I don't think it matters, even with infinite pockets. Let's say 3 is the max number of losses we will accept. 50-50 coin flip, start of betting a dollar. 7 out

I don't think it matters, even with infinite pockets. Let's say 3 is the max number of losses we will accept. 50-50 coin flip, start of betting a dollar. 7 out of 8 times we win a dollar. 1 out of 8 time we lose a dollar, double lose 2 dollars, double, lose 4 dollars quit. (7/8)(1) + (1/8)(-7) = 0 Generally: n = number of losses before quit. E(x) = (1-.5^n)(1) + (.5^n)(-2^n+1) = (1) - (1/2^n) - (2^n-1)/(2^n) = (2^n)/(2^n) - 1/(2^n) - (2^n+1) / (2^n) = (2^n -1 - 2^n + 1) / (2^n) = 0 So with a 50-50 our expected value is 0 even with an infinite bankroll. Which makes sense, there is no way to transform a series of neutral or negative expected value bets into a positive expected bet by combining them.