Let $p$ be a probability in $]0,1[$, and let $(X^p\_i)\_{i \geq 1}$ be a i.i.d. family
of variables with law $P(X=1)=p, P(X=-\frac{p}{1-p})=1-p$ (so that $E(X)=0$). 
Set $S^p\_n=\sum\_{k=1}^{n} X^p\_k$ for $n\geq 1$, let $e(p,n)$ denote the probability 
that at least one of $S^p\_1,S^p\_2, \ldots ,S^p\_n$ is positive (which means
that the random walk reaches a positive value at least once during the first
$n$ steps ). It is well known that for a fixed $p$, $e(p,n) \to 1$ when $n\to +\infty$.


Let $p\neq q$ be two probabilities. For large enough $n$, shall we have
$e(n,p) \lt e(n,q)$ or $e(n,p) \gt e(n,q)$ ?



Comparing hitting probabilities for two different random walks
For the first question: yes, suitably scaled, the sequence $S^p\_n$ tend to Brownian motion, which is positive infinitely often with probability 1. More precisely, for any fixed $p$, $e(n,p)$ will be asymptotically roughly $1-C\_p n^{-\frac12}$.


As for the second question: it really depends on the range of $p$, $q$ and $n$. For fixed $n$ it is not monotone in $p$: for example, when $p$ is close to 1 we have $e(n,p)=1-p$, so it looks like it's decreasing, but when $p\approx \frac12$ we have a jump in $e(2,p)$ from around $\frac14$ to around $\frac12$.


The thing is that you have two contributing factors here: the probability of crossing 0 in the first few steps (how few depends on $p$) which is mostly a combinatorial question with incontinuity points, and the longer range behavior which is determined by the constant $C\_p$ above. I guess that if you started the random walk at some distance from 0, you'd get mostly the latter factor.


