Mathematical Proof that Progressions will never Overcome a Negative Expectation Game

[aslan]

I have been thinking about it in another way.

Let's set a group of infinite number of players all using the Martingale strategy. Whoever wins one unit stops playing. Among this group of infinite number of players, there must be one player who never wins as he approaches playing infinite number of hands. The amount of money he is losing is infinite, more than the total win from all winners combined. As a result of this unluckiest player in the world (hopefully not me), the whole group will end up in the red.

:laugh::laugh:

Infinite thanks for that!

Or conversely, the others with infinite bankrolls can take up a collection to pay all his living expenses while he loses forever more. Sort of like an infinite dole out or spreading of the wealth. (Note to self: Send a message to the Obama administration.)

By the way, just when do you approach infinity? I figure at any point along the way you are still infinitely far from the end. :laugh:

 

[zengrifter]

Let's set a group of infinite number of players all using the Martingale strategy. Whoever wins one unit stops playing. Among this group of infinite number of players, there must be one player who never wins as he approaches playing infinite number of hands. The amount of money he is losing is infinite, more than the total win from all winners combined. As a result of this unluckiest player in the world (hopefully not me), the whole group will end up in the red.

Now that is the sort of out-of-box thinking we like in the Voodoo zone.
All we need to do is fire the unlucky ones before they drag down the whole team! zg

 

[k_c]

It's splitting hairs, but the proof shows that for any finite bankroll (regardless of how large) it's an unwinnable system. If the bankroll is infinite (what I meant by unlimited, and presumably what the others meant), you can win any arbitrary amount you want, by expanding your bets by any ratio desired (i.e. triple after a loss). You're guaranteed to be ahead at some point, after which you can stop with whatever win you choose.

This does not, however, asymptote to a +EV for infinite trials.

My point of view is that what is being shown is that if you start and commit to play a martingale to the bitter end then there is a negative expectation regardless of size of bankroll.

Once you run out of funds with the original bankroll it is mathematical cheating to say you still can succeed. If that's the case then all you need to do is secure another unlimited bankroll, which would need to be at least large enough to cover all present losses plus 1, and continue playing. This could be done an infinite number of times. Each of those new unlimited bankrolls also has a negative expectation and your first bet with the new bankroll will be a humongous amount. All you need is one win but it may never come. Saying that it must come is akin to the gambler's fallacy. All of the previous failed unlimited bankrolls do not make the next unlimited bankroll any more likely to succeed than the first.

btw I'm not greedy and would be happy with just one infinite bankroll.

 

[johndoe]

My point of view is that what is being shown is that if you start and commit to play a martingale to the bitter end then there is a negative expectation regardless of size of bankroll.

And I agree with you.

Once you run out of funds with the original bankroll it is mathematical cheating to say you still can succeed. If that's the case then all you need to do is secure another unlimited bankroll and continue playing. This could be done an infinite number of times. Each of those new unlimited bankrolls also has a negative expectation. All you need is one win but it may never come. Saying that it must come is akin to the gambler's fallacy. All of the previous failed unlimited bankrolls do not make the next unlimited bankroll any more likely to succeed than the first.

btw I'm not greedy and would be happy with just one infinite bankroll.

It's not "mathematical cheating" - having an unlimited (infinite) bankroll was a specific condition posed by the question. I'm just defending Thorp's (apparent) statement that if there was no bankroll limitation, you could play Martingale until you have a single win, and quit while you're ahead. And that single win is guaranteed to come eventually; its probability is approaches unity as the number of hands goes up.

The "cheat" in this case is that there's a bound set on the win (once), but no bounds on anything else ($, hands). Of course you'll hit positive at some point, because there's no other limitation ("end") to the problem, mathematically, until that happens.

Ultimately, it comes down to only this: Is the probability of (at least) a single win, given as many hands as it takes to get one, unity? Of course it is - by definition of the problem. ("as many hands as it takes")

(Hell, I'd settle for only half of an infinite bankroll!)

 

[k_c]

And I agree with you.



I'm just defending Thorp's (apparent) statement that if there was no bankroll limitation, you could play Martingale until you have a single win, and quit while you're ahead. And that single win is guaranteed to come eventually; its probability is approaches unity as the number of hands goes up.

That statement doesn't seem to make sense for a negative EV game.. If true you could play a martingale and quit n number of times for a guaranteed n unit profit. I think the statement works for positive EV but falls apart for negative EV.

 

[zengrifter]

That statement doesn't seem to make sense for a negative EV game.. If true you could play a martingale and quit n number of times for a guaranteed n unit profit. I think the statement works for positive EV but falls apart for negative EV.

Take another stab at providing a math-proof that unlimited BR cannot prevail with martingale.
I think the martingale will always prevail, without limits. Is that a paradox? zg

 

[psyduck]

By the way, just when do you approach infinity?

Just keep playing without bathroom breaks.

I figure at any point along the way you are still infinitely far from the end. :laugh:

What end?

