Mathematical Proof that Progressions will never Overcome a Negative Expectation Game

IMHO, one of the flaws in the argument against it working seems to be that if it doesn't work in one case, no matter if that case is hugely improbable, then you say it doesn't work. In that case, neither does card counting work, since there is the possibility that there is one case in which the card counter never wins a single hand.

When did we stop analyzing for probability, and start analyzing for certainty? Maybe the question should be, "What is the probability that a person could win with deep pockets and no house limit using a martingale over a year's time?" Stated another way, what are the odds that a run of losses could take place to wipe out one's theoretical bankroll over a year's time? Then, instead of simply dismissing the martingale as something that doesn't work, we can evaluate it as something that works x% of the time. If x% is less than 1%, is it any different than using a similar RoR in the case of card counting? It may not be as lucrative, but it should be a winner nevertheless. I wonder what the win rate per thousand hands would be?

I did not say that the odds of winning a trial are zero. I said it is possible that you won’t win any trials. That’s a trivial proof.
OK, I state by fiat that I will be ahead at some point, no matter what the math says.

Martingale proponents say they can't lose. That is false.

Sorry, no one said any of that.

Martingale proponents say they can't lose. That is false.

Agreed, in the real cases.

However, given that one has a nonzero probability of winning any given hand of blackjack, and given an infinite number of hands and an infinite bankroll, the probability of never winning at least one hand is 1.

I'm trying not to get sucked into this thread, so let me just say to everyone...

Is it seedless?

Obviously not. Cat's that have been spayed aren't that lively.

I'm trying not to get sucked into this thread, so let me just say to everyone...

Martingale proponents say they can't lose.

Thanks iCount.

These can't both be correct.

If you could point out the specific logical fallacy in my earlier post, that would be helpful.

I suspect that the notion of calculating "expectation" is misapplied here, since the condition for stopping the series is explicit (a win), which contradicts the concept of an "expected" value. Your equations describe the expected win/loss for some number of trials, infinite or not, but without that specific stopping condition. But there could be some other contradiction.

You simply cannot assume that you will win a hand before losing an infinite amount. You can't remove one of the possible outcomes from the analysis.

You simply cannot assume that you will win a hand before losing an infinite amount. You can't remove one of the possible outcomes from the analysis. Yes, the chance that you will lose every hand is infinitesimal. But, all possibilities are infinitesimal. The chance of losing the first and then winning every other hand, or every third hand -- all possible event strings have an equally infinitesimal chance. You aren't allowed to pick one and say it won't occur.

Martingale proponents say they can't lose.

A funny statement.



It goes both ways. If it requires an infinite number of hands to ensure a win of 1 unit, then the gain is 1/infinity. Which is zero. Same as saying .999999... is one.

You can't accept parts of the implications of infinity and ignore other parts.