weaselman
weaselman
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January 24th, 2011 at 6:56:39 AM permalink
Quote: Wizard

My argument that the Martingale doesn't work even with an infinite bankroll is inductive in nature. Let's take single-zero roulette and no maximum bet. For any finite bankroll you will eventually bust out. Let M(b) be the ratio of the expected loss to expected total amount bet for a bankroll of b. Simulations will show that M(b)=5.26% for b=1, 2, 3, ... googleplex. Since adding one more unit doesn't change the ratio, I claim that by induction we can say that M(b)=5.26% for any b, including infinity.

I hope jfalk sees this thread. We've argued about this years ago, and never came to an agreement about it.



The induction is applicable to the set of natural numbers. What you have actually proven here is that for any bankroll size that can be expressed as a natural number, martigale does not work. This says nothing about infinite bankrolls though, since infinity is not a natural number (strictly speaking, it's not a number at all, but some more concrete cardinals - like aleph-null for example - can be thought of as numbers in a certain extended sense of the word, but still, they are not naturals).
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WizardofEngland
WizardofEngland
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January 24th, 2011 at 7:06:37 AM permalink
But if your bankroll is 'infinity', do you actually win by getting +1 to your bankroll?

Surely the +1 is insignificant to the point that it does not exist when you add it to your infinate bankroll
http://wizardofvegas.com/forum/off-topic/general/10042-woes-black-sheep-game-ii/#post151727
DJTeddyBear
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January 24th, 2011 at 7:30:16 AM permalink
Quote: WizardofEngland

But if your bankroll is 'infinity', do you actually win by getting +1 to your bankroll?

Surely the +1 is insignificant to the point that it does not exist when you add it to your infinate bankroll

+1 is only insignificant in the imaginary world where an infinite bankroll exists.
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weaselman
weaselman
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January 24th, 2011 at 7:32:08 AM permalink
Quote: WizardofEngland

But if your bankroll is 'infinity', do you actually win by getting +1 to your bankroll?

Surely the +1 is insignificant to the point that it does not exist when you add it to your infinate bankroll



This is a good question - how to define a "win". Perhaps, you can measure changes to the casino's bankroll (which can still be finite), and declare a "win" if it decreases.

Quote: DJTeddyBear

+1 is only insignificant in the imaginary world where an infinite bankroll exists.



That's the one we are talking about though, isn't it?
"When two people always agree one of them is unnecessary"
Jufo81
Jufo81
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January 24th, 2011 at 7:41:19 AM permalink
Quote: weaselman

Let's not get into non-countable sets theory. We'd need at least some calculus and advanced algebra toaccurately define the terms.
In the countable (discreet) case, "improbable" and "impossible" mean the same thing.



Even in countable sets "improbable" (Probability of zero) and impossible are not the same thing. On roulette to get an infinite sequence of reds without ever getting black is "improbable" but still possible with probability zero. On the other hand spinning a yellow number is impossible.
weaselman
weaselman
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January 24th, 2011 at 7:49:17 AM permalink
Quote: Jufo81

Even in countable sets "improbable" (Probability of zero) and impossible are not the same thing.



yes, it is.

Quote:

On roulette to get an infinite sequence of reds without ever getting black is "improbable" but still possible with probability zero.



"Possible with probability zero" == "Impossible"

Quote:

On the other hand spinning a yellow number is impossible.


If you mean that the yellow number is outside of the domain, yes, this is correct ... And applies equally to the "infinite streak" event - these two are equally impossible.

"Impossible" means, that something will definitely not happen within any finite amount of time, no matter how long you wait for it.
"When two people always agree one of them is unnecessary"
P90
P90
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January 24th, 2011 at 7:56:23 AM permalink
Quote: weaselman

"Possible with probability zero" == "Impossible"


When working at infinity, the usual policy is to work with limit function. So there it is never actually zero, and there are multiple (infinitely so) "levels of infinity", beating one another. So it's well possible to arrive to a meaningful result even dividing infinity by infinity.
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FinsRule
FinsRule
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January 24th, 2011 at 7:56:33 AM permalink
Unsurprisingly to everyone I'm sure, I am not convinced. If this was Yahoo! Answers, I think I would pick the Wizard's as being the best, but the inductive answer just doesn't "feel" right.

I think I have thrown too many people off by bringing gambling/roulette/making money into the picture.

This is the question:

If betting systems do not work then the following is true: If a coin is built that has a 50.0000000001% chance of landing on "Tails", there is a possibility that "Heads" will never show up in an infinite number of tosses. But with that same coin, it's impossible that "Tails" will never show up in an infinite number of tosses.

How does that .000000002% difference between the likelihood of events mean that one can occur infinitely, and the other cannot?
HKrandom
HKrandom
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January 24th, 2011 at 7:56:45 AM permalink
The odds of winning 100% of your bankroll (therefore doubling it) before busting are a under 50% no matter what you do.
The odds of winning 50% of your bankroll before busting are under 66.6% no matter what you do.
The odds of winning 10% of your bankroll before busting are under 90% no matter what you do.
The lower percentage of your bankroll you are trying to win, the higher the odds of winning are but if you do it over and over again the odds get smaller. Using martingale until you double your bankroll has a smaller success chance than betting everything on one hand since the martingale exposes more money to the house edge. If your goal is to play once and leave then it might be ok but if you want a decent return on your investment you would have a higher chance of getting it by betting bigger units and doing less progression.
HKrandom
HKrandom
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January 24th, 2011 at 7:58:27 AM permalink
Quote: FinsRule

Unsurprisingly to everyone I'm sure, I am not convinced. If this was Yahoo! Answers, I think I would pick the Wizard's as being the best, but the inductive answer just doesn't "feel" right.

I think I have thrown too many people off by bringing gambling/roulette/making money into the picture.

This is the question:

If betting systems do not work then the following is true: If a coin is built that has a 50.0000000001% chance of landing on "Tails", there is a possibility that "Heads" will never show up in an infinite number of tosses. But with that same coin, it's impossible that "Tails" will never show up in an infinite number of tosses.

How does that .000000002% difference between the likelihood of events mean that one can occur infinitely, and the other cannot?



This answer is inaccurate. Both have a chance of never showing up in an infinite number of tosses; however the chance of tails never showing up is smaller.

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