There may be specific terminology for this, but I would call the process a one-dimensional random walk whose step size varies according to a Categorical distribution.* Perhaps this is simply called a Categorical random walk, like how when the step size is pulled from a Gaussian distribution that's called a Gaussian random walk. (*Or maybe formally this would be split into two Categorical distributions? Ie, a probability p of taking a positive step drawn from one distribution, and probability 1-p of a negative step drawn from the other, where p is the sum of the probabilities from the first distribution. Either way it's the same idea.)
Since you're taking a course on Markov Processes (good choice!), I should also mention that this is a Markov chain.
Quote:
I'm a bit puzzled by the paper in Bruce's post, I don't really understand what exactly they're doing there. I also don't really understand how I should choose the parameters of a Bernoulli distribution to represent my winrate and standard deviation.
The Ethier/Khoshnevisan paper? I think Bruce's English explanation is sufficient if one doesn't need full proof. Feller vol 1 is referenced in their paper and that's a book worth owning.
The goal there was to estimate the chance of going broke with a given bankroll, WR and SD. Ideally we'd know our exact monstrous distribution, but even if we did, we'd need computer brute force to figure out our ruin % because we couldn't derive a usable formula from that distribution. By contrast, the ruin probability concerning a series of coinflips (with a single payout) is much simpler, allowing for a friendly formula (Bruce's Eq 1). So by modeling each hand as a coinflip (instead of a 300-sided die roll with 300 different payouts), we'd be able to use the coinflip ruin formula.
Each coinflip, the gambler bets the same amount b and has the same probability p of winning. To make this analogous to a poker hand, we choose the p and b that would cause the coinflip EV to match our WR and the coinflip variance to match our observed variance. These turn out to be:
b = √(WR² + SD²)
p = ½ + WR/[2√(WR² + SD²)]
Then we can substitute those into Eq 1 which was r = (q/p)^B/b. If it's the bankroll B you're trying to solve for, continue on with the derivation of a simpler exponential formula, but if you only care about r, that's optional.
That's not 100% exact (even if the WR and SD are) since it's a vast simplification of the true underlying process (the many-sided die or Many-Faced Poker God), but accurate within fairly tight bounds established in the Ethier/K paper.