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Thread: brandnewtobj: Martingale System

  1. #1
    brandnewtobj
    Guest

    brandnewtobj: Martingale System

    Why does the Martingale System not work?

    The articles that I've read on it say it is because you will hit the table maximum. But, if I hit the table maximum, can't I just get up and find another table with a higher maximium? Isn't it just a matter of having enough money?

  2. #2
    Parker
    Guest

    Parker: Re: Martingale System

    > Why does the Martingale System not work?

    > The articles that I've read on it say it is because
    > you will hit the table maximum. But, if I hit the
    > table maximum, can't I just get up and find another
    > table with a higher maximium? Isn't it just a matter
    > of having enough money?

    We live in a finite world. Sooner or later you hit the highest maximum or run out of money.

    Bottom line is that no betting progression will turn a negative expectation game into a positive expectation one. Play long enough, and the house edge catches up to you.

  3. #3
    victoria
    Guest

    victoria: Re: Martingale System

    > We live in a finite world. Sooner or later you hit the
    > highest maximum or run out of money.

    > Bottom line is that no betting progression will turn a
    > negative expectation game into a positive expectation
    > one. Play long enough, and the house edge catches up
    > to you.

    Martingale does work if you have just two little things working for you, but you must have both.
    A casino with no limits
    A bankroll also with no limits.

    So if you have both go for it, if not forget it.


  4. #4
    Parker
    Guest

    Parker: What I always wondered . . .

    > Martingale does work if you have just two little
    > things working for you, but you must have both.
    > A casino with no limits
    > A bankroll also with no limits.

    > So if you have both go for it, if not forget it.

    If someone could possibly have an infinite bankroll, what would be the point of playing at all?

  5. #5
    brandnewtobj
    Guest

    brandnewtobj: Re: What I always wondered . . .

    > If someone could possibly have an infinite bankroll,
    > what would be the point of playing at all?

    In theory one doesn't need to have an infinite bankroll, because, in theory, you won't experience an infinite number of losing hands. There is a limit to the number of losing hands in a row. What is it? That's the question.

  6. #6
    Norm Wattenberger
    Guest

    Norm Wattenberger: Martingale fails even with no limits

    It really has nothing to do with limits. The fact is you have .5% against you on every bet you make. The amount that a Martingale player on average will lose is .5% of the amount bet. Period.

  7. #7
    victoria
    Guest

    victoria: Re: Martingale fails even with no limits

    > It really has nothing to do with limits. The fact is
    > you have .5% against you on every bet you make. The
    > amount that a Martingale player on average will lose
    > is .5% of the amount bet. Period.

    Norm, if you play a martingale starting at $5 and quit each and every time that you are up $5, you will land up a winner. You might have to bet 100 million at some point to land up $5 ahead, but you will be ahead if you leave after that win.
    The house edge percentage only seems to make sense but only with table limits and player loss limits and over his lifetime of loss, he should not average .5% but average (since the vast majority of players playing martingale do not play perfect basic strategy) whatever negative amount his play is times the amount bet.

    Victoria

    Victoria

  8. #8
    Magician
    Guest

    Magician: It depends what you mean by "fail"

    > It really has nothing to do with limits. The fact is
    > you have .5% against you on every bet you make. The
    > amount that a Martingale player on average will lose
    > is .5% of the amount bet. Period.

    Not so fast. It depends what you mean by "fail".

    The Martingale player's expected win after n hands is indeed -0.5%*n. However, after any winning hand, the player is at a new all-time high. With an infinite bankroll, the player requires an infinite losing streak in order to be wiped out. The probability of this is zero (which is not to say that it is an impossibility). The probability of the player winning a hand at some point in the future, and thus hitting a new all-time high, is one (which is not to say that it is a certainty). Is a system which has a probability of one of improving your bankroll a failure?

