two conditions
No house limits
player has unlimited funds
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two conditions
No house limits
player has unlimited funds
SBR Founder Join Date: 7/20/2005
Many have said it cannot work no matter what
Just wanted to clear up the uncertainity
SBR Founder Join Date: 7/20/2005
Can you prove it mathematically whilst avoiding falling foul of a gamblers fallacy?
SBR Founder Join Date: 8/10/2005
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.
SBR Founder Join Date: 8/28/2005
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.
SBR Founder Join Date: 7/21/2005
SBR Founder Join Date: 8/28/2005
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.
SBR Founder Join Date: 7/21/2005
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.
SBR Founder Join Date: 8/28/2005
Theoretically only a player's mortality would stand in the way of success.
SBR Founder Join Date: 8/22/2005
Exactly.
Hey why don't we give the player an infinite life-span too?
Or maybe that's just crazy.
SBR Founder Join Date: 7/21/2005
Thanks for the confirmation, it does work. I bet in AC you can use like $200k and beat the casino using this method.
SBR Founder Join Date: 7/20/2005
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 variable in question (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.
- Total Bankroll = 2n - 1
limn → ∞ 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
limn → ∞ 1-52.632%n = 1So 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
limn → ∞ 52.632%n = 0So 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
limn → ∞ $1 = $1So 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:
limn → ∞ $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
limn → ∞ 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
limn → ∞ $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)
limn → ∞ ($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.
SBR Founder Join Date: 8/28/2005
SBR Founder Join Date: 8/10/2005
It "works" because the none of the conditions will ever be true.
CHARITY DONOR
11/27/2011
$25 donation
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.
SBR Founder Join Date: 7/20/2005
SBR Founder Join Date: 8/28/2005
SBR Founder Join Date: 8/28/2005
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.
CHARITY DONOR
11/27/2011
$25 donation
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).
SBR Founder Join Date: 8/28/2005
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."
SBR Founder Join Date: 7/21/2005
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?
CHARITY DONOR
11/27/2011
$25 donation
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.![]()
SBR Founder Join Date: 12/14/2005