[jjgold]

two conditions

No house limits
player has unlimited funds

[Willie Bee]

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]

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

[jjgold]

Many have said it cannot work no matter what

Just wanted to clear up the uncertainity

[tacomax]

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

[Ganchrow]

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]

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]

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]

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]

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]

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]

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

[Mudcat]

Exactly.

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

Or maybe that's just crazy.

[jjgold]

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

[Ganchrow]

It always seems that misconceptions abound when discussing Martingale. This thread is no exception.

The claim that somehow Martingale "works" for a player with a large enough bankroll (and a sufficiently risk-neutral casino) is no less than completely inconsistent with the straightforward mathematics that underpin the system.

First off, check out the following table which shows some Martingale stats for players with various starting bankrolls. The table assumes the player is playing an even odds bet on an American Roulette wheel (single game win probability of 47.368%), and is using a Martingale size of $1.

Total Bankroll	Martingale Size	Win Probability	Loss Probability	Potential Win Amount	Potential Loss Amount	Total Expected Loss
$1	1	47.3684%	52.6316%	$1	$1	-$0.05
$3	2	72.2992%	27.7008%	$1	$3	-$0.11
$7	3	85.4206%	14.5794%	$1	$7	-$0.17
$15	4	92.3266%	7.6734%	$1	$15	-$0.23
$31	5	95.9614%	4.0386%	$1	$31	-$0.29
$1,023	10	99.8369%	0.1631%	$1	$1,023	-$0.67
$32,767	15	99.9934%	0.0066%	$1	$32,767	-$1.16
$1,048,575	20	99.99973%	0.00027%	$1	$1,048,575	-$1.79
$33,554,431	25	99.999989%	0.000011%	$1	$33,554,431	-$2.61
$1,125,899,906,842,620	50	99.9999999999988%	0.0000000000012%	$1	$1,125,899,906,842,620	-$12.00
$1.268 × 1030	100	(1-1.332×10-26)%	1.332 × 10-26%	$1	$1.268 × 1030	-$167.90
$1.809 × 1075	250	(1-2.049×10-72)%	2.049 × 10-72%	$1	$1.809 × 1075	-$370,763.96
$3.273 × 10150	500	(1-4.200×10-142)%	4.200 × 10-142%	$1	$3.273 × 10150	-$137,466,652,005.42
$5.922 × 10225	750	(1-8.606×10-212)%	8.606 × 10-212%	$1	$5.922 × 10225	-$50,967,817,073,015,500.00
$1.072 × 10301	1000	(1-1.764×10-281)%	1.764 × 10-281%	$1	$1.072 × 10301	-$18,897,080,413,852,900,000,000.00
$2n - 1	n	1-52.632%n	52.632%n	$1	$2n - 1	$1 - (2 × 52.632%)n

Now in mathematics it's not actually proper to talk about what happens at a bankroll of infinity, insofar as infinity is really a limiting concept rather than an actual number. What we can talk about however, is what starts happening as bankroll grows arbitrarily large. This is accomplished by taking what's known as the limit of a function as the independent variable (in this case, the bankroll) gets really, really, really large (i.e., grows towards infinity).

So let's look at what happens to each of the following as Martingale size (the number of permitted rounds of Martingale, we'll call it n) approaches infinity.

1. 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.
2. 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.
3. Loss Probability = 52.632%ⁿ

   lim_{n → ∞} 52.632%ⁿ = 0

   So as max allowable M