I Play perfect Basic Strategy and with counting cards I estimate my odds are 50/50 with house. I am trying to compare 2 concepts. I determined 1 Standard Deviation is 10 bets. For this, Heads is a win, tails a loss.
- Flipping a coin, how often will I get 1 SD (10 Bets aka "Heads") ahead, before seeing 3 SD (30 Bets) (tails) of Loss. With 10 bets a winning game and 30 losing bets, a losing game.
Do you see a better optimization of this theory? FYI: play is about 64% in the 1SD area, and in 2 SD area approx 93% of time, and within 3 SD approx 97% of time, which would seem to indicate that I will only see that out of 3SD loss 3% of time.
In practice, not knowing the math to optimize this, I win about 14 of every 15 games. (With counting I know I have up to a 2.3% edge over the house, but I must increase my bets with the count, and wonder if keep my play at the 50/50 level, can my betting strategy win for me, I could do this without getting barred from play.
If every bet is one unit and you really do have a 50% success rate on every bet, then you have a random walk with expected value of zero. Under these conditions your chance of getting to +10 before -30 is given by $${{30} \over {30+10}}=0.75$$
Your situation is different because of complicating factors like doubling down, blackjacks, splitting pairs, and the surrender option, if it is offered.
However, if your expected value is zero then these do not matter and your basic strategy will not have a positive expectation. Your betting strategy is irrelevant. Your overall expectation is the sum of the expectations of each bet, and a sum of zeros is zero no matter what you bet on any individual hand.
Read the Wikipedia article on "Random walk," especially the one-dimensional case.