 

[zengrifter]

Take another stab at providing a math-proof that unlimited BR cannot prevail with martingale.
I think the martingale will always prevail, without limits. Is that a paradox? zg

As an afterthought, I suspect that a math-proof that martingale will always work, without limits, would be the case. zg

 

[iCountNTrack]

And that single win is guaranteed to come eventually; its probability is approaches unity as the number of hands goes up.
you'll hit positive at some point, because there's no other limitation ("end") to the problem, mathematically, until that happens

I am sorry but this is not true that is basically gambler's fallacy even if you lost 100 hands in a row the probability of winning on the very next hand is still the same

 

[johndoe]

That statement doesn't seem to make sense for a negative EV game.. If true you could play a martingale and quit n number of times for a guaranteed n unit profit. I think the statement works for positive EV but falls apart for negative EV.

It does work though, per the reasons above. However, in the interim, the potential losses are unbounded.

 

[johndoe]

I am sorry but this is not true that is basically gambler's fallacy even if you lost 100 hands in a row the probability of winning on the very next hand is still the same

It's very different than the gambler's fallacy. You're right that the odds don't change from hand to hand, but as long as the odds of winning are anything other than zero, you'll absolutely win a hand eventually, given enough hands.

Since (with infinite bankroll) you don't care about the cost of reaching this point, you'll always win in the end. It doesn't even matter what the EV is, as long as it's not zero.

Think of it this way: What's the probability of flipping a fair coin to heads, at least once, given an infinite number of trials? It's absolutely unity.

 

[zengrifter]

Since (with infinite bankroll) you don't care about the cost of reaching this point, you'll always win in the end. It doesn't even matter what the EV is, as long as it's not zero. Think of it this way: What's the probability of flipping a fair coin to heads, at least once, given an infinite number of trials? It's absolutely unity.

Therefore a martingale is a winning strategy EXCEPT when muzzled by 'artificial limitations.'
In its "natural unbounded state" a martingale will always prevail over the housEdge. zg

 

[k_c]

Take another stab at providing a math-proof that unlimited BR cannot prevail with martingale.
I think the martingale will always prevail, without limits. Is that a paradox? zg

By employing a martingale in a negative EV game you will wind up with less than you started with in the long run regardless of bankroll size. If you are truly committed you will eventually be bankrupt regardless of bankroll size.

In a positive EV game an unlimited bankroll using a martingale is a virtual guarantee that you will be able to grow your bankroll 1 unit at a time forever.

I would be surprised to learn that Thorp said that a martingale virtually guarantees a 1 unit win in a negative EV game. If he said this it is wrong.

I wouldn't be surprised if he said that a martingale virtually guarantess a 1 unit win in a positive EV game. This is right.

 

[zengrifter]

Therefore a martingale is a winning strategy EXCEPT when muzzled by 'artificial limitations.'
In its "natural state" a martingale will always prevail over the housEdge. zg

Therefore the title of this thread is false >>

Mathematical Proof that Progressions will never Overcome a Negative Expectation Game​

 

[psyduck]

nothing special about martingale

With a large enough (not infinite) bank and without any other limiting factors, ANY betting system works as well as martingale if the ONLY goal is to quit when a profit (regardless of how small) is made.

Assuming an infinite number of players all play martingale, all it takes to prove the betting system does not work is to have one player who never wins (losing infinite amount of money). Among the infinite number of players, one such never-win player will exist.

 

[iCountNTrack]

Therefore the title of this thread is false >>

Mathematical Proof that Progressions will never Overcome a Negative Expectation Game​

No offense ZG but i think you should stick to conspiracy theories.

 

[zengrifter]

No offense ZG but i think you should stick to conspiracy theories.

Truly. zg

 

[zengrifter]

With a large enough (not infinite) bank and without any other limiting factors, ANY betting system works as well as martingale if the ONLY goal is to quit when a profit (regardless of how small) is made.

There is no quiting with limitless BR -
- the martingale will go on making money forever in a negative EV. zg

 

[zengrifter]

I would be surprised to learn that Thorp said that a martingale virtually guarantees a 1 unit win in a negative EV game. If he said this it is wrong.

Yes, Thorp, Griffin, Wilson, Epstein, Vancura. And not just a single 1u and quit.
The martingale will win forever. It would be hard to find a authority who would
not agree.

KC - take another stab at the math-proof.
It would be a real feather in your cap to trump the gambling Ph.Ds and Sc.Ds! zg

 

[aslan]

With a large enough (not infinite) bank and without any other limiting factors, ANY betting system works as well as martingale if the ONLY goal is to quit when a profit (regardless of how small) is made.

Assuming an infinite number of players all play martingale, all it takes to prove the betting system does not work is to have one player who never wins (losing infinite amount of money). Among the infinite number of players, one such never-win player will exist.

Sounds like pretty good odds to me. With counting you're not guaranteed to win within your lifetime either.