Just curious if all math geniuses here could answer a legitimate "negative progression" question seriously.

I know progressions can not beat the house usually (in spite of my nick name here, I try to count using a simple unbalanced KO method) but I was wondering :

Suppose there existed a blackjack game with EXTREME spread between minimum and maximum bet, lets say a penny minimum and ten thousand dollars max. Is there some theoretical mathematical equivalent to a card counting strategy with some fairly low risk-of-ruin, to playing it by negative progression Martingale method?

Example, an extreme progression of .01 .02 .04 .08 .16 .32 .64 1.28 2.56 5.12 12 25 50 125 250 500 1000 2500 5000 10000 would mean you could win only one hand out of 20, and still slowly grind out a profit.

Would the odds of losing 21 hands in a row be a less frequent tail-event (statistically low likelihood) than losing your entire bankroll with card counting and, say, 1% risk of ruin (also statistically low likelihood)?

Also given the fact that even a normal basic strategy player can expect a blackjack about once per 21 hands, wouldn't a neg progression which can survive loss streaks of 20 hands in a row mean that this eventual expected blackjack might push you into positive EV territory? Other games like red-black on roulette which do not have the player bonus 3:2 payout would not fit this: but maybe it works for blackjack?

By the way, it is not only an academic question.

I actually know of a currently available, playable game, with just such an extreme min and max offered. It is not countable unfortunately. I am trying to figure out the best way to attack it, if it is worth playing at all.