Re: Probability of going broke
If the bet size is 1000, then your winrate is
m = .5(1100) + .5(-1000) = 50
and your variance is
s^2 = .5(1100^2) + .5(1000^2) = 1,105,000.
You can use the risk of ruin formula
-(s^2/(2m))log(p) = b
to find that for a banroll of b = 1,000,000, your risk of ruin is
p = e^{-(2bm)/s^2}
= e^{-100,000,000/1,105,000}
= 5 * 10^{-40},
which is astronomically small.
You can do something similar for other bet amounts, but there is a danger. The risk of ruin formula is derived by assuming that your bankroll follows a Brownian motion with drift. Since a random walk converges to a Brownian motion as the step size goes to zero, this approximation will be good when the bet size is much smaller than the bankroll. But if the bet size is comparable to the bankroll, this is no longer a good approximation. Let me just warn you: getting an exact answer in the discrete setting would be an extremely daunting task. Good luck to all of you who would like to try it.
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