Martingale Works 100%

Willie Bee · 07-14-07, 05:58 PM

Originally posted by jjgold

two conditions

No house limits
player has unlimited funds

See JJ, that second one has always been the toughest on me. But good luck with it.

DrunkenLullaby · 07-14-07, 05:59 PM

JJ has seen infinity...and it is good.

jjgold · 07-14-07, 06:37 PM

Many have said it cannot work no matter what

Just wanted to clear up the uncertainity

tacomax · 07-14-07, 06:39 PM

Can you prove it mathematically whilst avoiding falling foul of a gamblers fallacy?

Ganchrow · 07-14-07, 06:48 PM

Originally posted by jjgold

two conditions

No house limits
player has unlimited funds

You really need to define "works" in this context.

Given a casino game with no house limits and a positive house edge (such as, say, roulette) as the player's bankroll approaches infinity, his probability of turning a profit with Martingale approaches 100%. However, when the player profits his bankroll will only increase by an infinitesimally small percentage, while his risk is of course infinite (he potentially risks losing his entire bankroll).

So what we have for the player is a probability approaching 100% of increasing bankroll by an percentage approaching 0% and a probability approaching 0% of bankroll decreasing by 100% (losing an infinite amount).

But anyway you slice it, when playing the Martingale the player expects to lose money. And the more money he has to devote to Martingale, the more he can expect to lose.

The HG · 07-14-07, 06:51 PM

The problem here, I think, is that jj is drawing a parallel between gambling and masturbation (which I didn't realize he had a problem with until the thread about it a few weeks ago), where a Martingale approach does in fact work, assuming the masturbator lives in a state with zero censorship and has unlimited bandwidth.

If you start with just one porn site, any site in fact, it can be chosen at random, if that site does not address a fetish of yours, if you click on two links from that site, and then two links from each of those two sites, and so on and so on, you will eventually land on a site that addresses a fetish of yours.

But this sort of thing does not work in gambling, even in theory, and it's important to understand why it works in one area but not the other, in order to do either one rationally.

Mudcat · 07-14-07, 06:55 PM

JJ is right given the two (absurd) conditions.

Basically it's like the old saying, "If my aunt had nuts she'd be my uncle."

It's true but I'm not sure how much value there is in discussing it in a serious way.

Ganchrow · 07-14-07, 07:08 PM

Originally posted by Mudcat

JJ is right given the two (absurd) conditions.

But he's really not right from the perspective of the player.

Although the player's probability of not losing approaches 100%, he's not actually increasing his bankroll when he doesn't lose. He's just ... standing still.

Mudcat · 07-14-07, 08:04 PM

I think you're over-complicating a bit there. Or maybe we're just working with slightly different definitions.

If the player keeps increasing his bet such that his win amount is always 5 bucks (or whatever he uses for the first bet of the series) more than what he's lost in that series of bets - and he can do that to infinity - then he's going to keep making his 5 bucks for every series of bets.

And If my aunt had infinite wheels she'd be an infinite bicycle.

Ganchrow · 07-14-07, 08:30 PM

Originally posted by Mudcat

If the player keeps increasing his bet such that his win amount is always 5 bucks (or whatever he uses for the first bet of the series) more than what he's lost in that series of bets - and he can do that to infinity - then he's going to keep making his 5 bucks for every series of bets.

As the player's bankroll approaches infinity, the probability of his winning $5 from any given Martingale round approaches 100%. (It doesn't equal 100% mind you, it just gets closer and closer as the player's bankroll increases without bound.)

So what happens when the player wins a Martingale round? Well he wins $5, which increases his bankroll by an amount approaching $5/$∞ = 0% (In other words ... his bankroll doesn't increase at all.) This happens with probability approaching 100%.

So what happens when the player loses a Martingale round? Well he loses an amount approaching $∞, which decreases his bankroll by 100%. This happens with probability approaching 0%.

So to put it another way ... if you win, then your bankroll stays the same (probability approaching 100%); and if you lose you lose an infinite amount (probability approaching 0%). Overall, no matter how large his bankroll, the player's expectation will always be negative, and what's more, the larger the player's bankroll .. the more the player expects to lose on average.

Razz · 07-14-07, 10:43 PM

Theoretically only a player's mortality would stand in the way of success.

Mudcat · 07-15-07, 07:36 AM

Exactly.

Hey why don't we give the player an infinite life-span too?

Or maybe that's just crazy.

jjgold · 07-15-07, 08:31 AM

Thanks for the confirmation, it does work. I bet in AC you can use like $200k and beat the casino using this method.

