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Topic: A progression that can't lose  (Read 13011 times)

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Offline AsymBacGuy

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A progression that can't lose
« on: May 11, 2016, 07:19:31 PM »
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  • We know that any progression will get the best of it whenever a zero equilibrium point will be reached within a fair amount of trials. Of course some progressions could do even better, that is getting the player a profit even when the W/L ratio is shifted toward the right.
    Notice that the well known D'Alambert progression will win 1 unit after the equilibrium is reached but not everytime as everything depends about the DISTRIBUTION of W and L.

    Here I'm talking about the almost absolute impossibility to lose our entire bankroll and this is a total different thing than stating that we will win easily. Nonetheless knowing that we won't lose in the longest possible runs isn't a vulgar accomplishment.

    I have to forcely consider a $100 standard unit bet and the total bankroll is $6600 (66 units).
    For simplicity we won't take into account the commission when applied.
    Remember that our goal is to reach at a given point a zero equilibrium point, meaning we want to get the W/L ratio = zero.
    Later more on that.

    Columns are: L deviations, betting amount in $, financial exposure, gain after the equilibrium will be reached

    0  $100           100   -
    1  $100 + $10 210 10
    2  $100 + $20 330 30
    3  $100 + $30 460 60
    4  $100 + $40 600 100
    5  $100 + $50 750 150
    6  $100 + $60 910 210
    7  $100 + $70 1080 280
    8  $100 + $80 1260 360
    9  $100 + $90 1450 450
    10 $100 + $100 1650 550
    11 $200 + $10 1860 660
    12 $200 + $20 2080 780
    13 $200 + $30 2300 780
    14 $200 + $30 2430 910
    15 $200 + $30 2760 910
    16 $200 + $30 2990 910
    17 $200 + $40 3130 1050
    18 $200 + $40 3370 1050
    19 $200 + $40 3610 1050
    20 $200 + $40 3850 1050
    21 $200 + $50 4100 1200
    22 $200 + $50 4350 1200
    23 $200 + $50 4600 1200
    24 $200 + $50 4850 1200
    25 $200 + $50 5100 1200
    26 $200 + $60 5360 1360
    27 $200 + $60 5620 1360
    28 $200 + $60 5880 1360
    29 $200 + $60 6140 1360
    30 $200 + $60 6400 1360
    31 $200 + $60 6600 1360

    We see that to lose our entire bankroll we need either a 5.56 sr negative deviation (like looking at 31 negative hands in a row, a 31 streak) or, most likely, a W/L gap of 31.

    Every roulette player knows that a gap between even chances could easily reach and surpass the W/L amount (btw a 31 streak is a very very very rare finding also at this game) but at baccarat we have a lot of ploys to find two opposite events that cannot reach the 31 negative (or less likely positive)value by any fkn means.
    Especially if we want to prolong the progression by another 10 or so steps. 

    So we know that adopting this slow progression we can't lose or, better sayed, that the probability to lose is really very very low, let's say almost impossible.

    And, wonder of wonders, with proper adjustments we may use it betting only the Player side, hence knowing that we won't pay a bit of commission.

    In a word, we can even regularly bet the unfavourable side knowing that we can't lose itlr.

    A further example why we have to play slowly and with a lot of patience.

    as.   

     

       

     

         

       

       
    Winners are simply willing to do what losers won't


    Offline Tomla

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    Re: A progression that can't lose
    « Reply #1 on: May 11, 2016, 08:20:33 PM »
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  • are you saying bet 1 unit 10x then 2 units 20x?

    Offline soxfan

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    Re: A progression that can't lose
    « Reply #2 on: May 11, 2016, 08:57:55 PM »
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  • There is an Armenian cat on a member only dice forum that has come up with a progression style that is, imho, nearly unbreakable. That's cuz it covers the mathematical expectation of a certain event popping within so many toss of the cubes. So, if the action at the dices table always play out according to expectation then he would never bust a progression. But there are practical reasons it's hard to play in that you need a minimum 10 thousands unit bankroll, and they stones to make the max bet of slightly more than 800 unit. And ya gotta have access to a joint that gives you a nice fat spread between min-mx bets, hey hey.
    It's a grind, baby, but a profitable grind, hey hey.

