Curious about a made-up paradox

This has little to do with random number generators.
As Tony K noted, this is a random walk where each step is $+1$ with probability $p < 1/2$ and $-1$ with probability $1-p$. Starting at $0$, the probability of ever reaching positive integer $m$ is $(p/(1-p))^m$. In your case, with $p = 0.46$ and $m = 200$, that probability is about $1.18 \times 10^{-14}$.

"From my understanding of statistics...": Your understanding is wrong. You are probably thinking of an unbiased random walk, where given any integer $n$, the probability of reaching $n$ at some point is 1. But this random walk is not unbiased, because the probability of losing is 0.54, not 0.5. So there is a non-zero probability (indeed, a very high probability) that the gambler never gets out of debt.

When you say he wins \$2 does that mean he gets back his \$1 plus another \$1? Or that he gets back his \$1 plus another \$2?

If he has -\$200 and bets \$1 (now having -\$201) and loses, he'd have -\$201 of course.

If he has -\$200 and bets \$1 (now having -\$201) and wins, would he then have -\$199 or -\$198?

If it's -\$199, then the probability of getting out of debt is very small as there is slightly more chance (p = 0.54) of the \$1 change away from zero (vs p = 0.46). As far as I know, calculating the expected number of events to reach an outcome that has a probability of < 0.5 does not make sense. Check out the Bernoulli distribution.

If it's -\$198 that changes things. That's where my stats gets a bit rusty and I'm can't remember if there's such a thing called a 'weighted' Bernoulli distribution or if that has a different name, but it makes it much more likely that your poor addict might get out of debt.

I know this doesn't answer the question fully but perhaps someone else can pick up from here?