Martingale for SNGs

[runner4life7]

i agree that there is a one in a huge number chance that you will lose every time and i know that this system cannot work, but with infinity tries you will win, that or I do not understand the concept of infinity, which is quite possible.

[Mr_J]

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What if you used the martingale system on both red and black? Then every time the wheel stops your winning one bet.

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Not sure what you mean??

[runner4life7]

i think what he means is like have them be two seperate systems. So first you bet $1 on each and if its red than you bet 2 on black and 1 on red the next time and so forth. I see this just reducing variance a little maybe? but obviously has the same downfall.

Edited to apologize for bumping, and didnt want to bump again, I got side tracked on this because i was a moron and tried something similar on BJ and while i know it fails, find it interesting.


re-edited to reply without bumping:

i did not mean because you can be ahead at some point its a winning game but rather that at some point in each individual thing you will win and hence make $1. As long as you always play till you get that win you will be ahead. This is different then just quitting when you are ahead, and is all pointless anyways because it cant happen so im finally done with this thread.

[curtains]

Cmon stop bumping this retardo thread, its on 4 pages already!! I take back what I said about the forum being better.

$1 on red
$1 on black

wheel spins: lands on black =>you win $2

$2 on red
$1 on black

wheel spins: lands on black=> you win $2

$4 on red
$1 on black

wheel spins: lands on red=> you win $8

$1 on red
$2 on black

wheel spins: lands on black => you win $4

you've wagered a total of $13 and won a total of $16

Net: $3
And you spun 3 times (discounting the first one which will be breakeven.) So you're EV should be $1 per spin.

[applejuicekid]

The sum of a series of -EV wagers will never be positive no matter how you sequence them. It is really silly to think that the order in which you make your bets will make you a winner. If you took every wager you made in a Martingale system and ordered them from largest to smallest and wagered them in that order would you still be winning?

Let's say you had an infinite bankroll. Each time you want to win a $1. You win $1 then repeat. Even if you had an infinite bankroll you will be making wagers lager than all the amount you have won more often then you will be up money.

Saying that you will be up at one point and thus it is a winning system is silly. Even with infinite bankroll and infinite tries you could never be up. The same argument could be said about any -EV game. If I play blackjack for an evening I am often times up at some point. Does this mean if I quit once I get up I have found a way to beat blackjack?

[pergesu]

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And you spun 3 times (discounting the first one which will be breakeven.) So you're EV should be $1 per spin.

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rofl you can't base your EV on the results of three previous trials. You can't base your EV on any results of previous trials. That's absolutely ridiculous.

[FlyWf]

And when it lands on O or OO?

[GrekeHaus]

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Cmon stop bumping this retardo thread, its on 4 pages already!! I take back what I said about the forum being better.

[/ QUOTE ]

Sorry to bump this retardo thread, but I have the answer to the question.

The problem with this question is that it is a divergent series. This is why people are confused and arguing about it. Given an integer N, you will at one point be up N units and at one point be down N units with probability 1. This is because no matter how well you've done so far, the probability that you will never go on a streak that will win or lose you that much is precisely 0.

Because of this, it is impossible to say anything about what will happen over an infinite amount of time. It is simply a question of the way you choose to group the terms when you sum them.

[Guthrie]

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this is a stupid debate because there is no such thing as an infinite bankroll.

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Congress would disagree.