Consider $$X(t)=\mbox{Number of wins} - \mbox{Number of losses}$$ for $t$ Bernoulli($\theta$) trials.
I calculated that $$P(X(t) = x) = {t \choose (t-x)/2} \theta^{\frac{t+x}{2}} (1- \theta)^{\frac{t-x}{2}}$$ i.e. (winning trials - losing trials) = $x$ and (winning trials + losing trials) = $t$.
If we let $S = \{X(t)\}_{t=0}^{T}$, my question then is:
For an arbitrary integer $x$, what is $P(x \in S)$ as $T \rightarrow \infty$?
Since $X$ is performing a random walk with drift $2\theta-1$ and $S_T$ is the range of $X$ up to time $T$, if $\theta=\frac12$ then $P(x\in S_T)\to1$ for every $x$, when $T\to\infty$. If $\theta\gt\frac12$ then $P(x\in S_T)\to1$ for every $x\geqslant0$, when $T\to\infty$. If $\theta\lt\frac12$ then $P(x\in S_T)\to1$ for every $x\leqslant0$, when $T\to\infty$.
By symmetry, the only remaining case is when $x\gt0$ and $\theta\lt\frac12$. Then $P(x\in S_T)\to h^x$ where $h=\lim\limits_{T\to\infty}P(1\in S_T)$. The usual one-step Markov recursion indicates that $h=(1-\theta)\cdot1+\theta\cdot h^2$ hence $h=\theta/(1-\theta)$.