Gambler's ruin, probability of loss, infinite turns A gambler starts with $\$1$ and bets $\$1$ every turn of a game, where he has the probability $p$ to obtain $\$2$ and $1-p$ to obtain nothing. If $p<1/2$, what is the probability he will eventually lose all the money?
I'm looking for a more elegant solution than what I've come up to:
The probability of not losing in $n$ turns is $\leq$ of the probability of having more winning turns than losing ones, which is
$$ P_n =\sum_{h \leq n/2} \binom{n}{h} p^h (1-p)^{n-h}. $$
Since $p < 1-p$ and  $\binom{n}{h}$ is maximum for $h = n/2$,
$$ P_n \leq \dfrac{n}{2} \binom{n}{n/2} \big( p (1-p) \big)^{n/2}. $$
From Stirling formula one gets that
$$ \binom{n}{n/2} \sim 2^n \sqrt{\dfrac{2}{n\pi}},$$
therefore
$$ \lim_n P_n \leq \lim_n \sqrt{\dfrac{n}{2\pi}} \big( 4 p (1-p) \big)^{n/2} = 0,$$
since $p(1-p) < 1/4$.
 A: Let $S$ be the probability that you never run out of money starting from $\$1$. Let $X$ be the probability that you do eventually run out of money, so $X=1-S$. 
Now consider the first bet made. There is a probability of $p$ that you don't immediately run out of money, and end up with $\$2$. We can consider having $\$2$ as having two $\$1$s, so the probability of never running out of money from this state is $1-X^2$, since if we were to run out, we would need to run out from both of our starting $\$1s$.
This gives
$$S=p(1-X^2)=p(1-(1-S)^2)$$
$$\rightarrow S=2pS-pS^2$$
Hence $S=0$ or $S=2-\frac{1}{p}$
When $p\leq\frac{1}{2}$ this gives the uninteresting result $S=0$, since $2-\frac{1}{p}\leq 0$. However, when $p>1/2$ we claim $S=2-\frac{1}{p}$, so you have a nonzero probability of $\textit{never}$ running out of money.
To see this, allow yourself to possibly have a negative amount, and note that with $p>\frac{1}{2}$, your expected amount of money approaches infinity. This means that you can only cross the $\$0$ mark finitely many times. Since at each step there is a nonzero probability to not cross the $\$0$ mark, and only finitely many possible steps where this is a possibility, the probability of never crossing the $\$0$ mark is nonzero.
Since the probability is nonzero, we must have $S=2-\frac{1}{p}$.
