Inspired by this question, I have been investigating the following problem:
Suppose we start with $1$ unit of currency and place a series of wagers where with each wager we gain a unit of currency with probability $p$ or lose a unit of currency with probability $1-p$. We stop when we either hit an upper limit $L$ (in which case we win the game) or $0$ (in which case we lose). In terms of $p$, what is the limit of the chance that we win this game as $L \to \infty$?
I have determined experimentally that the answer seems to be:
$$ \lim_{N\to\infty} P_{\textrm{win}} = \begin{cases} 0 & \textrm{, if $p\leq0.5$}\\ 2-\frac{1}{p} & \textrm{, if $p > 0.5$} \end{cases} $$
What I don't know is why. I'm pretty sure I can prove that the limit should be $0$ for $p = 0.5$ (and therefore also for any smaller $p$), but the result for larger $p$ I find a bit surprising, and don't know how to prove or justify beyond "that's what the simulations show".
From comment:
Huygens's Gambler Ruin result for $p\not =\frac12$ would suggest
$\lim\limits_{L \to \infty} \dfrac{1-(\frac{1-p}{p})^1}{1-(\frac{1-p}{p})^{L+1}}$ which is $\lim\limits_{L \to \infty} \dfrac{2-\frac1p}{1-(\frac{1-p}{p})^{L+1}}$
and, for $p \gt \frac12 \implies \left(\frac{1-p}p\right)^{L+1} \to 0$, is $2−\frac1p$ as your simulation suggests