zengrifter
Banned
Ain't that the truth! zgSounds like pretty good odds to me. With [PERFECT] counting [IN A DREAM GAME] you're not guaranteed to win within your lifetime either.
Mathematical Proof that Progressions will never Overcome a Negative Expectation Game
- Thread starter iCountNTrack
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aslan
Well-Known Member
But if you first hit the possibility that you will never be behind, it won't matter whatever other possibilities could have existed. So, it doesn't have to happen to everyone.
You can't ever be bankrupt if you have an infinite bankroll. That's my whole point, and how my (and Throp's) scenario is different than yours. The longrun is not infinite hands - you only play until you win once.
He says it on p.114-116 of "Mathematics of Gambling", while also noting the ridiculousness of the strategy. He even confirms that "The player should arrange from the start to have unlimited credit, reasonably pointing out that since he must eventually win he's sure to pay it off!" (p.116) He then goes on to say how house limits and credit limits prevent this from happening.
How about this gets re-cast as "no table limits" and "unlimited credit available" instead of an infinite bankroll? It's the same thing, but might be easier to swallow as a sure (eventual) win.
psyduck
Well-Known Member
Using martingale, how do you hit the possibility that you will never be behind?
zengrifter
Banned
The paradox, therefore, is that in an infinite universe (which we all exist in BTW) a negative progression -EV bettor will ALWAYS win and a +EV bettor will ALWAYS ultimately lose? Don't that beat all! zg
zengrifter
Banned
I originally stated it as NO UPPER BOUNDARIES.
I also said: "...progressions DO work, provided there are --
no arbitrary limits on capital and bet-sizing".
psyduck
Well-Known Member
Without any limit and you play infinite number of sessions (let's call each time you succeed with martingale and win one unit a session). Then what. Eventually you run into a session in which you never win and lose infinite amount of money. That eliminates all your wins combined and leaves you in the red. End of story.
(Of course, if you play only for the purpose of winning one unit and then quit, any system will "work".)
QFIT
Well-Known Member
I can't understand why you keep saying this. Clearly in the set of possibilities, there is a member where this is not true.
Yep, you're correct, and this is not the case that k_c's proof addressed.
If your odds of winning are nonzero, the probability of no wins "forever" is exactly zero. Is it not? "No wins for an infinite time" is not in the set that includes any possibility of a win.
Therefore, you are 100% guaranteed to have a win eventually.
And if you can ever have a win, with no table limits and infinite credit, a suitable progression will always let you win as much as you care to. Eventually.
zengrifter
Banned
He's taking another stab at it as we speak. zg
QFIT
Well-Known Member
Sorry, I believe you are missing the point. We are talking about infinity. Yes, the possiblility exists, in fact must happen, that you will win every hand. And, you will also lose every hand. Such is infinity. That does not mean that you can be assured of a win. Infinity is different. That's why we call simple arithmetic involving infinity "undefined." You can't use simple arithmetic.
Put simply, with n trials, you have negative EV. with n+1 trials, you have exactly the same negative EV. So, with infinite trials, you have exactly the same negative EV.
aslan
Well-Known Member
If you play long enough you will eventually hit it, unless you hit the possibility that you never get ahead first.
Of course, you can never know whether you hit it. 
All of this is silly mathematical doodling. A martingale player can quit any time he wants. The odds are hugely in his favor that he will have innumerable (so to speak) opportunities to quit winners throughout his lifetime. But strictly speaking, he's playing a -EV game. So what? If he in all probability never loses, who cares if he's playing a -EV game? You can't force him to play an infinite number of hands--he couldn't do it even if he tried. Silly. While they're arguing how many angels can dance on the head of a needle, he's cashing in his martingale winnings. In a finite sense, it's about as close to a sure thing as you can get.
No, I completely understand what you're saying, I just don't agree.
If (and only if), in infinite trials, there are no wins at all, then the probability of a win is exactly zero. That's what the probability being equal to zero means.
But in the case we're considering, a stipulation is that the probability is winning is greater than zero.
Therefore, the possibility of losing every hand forever is not part of this problem. That's a condition we set at the start. Obviously a martingale can't beat a game you can never ever win once. I think this is where you're hung up.
If you can win even one bet, ever, and have unlimited credit and no table limits, a martingale will always win. It doesn't even need to be a martingale; the next bet just has to be greater than the accumulated losses.
