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.

With an infinite bankroll there is no reason to stop at three losses. In an infinite series of fair coin flips there will be deviations in the distribution around the mean. A gambler can simply choose to stop at any point of excess “wins” as long as he isn’t stopped out due to exceeding his bankroll.

The crazy thing about infinite pockets is that the strategy will favour the gambler even if the odds are 1-99 (or any nonzero probability of winning).

Using expected value for assess winning strategies don’t work with infinite bankroll.