    With a finite bankroll, there exists some finite losing streak that will wipe out the player. Unlike before, the probability of this streak happening eventually is one. In this case, the probability of reaching a new all-time high before being wiped out is about 49.75%. The more times the player tries to increase his bankroll, the more likely he is to be wiped out.

    In the sense of "probability of improving your bankroll", Martingale "succeeds" with no limits but "fails" otherwise. Of course, a system with positive expectation would "succeed" either way.

  9. #9
    Norm Wattenberger
    Guest

    Norm Wattenberger: Re: It depends what you mean by "fail"

    > With an infinite
    > bankroll, the player requires an infinite losing
    > streak in order to be wiped out. The probability of
    > this is zero (which is not to say that it is an
    > impossibility).

    Right. But, the amount lost is infinite. And what is infinity times zero? You must add that into the sum of all losses when computing the overall EV.

    If you have an infinite number of trials, all things that can happen must happen. Since an infinite losing streak can happen - it will happen given infinite trials. So, to compute the EV you calculate the Limit of -0.5%*n as n approaches infinity. Clearly this is a negative number since n is positive and a positive times a negative is always negative.

  10. #10
    brandnewtobj
    Guest

    brandnewtobj: Re: Martingale fails even with no limits

    > It really has nothing to do with limits. The fact is
    > you have .5% against you on every bet you make. The
    > amount that a Martingale player on average will lose
    > is .5% of the amount bet. Period.

    Nothing to do with limits? The probability of losing every hand you play from now to forever is 0. Therefore, with no limits, the probability of losing with the Martingale System is 0.

  11. #11
    Norm Wattenberger
    Guest

    Norm Wattenberger: Calculation of the absurd

    You must be very careful when you throw infinity into any calculation. The probability of losing every hand is not zero. It is infinitesimal. Do you think this an absurd distinction? Well is it any less absurd than the concept of an infinite bankroll or infinite time? You can't allow one absurd sounding attribute of infinity into your calculation while ignoring another.

    > Nothing to do with limits? The probability of losing
    > every hand you play from now to forever is 0.
    > Therefore, with no limits, the probability of losing
    > with the Martingale System is 0.

  12. #12
    Magician
    Guest

    Magician: Re: It depends what you mean by "fail"


    > Right. But, the amount lost is infinite. And what is
    > infinity times zero? You must add that into the sum of
    > all losses when computing the overall EV.

    Yes, I see what you are saying. But the "infinity times zero" bit reminded me of a Slashdot article (link below).

    > If you have an infinite number of trials, all things
    > that can happen must happen. Since an infinite losing
    > streak can happen - it will happen given infinite
    > trials. So, to compute the EV you calculate the Limit
    > of -0.5%*n as n approaches infinity. Clearly this is a
    > negative number since n is positive and a positive
    > times a negative is always negative.

    I did concede that the EV was negative in my first sentence and proceeded to show that despite this you had a probability P=1 of getting ahead. I then suggested that a system with this property might not deserve to be called a failure.

    But I've decided that in fact EV is all that matters. If you start with an infinite bankroll and make finite bets there are only two possible outcomes: either you still have an infinite bankroll or you have nothing (i.e. "fail"). To fail you have to play an infinite number of hands and you have to have a negative EV. So I'm agreeing with you now (and I should have known better).

    Anyway, no more absurdness from me. If I had an infinite bankroll and infinite time I wouldn't spend it playing blackjack.



  13. #13
    brandnewtobj
    Guest

    brandnewtobj: Re: Calculation of the absurd

    > You must be very careful when you throw infinity into
    > any calculation. The probability of losing every hand
    > is not zero. It is infinitesimal. Do you think this an
    > absurd distinction? Well is it any less absurd than
    > the concept of an infinite bankroll or infinite time?
    > You can't allow one absurd sounding attribute of
    > infinity into your calculation while ignoring another.

    I stand corrected, the probability of losing every hand one plays from now to eternity is > 0. It could happen. So one could still end up a loser using the Martingale System even with the infinite bankroll and no limit on the bet size.

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