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Roulette EC: no less than 380hit in 1000spins?!

Discussion in 'Roulette Forum' started by beat-the-wheel, May 4, 2015.

  1. beat-the-wheel

    beat-the-wheel Member Founding Member

    Joined:
    Feb 1, 2015
    Likes:
    1
    It very interesting probability fact,
    that being posted by ,
    one of the greatest mind of gambling in the world,
    namely, Albalaha, recently .

    ================
    Albalaha says:

    quote:EVERY 100 spins may not have at least 38 wins but as a whole, surely in 1000 spins u will have more than 380 wins. Think of something that can win u in 380/1000, in any manner. Then u can play in sessions of 100 spins or 200 or any length, for that matter.
    unquote...
    -------------------------

    I am surprised, as I never read this kind of fact, anywhere in the net.

    While all the time we argued about the stale EDGE, the toward-the-mean and all the mathematically facts by the math proffesors, we begin to FORGET that the fact of "LONG SPINS",

    ....which we conveniently IGNORED!
    or just do not see that way....
    instead of the stale 'edge, bla,bla,bla...
    since first casino open shop!

    And no one seems to THINK in that way.

    There never less than 380hits in 1000spins!

    When Albalaha say, never less than 380hits/1000spins EC, no one can dispute that!

    It have the casino EDGE factoring in.
    It have the MEAN, factoring in.
    it have all the bla-bla-blah factoring in!

    No dispute .PERIOD.


    And when he says, the long spins,
    could be bet as many mini seasons,
    no one even asked, HOW?
    ==============


    as I understand it.
    What I think...

    if...
    1000spins divided by 20spins round, =50 sets.
    Thus each 20spins probability=0/20 worst to 20/20 best.

    Thus 50 sets of 20spins will have no less than 380hit combined.

    And averages 50set of 20spin, each set of 20spins will have 7.6 hit /20spins.

    now every gentlemen who worth his salt, sugar, and spice,
    had now learned something different.
    Something different to ponder.


    If 20hit in 20spins...
    thus minimum 380hit/20spins,
    20spins to have 20/20,
    as best=19 max,
    which of course=impossible.

    If u do not understand,
    what being said by Albalaha,
    then u still in the early stage of learning
    the facts of gambling.

    and as usual, I am not selling anything
    or try to make any single cent from anyone here,
    or affiliate with anyone.
    What u post here reflect your
    very own temperament, EQ,skill , idiotically,and understanding.
     
  2. beat-the-wheel

    beat-the-wheel Member Founding Member

    Joined:
    Feb 1, 2015
    Likes:
    1
    Now, if we fully understand that in next 1000spins
    will have no less than 380hits,
    then HOW are we going to bet for profit?

    1000spin divided by 20spins =50sets of bet.
    380/50=7.6hits. averages.

    max hit per 20spins=20hits.=20/20.
    min. hit per 20spins=NONE=0/20.

    Cut losses are necessary,
    Thus In the 50 round of bet,
    there will be profit, fabulous profit,
    loss,horrible losses,
    and whatever staking u applied,
    at the end the combined profit
    must more than the combined losses...

    We see that the averages hit,
    if 380/1000 do happen, is 7.6

    If we take 7/20 as a line of loss and profit,
    then how are we going to profit?

    This mean we must win X if >7/20.
    and lose Y if <7/20.
    [we take 7hit/20 as threshold for profit/loss]

    And Albalaha hint that Y losses,
    should be able to be recoup by two to three X win.

    So in NEXT 50 round of bet ,
    we will see that around 15 will produce losses,
    and the other 35 will produce profit,
    and at the end will produce nett profit...ONLY IF we
    know how to bet them!

    ================
    we may bet 1000, 2000,3000 etc.
    and 1000 may be divided by 20,15,10,5 spins...as your preference.