Ganchrow · 07-15-07, 03:46 PM

Total Bankroll = 2ⁿ - 1
lim_{n → ∞} 2ⁿ - 1 = ∞
So as max allowable Martingale trials approaches infinity, total bankroll necessary would also approach infinity. This should not be surprising.
Win Probability = 1-52.632%ⁿ
lim_{n → ∞} 1-52.632%ⁿ = 1
So as max allowable Martingale trials approaches infinity, the probability of coming out ahead approaches 100%. So in other words, the more money you can devote to Martingale, the more likely you are to win.
Loss Probability = 52.632%ⁿ
lim_{n → ∞} 52.632%ⁿ = 0
So as max allowable Martingale trials approaches infinity, the probability of ending up a loser approaches 0%. So in other words, the more money you can devote to Martingale, the less likely you are to lose.
Potential Win Amount = $1
lim_{n → ∞} $1 = $1
So no matter how large your bankroll, winning a single round of Martingale will pay off $1. This means that your bankroll would increase by the following percentage in the event of a win:
lim_{n → ∞} $1 / ($2ⁿ - 1) = 0%
So as max allowable Martingale trials approaches infinity, a win will increase bankroll by a percentage approaching 0%.
Potential Loss Amount = 2ⁿ - 1
lim_{n → ∞} 2ⁿ - 1 = ∞
So as max allowable Martingale trials approaches infinity, the possible loss that the player faces also approaches infinity. This corresponds to 100% of bankroll.
Total Expected Loss = $1 - (2 × 52.632%)ⁿ
lim_{n → ∞} $1 - (2 × 52.632%)ⁿ = - $∞
So as max allowable Martingale trials approaches infinity, the expected (or average) loss approaches infinity. This means that even though the probability of actually losing a Martingale round is approaching zero, the expectation from playing that round is still infinitely negative.
Total Expected Loss as Percent of Bankroll = ($1 - (2 × 52.632%)ⁿ) / (2ⁿ - 1)
lim_{n → ∞} ($1 - (2 × 52.632%)ⁿ) / (2ⁿ - 1) = 0%
So as max allowable Martingale trials approaches infinity, the expected (or average) loss as a percentage of total bankroll approaches zero. This means that even though the expected loss may be infinite in raw dollar terms, the expected percent impact on bankroll becomes increasingly negilible.

So in other words, if you ran the Martingale process with an infinite bankroll an infinite number of times (meaning approaching infinite in each case), then your average loss per Martingale round would be also be (approaching) infinite.

Saying that all the Martingale needs to "work" is an absence of house limits and unlimited access to funds, isn't just unrealistic, but also mathematically incorrect. Now that doesn't mean that given a casino's realistic risk preferences, it'd grant a player unlimited bet size ... but even if the casino did, it would be hard to consider that somehow "working" for the player when you consider that the player's bankroll will only increase by a de minimis percentage with each successful round and will drop by 100% the tiny percentage of the time that the Martingale failed.

tacomax · 07-15-07, 11:55 PM

Originally posted by The HG

The problem here, I think, is that jj is drawing a parallel between gambling and masturbation (which I didn't realize he had a problem with until the thread about it a few weeks ago), where a Martingale approach does in fact work, assuming the masturbator lives in a state with zero censorship and has unlimited bandwidth.

If you start with just one porn site, any site in fact, it can be chosen at random, if that site does not address a fetish of yours, if you click on two links from that site, and then two links from each of those two sites, and so on and so on, you will eventually land on a site that addresses a fetish of yours.

But this sort of thing does not work in gambling, even in theory, and it's important to understand why it works in one area but not the other, in order to do either one rationally.

Although the probability of finding a site which addresses your fetish approaches 100% the more sites up to infinity you visit, it still doesn't guarantee that you'll find the one to address your fetish. Looks like you're a victim of the wanker's fallacy.

wrongturn · 07-16-07, 11:51 AM

It "works" because the none of the conditions will ever be true.

jjgold · 07-16-07, 01:46 PM

Ganch always sways everything to the negative giving the player no hope of ever winning at anything. All his computations actually do prove the player cannot win squashing though the American Dream of beating sports. Most of us here have nothing but do have slight hope that we can beat sports and maybe buy a new car or house but no he has to put the negative damper on it.

Ganchrow · 07-17-07, 08:20 AM

Originally posted by wrongturn

It "works" because the none of the conditions will ever be true.

No. This is exactly the sort of misconception to which I had alluded.

