Mathematical Proof that Progressions will never Overcome a Negative Expectation Game
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We can put some limits on it. For example, given 20 years more of life and a bankroll that is not unlimited, but extremely large. All this infinity nonsense puts everything out of range. For all practical purposes, a very large bankroll and no bet limits cannot be beat. If the genie grants me this, when can I start? 
Its paradoxical, or at least asymptotic - that without upper boundaries the martingale can always cheat 'death' or 'gravity' so to speak with another bet -the housEdge only prevailing as the trials converge upon infinity.
Oops, there's that word infinity again. Brother Shad and Brother Sage, you may take it form here. zg
Oops, there's that word infinity again. Brother Shad and Brother Sage, you may take it form here. zg
aslan said:
We can put some limits on it. For example, given 20 years more of life and a bankroll that is not unlimited, but extremely large. All this infinity nonsense puts everything out of range. For all practical purposes, a very large bankroll and no bet limits cannot be beat. If the genie grants me this, when can I start?
Martingale doesn't work. Period.
Let's assume martingale works for a negative EV game.
Worst possible EV in a single payout 1:1 odds game is -100% (sure loss.) Even you should agree that a martingale can't possibly work when there is zero probability of winning even once. What if EV = -99.9999999999999999999%? Now martingaler has at least a miniscule chance of winning so given enough time and bankroll he'll eventually experience a win. Would martingale work in this case?
In order for martingale to succeed access is needed to funds to cover all cumulative losses plus 1 for next bet. A martingale failure means that no matter how large initial bankroll is then there will eventually be a point where martingaler will be unable to pony up the next bet and will be out a humongous sum that is more than his incidental wins.
Worst possible EV in a single payout 1:1 odds game is -100% (sure loss.) Even you should agree that a martingale can't possibly work when there is zero probability of winning even once. What if EV = -99.9999999999999999999%? Now martingaler has at least a miniscule chance of winning so given enough time and bankroll he'll eventually experience a win. Would martingale work in this case?
In order for martingale to succeed access is needed to funds to cover all cumulative losses plus 1 for next bet. A martingale failure means that no matter how large initial bankroll is then there will eventually be a point where martingaler will be unable to pony up the next bet and will be out a humongous sum that is more than his incidental wins.
I don't know how to put it. Even an unlimited bankroll won't work. An unlimited bankroll is finite in the sense that once you start it is fixed. If you add to it after that then that is mathematically cheating. However it is still unlimited since it can be arbitrarily large. A martingale failure means any arbitrarily large bankroll will eventually be insufficient.
I find that hard to accept, Qfit. Not to be argumentative, but 99.999% of the time is good enough for me (unless my immortal soul is at stake). It's the same with probabilities, that is, nothing is for mathematical certainty--a person may not reach the long run in twenty lifetimes, but for practical purposes we use probability theory for card counting. A run of 20 losses at an even money bet is about a million to one shot, I believe. A run of 30 consecutive losses at an even money bet is about a billion to one. It sounds to me like one could use a martingale at an even money bet with no more risk than crossing the street at a traffic light. Where am I going wrong? I'm assuming, say, a hundred billion dollar bankroll at a $100 min game. lol Hmmm! Pretty boring stakes for a multi-billionaire. :laugh:
If that's how you're defining "unlimited", then you're correct. I was defining it as "infinite"; that's how I took it to mean from Thorp and the others that (supposedly) claimed it "could work". When finite, arbitrarily large vs. arbitrarily small doesn't really matter; it's just a matter of time and it's only a scaling factor.
Of course, QFIT is absolutely correct that the discussion really is entirely pointless.
Of course, QFIT is absolutely correct that the discussion really is entirely pointless.
Nonlocality in quantum theory is a very real (though grossly misunderstood) effect, which certainly merits discussion (though perhaps not here).
I'd say Thorp and Wilson pointing out that with an infinite bankroll you can "win" is still totally pointless. Just because such statements came from well-known figures in the field does not make them important or useful.
I'd say Thorp and Wilson pointing out that with an infinite bankroll you can "win" is still totally pointless. Just because such statements came from well-known figures in the field does not make them important or useful.
No one in this thread suggested that it was important or useful that negative progressions can always prevail in the absence of upper boundary. Right? no one suggested otherwise?
I must say, KC really had me going when he threw out that proof that even without limits a martingale could not prevail in the end. I conceded! z:laugh:g
I must say, KC really had me going when he threw out that proof that even without limits a martingale could not prevail in the end. I conceded! z:laugh:g