choose{r}$

Another way of phrasing the problem (which is the context in which I would like to solve it) is: Suppose you are at a distance n from a goal, and with probability p you move towards it, and 1-p away from it. What is the expected number of steps you will need to reach the goal?

I don't know how to evaluate any of those summations. I only need an approximate answer - as a function of $p$ and $n$. Is there any other way than by using Stirling's formula for the binomial coefficients (which gets quite messy)?