(As for infinite trials, we aren't considering that case either: trials stop at a win, which is always in a finite number of trials.)
fin
What we seem to have is a battle of dueling infinities.
Case A
On one hand it was shown that for an unlimited bankroll a negative EV game's martingale losses will approach infinity. In other words no matter how large a bankroll is defined it still won't be enough and the martingale eventually fails.
Case B
On the other hand is the claim that there is a bankroll big enough that it will never be depleted no matter what, even if martingaled when EV = -99.999999999...infinite number of 9s%.
Case A proves Case B to be false. However proponents of Case B refuse to believe that so there seems not much left to say.
What we seem to have is a battle of dueling infinities.
Case A
On one hand it was shown that for an unlimited bankroll a negative EV game's martingale losses will approach infinity. In other words no matter how large a bankroll is defined it still won't be enough and the martingale eventually fails.
Case B
On the other hand is the claim that there is a bankroll big enough that it will never be depleted no matter what, even if martingaled when EV = -99.999999999...infinite number of 9s%.
Case A proves Case B to be false. However proponents of Case B refuse to believe that so there seems not much left to say.
Sorry, I don't believe it because you are still mistaken. Case A does not prove Case B to be wrong. Case A requires an a priori specification of bankroll size, while Case B defines it to be sized enough to achieve one win.
One very clear error of yours: -99.9999.. is exactly equal to -100. (proof) So you can't have that as a condition of the problem, since there must be non-zero odds of winning. There is no battle of infinities. This must be where you're hung up.
Once again, the problem statement, simplified:
1. I can make any sized bet I want.
2. Every bet is chosen such that, if it wins, I come out ahead overall.
3. I have a non-zero chance of winning a bet.
Any problem with those conditions?
4. If I have a non-zero chance of winning a bet, eventually, I will win it, per my response to QFIT. That's by definition. If I never have a chance to win it, P(win)=0, which violates the condition of the problem.
5. Therefore, if I ever win a bet, I will come out ahead.
6. Since I will always eventually win a bet, I will always (eventually) come out ahead.
Where is the logical violation in the statements above?
I have no problems with people correcting old luminaries, but I'm sticking with Thorp on this one, because he was right. You've done a good job proving your case for a finite bankroll, but the proof does not apply to an infinite (unlimited) one, where the bankroll is always big enough to allow at least one win.
QFIT
Well-Known Member
You cannot make this stipulation. It's like saying that arsenic won't kill you if we make the stipulation that arsenic won't kill you. Of course a system will work if you stipulate that it will work.
I can absolutely make this stipulation, and it's a requirement when assessing the results of any game of chance. If the game can't ever be beat (for example, I only bet "37" on a roulette wheel, or always bust my hands), then I'll always lose no matter what I do. But that's not the case we're considering. The odds of winning must be nonzero or it's not a game of chance.
I didn't make the stipulation that "the system" works. I only made the stipulation that the odds of winning any given trial are not zero. Is that such a stretch? In what game do you always have zero probability of winning?
QFIT
Well-Known Member
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.
This is the way these Martingale debates always go:
Martingale works, just double your bet when you lose and you can’t lose.
No, a losing streak will break you.
OK, no table limits.
Still won’t work.
OK, both I and the bank have $100,000,000.
No, still won’t work.
OK, we BOTH have infinite money.
No, you’ll die.
OK, I will live forever.
No, still won’t work because it’s possible you’ll never be ahead
OK, I state by fiat that I will be ahead at some point, no matter what the math says.
OK, if you state in the rules that you will be ahead, you will be ahead. But, that ain’t a Martingale. (I want to play that game.)
This is how far you must twist reality in order to make Martingale a winning strategy.
This is the way these Martingale debates always go:
Martingale works, just double your bet when you lose and you can’t lose.
No, a losing streak will break you.
OK, no table limits.
Still won’t work.
OK, both I and the bank have $100,000,000.
No, still won’t work.
OK, we BOTH have infinite money.
No, you’ll die.
OK, I will live forever.
No, still won’t work because it’s possible you’ll never be ahead
OK, I state by fiat that I will be ahead at some point, no matter what the math says.
OK, if you state in the rules that you will be ahead, you will be ahead. But, that ain’t a Martingale. (I want to play that game.)
This is how far you must twist reality in order to make Martingale a winning strategy.