Whether or not the preconditions set forth by JJ were true, it would be very difficult to consider Martingale "working" for the player in any meaningful sense.

WileOut · 07-17-07, 10:16 AM

It wouldn't work for the player with unlimited money because to the player with unlimited money, winning any amount of money means nothing. Because the player already has "unlimited money", adding anything to his BR adds 0% to his bankroll.

DrunkenLullaby · 07-17-07, 10:21 AM

Originally posted by Ganchrow

No. This is exactly the sort of misconception to which I had alluded.

Whether or not the preconditions set forth by JJ were true, it would be very difficult to consider Martingale "working" for the player in any meaningful sense.

Ganch, I think that all that wrongturn probably meant is that, for example, you can prove that if 1+1=3 then 1+1+1+1=6, but that such a proof is not terribly useful.

Tchocky · 07-17-07, 10:31 AM

"unlimited funds" is really a vague term. I don't think Bill Gates could keep losing money every day without it eventually catching up to him.

Ganchrow · 07-17-07, 10:48 AM

Originally posted by DrunkenLullaby

Ganch, I think that all that wrongturn probably meant is that, for example, you can prove that if 1+1=3 then 1+1+1+1=6, but that such a proof is not terribly useful.

The point is that even if you accept JJ's premise, it doesn't logically imply his conclusion.

It's not an issue of unrealistic assumptions, it's an issue of mathematical fallacy.

DrunkenLullaby · 07-17-07, 10:53 AM

Originally posted by Ganchrow

The point is that even if you accept JJ's premise, it doesn't logically imply his conclusion.

It's not an issue of unrealistic assumptions, it's an issue of mathematical fallacy.

Gotcha.

wrongturn · 07-17-07, 11:34 AM

Ganch, your analysis is really deep. Just a question, if (3) Loss Probability is 0, instead of infinitly approaching 0, then JJ's conclusion is right? Of course that can't happen in reality, probably useless in discussion.

Ganchrow · 07-17-07, 11:47 AM

Originally posted by wrongturn

Just a question, if (3) Loss Probability is 0, instead of infinitly approaching 0, then JJ's conclusion is right?

Well loss probability is never actually zero. The idea is that loss probability approaches zero as one's bankroll approaches infinite.

But it still isn't quite as simple as that -- even though loss probability is indeed approaching zero, expected (that is average) loss still approaches infinity. And to make matters worse, when the player does win he's not actually adding to his bankroll at all (his bankroll is already infinite, hence increasing it by a finite amount is not possible).

Mudcat · 07-17-07, 12:10 PM

What a concept. A dude with an infinite bankroll spending his infinite lifespan trying to make a second infinity dollars 5 bucks at a time.

It feels like there's a Twilight Zone episode in there.

Kind of reminds me of a joke we recovering addicts sometimes tell. A genie grants an alcoholic 2 wishes. For his first wish he says he wants a neverending bottle of whiskey.

POOF! A big, unbreakable bottle full of amber liquid appears. The alky drinks it down, likes it. And then magically the bottle refills itself. Fantastic.

The genie asks what he wants for his second wish.

"I want another one of those."

wrongturn · 07-17-07, 12:19 PM

In defense of JJ, we, the losing gamblers, are not interested in taking $5 a time from casino, are not interested in increasing our bankrolls. We simply want to completely crash the casino and sportbooks one by one by using this system just for the fun of doing it. Right?

AgainstAllOdds · 06-04-08, 06:42 PM

Martingale can be hard to use sometimes. I have found its not worth it in the long run.

mathdotcom · 06-04-08, 06:44 PM

Youre an idiot for even trying

Dark Horse · 06-04-08, 06:47 PM

Didn't read all posts, so probably somebody already mentioned the obvious: nobody in his right mind is going to bet 100K to get his original 100 bucks back.

We're back at streaks. The true key to money management.

SlickFazzer · 06-04-08, 06:48 PM

Originally posted by jjgold

two conditions

No house limits
player has unlimited funds

Some impressionable young gambler will follow your words JJ, thinking they have no house limits and that they can come up with unlimited funding.

And yes, they will be barreled in by the 4th of July.

VegasDave · 06-04-08, 06:50 PM

Assuming you had infinite wealth, you wouldn't really have to worry about martingales or anything else for that matter, would you?

crisp · 06-04-08, 06:55 PM

why would somebody bump this thread from a year ago?

staf · 06-05-08, 10:39 AM

I prefer a modified D'Alenbert system for sports gambling.