Question: A particle initially stands at $0$ in $\mathbb{N}$. Every second, it jumps one unit towards $-\infty$ or $+\infty$ with probability $\frac{1}{2}$ each.

What is the probability that it reaches $+1$?

Answer: 
The probability to hit $1$ after $1$ step is $\frac{1}{2}$, encoded by the string $+$; the probability to hit $1$ after $3$ steps is $\frac{1}{8}$, encoded by the string $-++$; the probability to hit $1$ after $5$ steps is $\frac{1}{16}$, encoded by the strings $-+-++$ and $--+++$. To compute the probability to hit $1$ after $2k+1$ steps, we just have to count how many binary strings of length $2k$, with $k$ zeroes and $k$ ones, have more zeroes than ones in every prefix. This is a well-known problem: the solution is given by the [Catalan numbers](http://en.wikipedia.org/wiki/Catalan_number):
$$ C\_k = \frac{1}{k+1}\binom{2k}{k}. $$
The probability to hit $1$ in the first $2N+1$ steps is so:
$$ \sum\_{k=0}^{N}\frac{1}{2^{2k+1}(k+1)}\binom{2k}{k}=\sum\_{k=0}^{N}\left(\frac{1}{4^k}\binom{2k}{k}-\frac{1}{4^{k+1}}\binom{2k+2}{k+1}\right)\\=1-\frac{1}{4^{N+1}}\binom{2N+2}{N+1}=1-\frac{1}{\sqrt{\pi(N+5/4)}}+O\left(\frac{1}{N^{5/2}}\right).$$
if you are allowed to wait forever: 1. 

There are lots of ways to do this. Suppose we ask the question hit 1 or -n first and stop when this happens. 

Then since hitting time is a finite stopping time, the value of the random walk stopped at the hitting is still a martingale and so has expectation 0. So probability of hitting $1$ is $n$ times that of hitting $-n.$ So the probability of hitting $1$, is $n/(n+1).$

The probability you want must be bigger than this for any $n$ so it must be $1.$