Curious about a made-up paradox I have thought up a paradox, that may already exist, but I do not know what it's called. It's bothering me though, so any help regarding solving it or proving it impossible would be appreciated.
In this paradox, you have a gambler. The gambler is \$200 in debt, but the gambler has a wealthy friend who lets the gambler bet using money he does not have to play a game. In this game, the gambler bets \$1, and a random number generator generates a number from 1 to 100. If the number roll over, or exactly 55, the gambler wins \$2. If the number rolls under 55 the gambler loses that \$1 he bet.
From my understanding of statistics, over a certain expected period of time, the gambler should hit the unlikely scenario to get out of debt using this method. However using my computer simulations, it seems to take more time than I can allow my simulations to run.
Is it possible to guess an expected number of times the gambler would have to play the game in order to get out of debt using some
mathematical model?
I am also concerned that the nature of random-number generators may make it impossible for the gambler to get out of debt, as random number generators could be heavily biased to avoid situations such as randomized decks to be fully sorted, or getting out of debt with negative debt and unlikely odds.
What I want to get out of this question is, how to explain why it's possible, how to calculate expected number of times the game has to be played to reach the goal, or why it's impossible, or some existing question I can try to study to better understand the problem.
 A: 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?
A: 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 $1.18 \times 10^{-14}$.
A: "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.
