Re: bankroll question
I can give you some guidance. Suppose your chance of losing each bet is p. That means you must win 1.0015/(1-p) times the amount bet when you win (assuming there is a fixed payoff for winning).
If p is 0.5, as in a fair coin flip, you win 2.003 times the stake when you win. If p is 0.999999, as in a lottery, you have to win 1,001,500 times the stake when you win.
After k bets of b each time, your expected profit is 0.0015*k*b and the standard deviation is 1.0015*[k*p/(1-p)]^0.5. For two standard deviations (about 97.7%) protection, you need a bankroll of:
2.003*[k*p/(1-p)]^0.5 - 0.0015*k*p
This expression has a maximum when k = (1.0015/.0015)^2/[p*(1-p)]. At that point it equals 669/(1 - p). So you need a bankroll of about 1,338 bets for p = 0.5 (coin flip) or 669,000,000 for a 1 in a million lottery.
This is not an exact result, but it's a decent rule of thumb.
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