    Say , if we choose 1000 divided by 10spins set,
    that will be 100 round to plat, and
    each 10spins,
    risk=0/10, to 10/10, with 3.8 averages per set,
    THUS how u gonna make huge profit and minimise losses,
    when each 10spin, with 3 or 4 hit as threshold will be crucial!
     
  3. beat-the-wheel

    beat-the-wheel Member Founding Member

    Joined:
    Feb 1, 2015
    Likes:
    1
    so if we bet 1000/10spins, we need to bet 100 round.
    And here my thought.

    if the averages=3.8spins/ 10spins,
    and we take 3/10 as threshold to LOSE,
    and 4/10 as Huge profit, then we need ..

    LOSS=X
    PROFIT=Y

    and the X should be recoverable by two to three Y....
    Then we on our way to win the whole 100 set-off bet.

    If we hit, the averages of , say, 480/1000,
    that the hits after the EDGE, bla-bla-bla,
    as stressed by the math professors,
    then we have 4.8/10 spins.

    Now we have understand why we cant win long term,
    as the averages is just 4.8/10
    in NORMAL occurrence,
    and with the variance, random, bla-bla-bla,
    we bound to lose!

    Many thanks to Albalaha!
    If not for him and his generous information,
    in his blog,
    we may not understand this way of betting.
    Albalaha, you're a genius!
     
  4. Jesper Svensson

    Jesper Svensson Member

    Joined:
    Apr 9, 2015
    Likes:
    1
    If 385 which is more than a third is the limit, a progression should overcome, if we add to the rules we can bet zero valu chips
    as soon we are a head. Frankly 615 loss in a row and then the winnings are easy to overcome. It sounds a bit strange
    after 615 losses (not in a row) say, the ball can only make winnings now for the last spins up to 1000 (bet max :)).
     
  5. beat-the-wheel

    beat-the-wheel Member Founding Member

    Joined:
    Feb 1, 2015
    Likes:
    1
    we do not try to bet from the very 1st spins to last 1000spins...
    Try to chop down the 1000spins into 50 session of 20 spins...
    thus 20spins times 50=1000spins.
    Then try to bet every 20spins to win X amount.
    Cut loss when losses = 2x...stop.
    start a new 20spins session...win =x, stop.

    1st. 20spins=lose=-2X.
    2nd 20spins=lose=-2x
    3rd 20spins=lose=-2x
    4th 20spins=win=X
    5th 20spins=win=X
    and so on...
    till 50th 20spins...

    at the end after all 50 session, being bet...
    all the fifty, -2X and X, add up produce a profit.

    of course the above example is an OVER-SIMPLIFIED!

    There are a progression, and skipping method.

    we must have a method where -2x losses recoup by two to three winning 20spins session.

    1000 will no less 380hit.
    380/50=7.6hit averages.
    thus 20spins will have 7.6 hit averages.
    think how to win X, in 7 hit.
    and lose -2X, if <7hit cut losses...

    thats a grail.....hopefully.
     
  6. Jesper Svensson

    Jesper Svensson Member

    Joined:
    Apr 9, 2015
    Likes:
    1
    Have a look at which is the longest run you ever see in 1000 spins. Follow the last and use a reversed Grand Martingale catching the losses.
    a run of 8 catch up 247 a run of 9 741. You will hardly see any 1000 spins without a run of seven. I have used it and never seen it. Not tested old spins of millions thought. With patients I think you will beat a NOZ every time. We will need a spread of at least 1000 units. Playable? maybe not, but would you lose?
     
  7. albalaha

    albalaha Active Member Founding Member

    Joined:
    Dec 29, 2014
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    122
    Occupation:
    player
    Location:
    India
    Martingale or grand martingale are good for nothing. Similarly, its reverse form is no guarantee to win. Even if we get a run of seven once in a thousand spins, we can not earn enough. Remember, positive progression is no guarantee of win. Before trying marty or grand marty we will have to makeup our minds for the attempts to try. Say, we look for a run of 7. It will give us a 6 step win only. Calculate yourself, what would happen in that case.
     


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