Exact formula

For a biased coin that pays one unit when you win, and one unit when you lose, let p = probability of a win, q = 1-p = probability of a loss, B = the loss rebate per units lost, and G = a +/- Goal (player will quit once reaching +G units or -G units, but not before), then the expectation for the strategy is:

E(+G/-G) = G x ((p/q)^G - 1 + B)/((p/q)^G + 1)

It's easy to show algebraically that this is correct for G = 0, 1, and 2 (the last requires a little power series trick). I used induction to show it holds for all positive integers.

However, as usual, I've reinvented the wheel. I did a little research and can now produce formulas for E(+G/-H) with G<>H, and EVEN (hold your breath) for uneven payoffs, such as single number roulette bets. I'm not posting these results yet, because I'm not sure they've ever been connected to the loss rebate problem, so I'm taking my time, making lots of chicken scratches on 8x11s, following Gauss' dictum "few but ripe."

Here are some results, to six digits, for your parameters, p=0.497, q=0.503, and B=0.1:

E(+1/-1) = 0.044300

E(+2/-2) = 0.077201

E(+3/-3) = 0.098705

E(+4/-4) = 0.108816

E(+5/-5) = 0.107541

E(+6/-6) = 0.094886

E(+7/-7) = 0.070861

I also checked some other strategies, like E(+5/-4), but E(+4/-4) is in the lead so far.

Since my theory and your sims don't match perfectly, it's incumbent on me to do a sim. I'll keep you posted on that within a week or so.

Griffin has a formula, in "Extra Stuff" for E(n), so a fairly complete solution to the rebate problem is there, even with a wager requirement. Play only if E(n) is positive, and once you've waged the required amount, continue play only if you are inside the optimum +/- boundaries.

Complete solution EXCEPT for the dirty little detail that blackjack has multiple payoffs. The above formula can be adjusted to work when ties are allowed (p + q <> 1), but not for doubles, splits, etc. For that, I think a sim is needed.

ETF