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Roulette 20 EC permutation

Discussion in 'Roulette Forum' started by beat-the-wheel, Apr 27, 2015.

  1. beat-the-wheel

    beat-the-wheel Member Founding Member

    Joined:
    Feb 1, 2015
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    1
    Hi respected gentlemen,
    and due respect to my Guru.

    ==================
    Anyone has the programming ability,
    to see, calculate
    how...

    If we bet,
    123456789,10,11,12,13,14,15,16,17,18,19,20 stop.
    progress add one, whether win or lose

    TWENTY spins of EC PERMUTATION,
    albeit ZERO.

    Then bet all 20spins permutation, and see how the loss, or win sum.

    say...
    first possible permutation=worst=20 losses in row
    1] LLLLLLLLLLLLLLLLLLLL=-210u=x1

    2nd possible..
    2]LLLLLLLLLLLLLLLLLLLw=-170u=x2

    3rd possible..
    3]LLLLLLLLLLLLLLLLLLww=-132u=x3

    4] and so on...

    then add up all x1,x2,x3.......to x10000
    and see how the win lose of total sum of x.

    ===========================
    this way, we could see, how worst win/loss perform,
    if we bet them as a season...
    say..
    first season,

    all losses!..in row
    1] LLLLLLLLLLLLLLLLLLLL=-210u losses!


    2nd season, best permutation=all win! in row

    2]wwwwwwwwwwwwwwwwwwww=+210 win!

    With a list of long permutation, u may see how to create a risk/reward
    method to bet sytematically.

    thanks.
     
  2. Jesper Svensson

    Jesper Svensson Member

    Joined:
    Apr 9, 2015
    Likes:
    1
    Could work many times! I do something like that. Even on straight up. Use pennies ad a chip every spin until hit, if not on plus go on until.
    Break long sleepers, win most of the time, but it may be a loss of 10000 (100 Euro). The winning can be more than double.
    Frankly, we can not win allways, but a brave and for the bet a suitable progression it may work.
    Do not use scary Money, if you are not able to lose them without much problem you will not win.
     
  3. albalaha

    albalaha Active Member Founding Member

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    Dec 29, 2014
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    Occupation:
    player
    Location:
    India
    @beat-the-wheel,
    Since wins and losses will be equal in all these steps taken together, a simple D'alembert or Oscar's grind will fetch you a net win somewhere but there is no guarantee that u will finish a net win too. If you get more of wins to begin with and later successive losses, that can lead you to a net loss, at the end.
     
  4. Jesper Svensson

    Jesper Svensson Member

    Joined:
    Apr 9, 2015
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    1
    Today I got 200 Euro. My starting bank, which I could lose was 100 Euro. That's a bankroll of 10000 cents. I used BV zero and set the fair-Control at 60. Then I bet one cent at one number. add a cent every spin until a hit. Every time I did 60 spins, I change the number, to make it difficult to know my number from the casino. The highest pile was 230 cent. On a hit, and not break even, I reduced the pile with 36 chips.
    If we hit sleepers dispite changing numbers every 60 spins, we ar in an unlucky game and will lose. I have managed long sleepers, which awake use to hit better. Playing cents is not very low-rolling using the method, you need at least 10000 chips, and even if it is cents, your bets are mostly Euros. When the numbers hitting close to each other, a longer sleeper will be managed.

    It may with a suitable bank, be better to start with more than one chip, in case of early hit.

    If we have not more than a third of a chanse to tripple the bankroll using a method, we can as well set all on NOZ doz in one spin.o_O

    Most of all progressing is variants of Martingale or d'Alembert. One can handle long losing streaks and the are sensitive to inbalance.
     
  5. beat-the-wheel

    beat-the-wheel Member Founding Member

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    Feb 1, 2015
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    Hi Albalaha,
    U say,
    .
    "Since wins and losses will be equal in all these steps taken together.."
    ============
    Thus, I think u mean, that if we bet "every spin of the permutation", then the result will always equalized, and with the green SABOT thrown in!, then there will disaster in waiting!
     
  6. albalaha

    albalaha Active Member Founding Member

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    Location:
    India
    We can expect an equilibrium theoretically in a no house edge game and there is an equilibrium in all my combinations taken together but practically, a counterpart of an EC can be ahead of other and may remain so, even after a few million spins.
    We may remain above equilibrium, at equilibrium or below equilibrium for very long time and can't guess that even in a game of no house edge.
     
  7. beat-the-wheel

    beat-the-wheel Member Founding Member

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    Hi Albalaha,
    Thus,
    since all the permutation, may not appeared/hit EQUALLY,even after million spins,
    then , as u said,

    "....may remain above equilibrium, at equilibrium or below equilibrium for very long time and can't guess that even in a game of no house edge."

    Then I think, the best method will be wait, skip and hit/ run..when the opportunity arise?
    Some sort of variance management?

    Please tell more.
    looking forward to it.
    Thanks.
     

  8. albalaha

    albalaha Active Member Founding Member

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    Occupation:
    player
    Location:
    India
    No trigger can guarantee you more wins than losses and the order in which wins and losses will appear.
     

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