Mathematical Proof that Progressions will never Overcome a Negative Expectation Game

[aslan]

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?

 

[QFIT]

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?

Sorry, no one said any of that.

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

 

[johndoe]

Here is where your logic breaks:

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.

Your claim, and your objection, are contradictory.

If the odds are not zero, as you allow, then there will be a win eventually, given enough trials. That's exactly what the math predicts, because we are not putting a bound on the number of trials. In fact, if you like, we're defining the number of trials as being that required to secure a single win. (That's not a "twist" either.)

The only "twists" here are the obvious ones of unlimited bankroll/credit, time, and table limits. Given those, and only those (entirely unrealistic) conditions, Martingale works with complete certainty, just like Thorp said.

 

[rrwoods]

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.

 

[aslan]

Sorry, no one said any of that.

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

Anyone can lose. What should be of interest to APs is what are the odds of winning or losing using various game strategies.

But it is a fun exercise, if no one takes any of it personally. And it does give an opportunity to exercise one's mathematical knowledge and abilities, of which I have next to none.

 

[QFIT]

Agreed, in the real cases.

A funny statement.

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.

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.

 

[Sonny]

EVERYTHING happens in infinity

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

 

[aslan]

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

Is it seedless?

 

[QFIT]

Is it seedless?

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

 

[aslan]

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

:laugh:

 

[iCountNTrack]

One More Attempt

 

[zengrifter]

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

​

And I do believe he means EVERYONE! zg

 

[zengrifter]

Martingale proponents say they can't lose.

I missed that one. zg

 

[johndoe]

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.

 

[k_c]

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.

In the expected value calculation there is no stopping point. Martingaler is required to continue betting larger and larger amounts for any number of successive losses weighted by the probability of each successive loss. It's just figuring EV for all possible bets based on intial wager of 1 unit the same as you would do in figuring overall EV when spreading at different true counts with differing probabilities of occurrence. The only difference is in blackjack the EVs are different at differing true counts but in the martingale the EV is fixed.

I'm sure icnt can probably make it more clear than I can.

 

[QFIT]

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.

 

[psyduck]

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.

That is the point I have been trying to make in all my previous posts.

 

[johndoe]

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.

But the chance of losing every (infinite) hand isn't infinitesimal, it really is zero. The only way an infinite number of hands can always lose is if the probability of winning any hand is exactly zero.

 

[zengrifter]

Martingale proponents say they can't lose.

They cannot lose in the absence of boundaries. zg

 

[rrwoods]

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.

What? Why? When I win one unit, isn't my gain one unit? Are you defining gain to be inversely proportional to time?