"Extreme Progression" Math?

[Martin Gale]

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.

[Norm]

No.

[Kat]

No.

Two letters... Has there ever been a simpler, more concise answer to a long-winded question ever, on the Internet?

[Smartpart]

it is not an academic question. ******

fixed

[Hsiaodi]

1 to 1,000,000 spread? I'd want to be a counter to play this game.

[DSchles]

Calculate your average bet for the progression. Multiply by the house edge, which you can't overcome. Multiply by the number of hands you'll play.

That's how much you'll lose.

End of story.

Don

[Three]

What are you going to do when you have to double or split the a hand? The HE is much higher if you don't take advantage of the players ability to double or split. To answer your question take the total amount you have bet at any point and multiply by the HE and you have your expected loss to that point assuming you move your bet independent of the shifting HE (which is a given for the martingale player) and play correct basic strategy which many martingale players do not because it makes the martingale messy.

[seriousplayer]

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.

Your EXTREME bet progression only works in a hypothetical world and not in reality.

1. If you have unlimited bankroll.
2. You will need to be able to weather the heavy losing streak without going broke.
3. By the casino limiting your bet spread.

The bad news is that in reality players play with a limited bankroll. The table max will be finite. In all casino games what matter is the house edge. Remember this all casino games have an integrated house edge that is safe from betting progressions. From probability point of view betting progressions does not estimate the likelihood of occurrence of an event. Therefore, betting progressions does not override the natural of probability like card counting and advantage play.

Bet progression does sometimes create a positive short term result.

[Norm]

If you have unlimited bankroll.

If you have an infinite bankroll, it is impossible to increase it.

[seriousplayer]

If you have an infinite bankroll, it is impossible to increase it.

You could say it that way since infinity + infinity = infinity.

[Nikky_Flash]

What about Infinity times Infinity?

[Norm]

What about Infinity times Infinity?

See http://en.wikipedia.org/wiki/Aleph_number

[Martin Gale]

Well geez sorry but all you guys have totally missed the point of what I was trying to ask.

Look, I know that progressions are not a long term way to beat the house edge. I said so specifically in my first post. That is not what I was asking. The answers about just multiplying total average bet by house edge yielding loss are not helpful.

I will try again. I was criticized for being long winded too but I suppose to get through to you all with the question I will have to be even more verbose. Sorry in advance if that is bad, LOL

Okay so look let's start with some basic premise to illustrate what I am trying to ask.

Assuming an infinite bankroll (yeah, never mind) and no table max limit, it's a fact that a bet progression will win. And the shorter the possible progression is, therefore, the more likely it is to lose in a shorter-run time period. So there is some mathematical way to model how much total risk to any given bankroll there is, when it's applied to a longer progression experiment versus a shorter progression experiment.

Okay so far?

Now, compared to the premise above we have card counting. Given any particular game's player advantage (rules, possible spread, etc) and the player's available bankroll funds, we can calculate a risk of ruin likelihood. This means even if the player does everything right, depending on RoR there is still some percentage chance (likelihood) that he will lose all his bankroll money anyway.

If that RoR is very high (small bankroll, bad game, whatever) it might even be considered still just simply gambling.

Still okay so far? Hope so.

Now then, considering these two scenarios isn't there some way to make a model of exactly how long of some given "extreme progression" might be roughly equivalent to some particular card counting experiment (with a similar risk % of the player losing all his bankroll).

This is what I am trying to figure out, but I do not have the actual math skills to do it alone.

And yes I am asking because believe it or not I have found an online casino that offers an incredibly small minimum bet (fraction of a penny) and as far as I have been able to tell so far NO MAXIMUM BET LIMIT AT ALL.

I believe this may be a mistake on the part of the online casino operator, and it may not last.

Obviously I am interested in this because if a reasonable statistical model for an "extreme progression" can show that it is comparable in overall risk to real-world bankroll, as it might be applied in some card-counting venture (with some "acceptable" RoR percentage?) then I will want to put some money into this, and exploit it before the error is corrected.

Also, obviously, even if a single human could only play this game for a penny a minute, some software robots could be set up to play it hundreds or thousands of times simultaneously. So it's not a worthless game even for 'low stakes', if it is beatable (or even simply just playable) with some known risk tolerance.

Once again please I am not an idiot and I know "progressions can't win" but that is not what I am asking in this post.

Please lets have some real answers, if it is possible here.

Maybe it was a mistake for me to post this in the disadvantage area since it is really not a voodoo question, I hope. Moderators please move the thread elsewhere if it would be better.

Thank you.