Mathematical Proof that Progressions will never Overcome a Negative Expectation Game

[aslan]

Can't agree.

First, unlimited bankroll is meaningless. This means that you have infinite money, the house also has infinite money, and that you both have infinite time. All three are impossible making the situation moot.

Secondly, if you have infinite money, how can you increase it? That is, how can you win?

Thirdly, the cost of money (i.e. interest rate) on an infinite bankroll is infinte. That means, to win, you have to make more than infinite money as your money is decreasing in value by inflation.

Fourthly, in an infinity of time, all things are possible. That is, it is possible that you will lose every bet. And, if it is possible, it must happen in an infinite number of trials.

The argument for infinity cuts both ways, but....

You wouldn't really need infinite bankroll to come out on top in 99.999999% of the time, just enough to get by a very large number of double ups. And hey, I don't anticipate playing for an infinite amount of time. I'd have no time left to spend it.

 

[aslan]

I don't think that's true zg. If you have to include unlimited capital in the equation, then you also have to include the inevitable occurance of never winning another hand before you die. Apply them to a negative EV game, and I think you must die before ending every single losing streak you can have. No??

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?

 

[zengrifter]

Sure, if you just decide to stop somewhere you'll always be able to claim a win at some point. But the series undoubtedly "converges" to EV as k_c shows.

Kudos to KC for trumping Thorp, Griffin, et al ! zg

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

 

[aslan]

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

When you start speaking asymptotically, you have already left the planet. For all practical purposes, convergence upon infinity should not enter a discussion of martingales, since it seldom comes up in the real world (<--that was a joke).

 

[zengrifter]

I don't think that's true zg. If you have to include unlimited capital in the equation, then you also have to include the inevitable occurance of never winning another hand before you die. Apply them to a negative EV game, and I think you must die before ending every single losing streak you can have. No??

Okay, we just added unlimited TIME as well. zg

 

[aslan]

Okay, we just added unlimited TIME as well. zg

They're just pulling your chain. :laugh:

 

[QFIT]

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?

Forget 20 years. It would take vastly more than all the money in the world to ensure you make it through an hour.

Martingale doesn't work. Period.

 

[k_c]

Although I usually agree with KC, this time I think he has bitten off more than he can chew. There is no way on God's green earth that a martingale can lose given no bet limits and unlimited bankroll. If a sure thing cannot beat a house edge, there's something wrong with your math.

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.

 

[johndoe]

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.

If the bankroll really is unlimited, that point never arrives.

For any finite bankroll, you're right.

 

[zengrifter]

If the bankroll really is unlimited, that point never arrives.

KC, I think, says that it does arrive, regardless. zg

 

[QFIT]

If the bankroll really is unlimited, that point never arrives.

If the moon is made of blue cheese, then battery acid is good for the heart.

Equally true.

 

[k_c]

If the bankroll really is unlimited, that point never arrives.

For any finite bankroll, you're right.

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.

 

[aslan]

Forget 20 years. It would take vastly more than all the money in the world to ensure you make it through an hour.

Martingale doesn't work. Period.

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:

 

[aslan]

If the moon is made of blue cheese, then battery acid is good for the heart.

Equally true.

I think that's green cheese. :grin::whip:

j/k :joker: Point taken.

 

[johndoe]

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.

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.

 

[zengrifter]

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".

Yes - unlimited means infinite. I said "without upper boundaries", meaning the same. zg

 

[zengrifter]

Of course, QFIT is absolutely correct that the discussion really is entirely pointless.

No more pointless than some science geeks debating the nonlocality of quantum theory.
No more pointless than for Thorp and Wilson to point out that a martingale will (not "could")
work theoretically sans all upper-boundaries. zg

 

[johndoe]

No more pointless than some science geeks debating the nonlocality of quantum theory.
No more pointless than for Thorp and Wilson to point out that a martingale will (not "could")
work theoretically sans all upper-boundaries. zg

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.

 

[shadroch]

Of course, QFIT is absolutely correct that the discussion really is entirely pointless.[/QUOTE]

Kinda like Glen Beck or Faux News,eh?

 

[zengrifter]

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