Drunken sailor's Random Walking A drunken walker is on $x=0$, and if $x<0$, he falls and he dies.(Once he gets position $x<0$, he dies permanently.)
There is $0<p<1$ chance to move right ($x \rightarrow x+1$), and $1-p$ chance to move left 
($x \rightarrow x-1$).
(1) After $n$th step, what is the probability he is still alive?
(2) I have no idea to how to define probability he is still alive after infinite iteration, since there are infinite state and I can't convince this probability will be limit of (1) when $n$ goes $\infty$. So how to define that probability precisely and why it will be limit of (1)?
 A: The generating function of the first hitting time $T$ of $-1$ starting from $0$ is $$E(s^T)=\frac{1-\sqrt{1-4pqs^2}}{2ps}=\sum_{n\geqslant0}\frac1{n+1}p^nq^{n+1}{2n\choose n}s^{2n+1},$$ where $q=1-p$. Thus, the probability to be still alive at time $2n+1$ is $$P(T\geqslant2n+3)=1-P(T\leqslant2n+1)=1-\sum_{k=0}^{n}\frac1{k+1}p^kq^{k+1}{2k\choose k}.$$ The probability to stay alive forever is $0$ if $p\leqslant\frac12$ and $1-(q/p)=(2p-1)/p$ if $p\gt\frac12$.
A: When $p=1/2$, the probability he is still alive after $2n$ steps is given by the Catalan numbers (see http://en.wikipedia.org/wiki/Catalan_number for the classical path-enumeration argument):
$$ p_{2n} = \frac{1}{4^n(n+1)}\binom{2n}{n} = O\left(\frac{1}{n^{3/2}}\right)$$
hence the probability to be alive drops to zero quite fast as a function of $n$. The @This is much healthier's comment explains pretty good why the probability to be alive forever is just $\lim_{n\to +\infty}p_n$. If $p<\frac{1}{2}$, then the probability to stay alive forever is clearly still zero. If $p>\frac{1}{2}$ and we remove the sea, the position of the drunk after $n$ steps is concentrated around $(2p-1)n$ with a variance equal to $4p(1-p)n$. 
Due to the Hoeffding's inequality, the number of $n$-steps paths that land before $0$ is exponentially small in $(2p-1)^2 n$. This gives that after roughly $\frac{1}{(2p-1)^2}$ steps toward the right, the probability of never falling back into the sea becomes closer and closer to one. A careful estimation should show that, if $p>\frac{1}{2}$, the probability to survive forever is greater than:
$$ C\cdot p_{\frac{1}{(2p-1)^2}}>\frac{C(2p-1)^3}{\sqrt{2\pi}}.$$