    Offline 21 Aces

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    Re: A progression that can't lose
    « Reply #3 on: May 11, 2016, 09:06:04 PM »
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  • The gain after the equilibrium will be reached looks incorrect.  If you bet $110 the 2nd bet and your total value it risk from the start with that bet is $210 and you win then you are at -$100 P&L (-$100 1st Bet + $110 2nd Bet = -$100).
    Life is something you dominate if you're any good. - Tom Buchanan

    Offline AsymBacGuy

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    Re: A progression that can't lose
    « Reply #4 on: May 11, 2016, 09:31:00 PM »
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  • are you saying bet 1 unit 10x then 2 units 20x?

    Nope. You have to start the progression whenever you get some losses and the first column (# of losses) dictates how mush to bet.
    You keep staying at the same level if you win up to the point where you get the zero equilibrium point, in that case you get a profit.
    If you lose, you keep track of the losses (again column #1) and act accordingly.

    Of course a perfect progression/regression should start at a higher standard bet working counterwise, let's say $200. Whenever you win you go down the same you would do with the L progression.

    When we win the house owns something from us as well as we expect something from it after some losses. 

    as.

     
       
    Winners are simply willing to do what losers won't

    Offline AsymBacGuy

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    Re: A progression that can't lose
    « Reply #5 on: May 11, 2016, 09:32:53 PM »
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  • There is an Armenian cat on a member only dice forum that has come up with a progression style that is, imho, nearly unbreakable. That's cuz it covers the mathematical expectation of a certain event popping within so many toss of the cubes. So, if the action at the dices table always play out according to expectation then he would never bust a progression. But there are practical reasons it's hard to play in that you need a minimum 10 thousands unit bankroll, and they stones to make the max bet of slightly more than 800 unit. And ya gotta have access to a joint that gives you a nice fat spread between min-mx bets, hey hey.

    Yep, limiting the bankroll should be of paramount importance at the cost to play a very slow game.

    as.
    Winners are simply willing to do what losers won't

    Offline AsymBacGuy

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    Re: A progression that can't lose
    « Reply #6 on: May 11, 2016, 09:38:04 PM »
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  • The gain after the equilibrium will be reached looks incorrect.  If you bet $110 the 2nd bet and your total value it risk from the start with that bet is $210 and you win then you are at -$100 P&L (-$100 1st Bet + $110 2nd Bet = -$100).

    ??

    210 is the loss amount if you lose both 1st and 2nd hand.
    If you are at a -210 level, you need to win two hands at level #3 ($200 + $30) getting a gross profit of $30. 

    No way the progression can be wrong, it was carefully studied more than one century ago.
    But unfortunately it was devised for roulette where it cannot work by obvious reasons.   

    as.



    Winners are simply willing to do what losers won't

    Offline 21 Aces

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    Re: A progression that can't lose
    « Reply #7 on: May 11, 2016, 09:59:55 PM »
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  • Ok I now understand you are going for two consecutive wins after the losses.  Then I don't understand the last column 'gain after the equilibrium will be reached'.  you are just trying to find two wins after a possibly huge series of losses?
    Life is something you dominate if you're any good. - Tom Buchanan

    Offline AsymBacGuy

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    Re: A progression that can't lose
    « Reply #8 on: May 11, 2016, 10:11:43 PM »
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  • As many times mentioned, a progression must hold up to the most likely variance features any game will provide. And talking about 50/50 games, variance cannot act other than on W/L or on/off outcomes.

    It's reasonable to put the variance limit into a deviation range of -25 and +25 (5 sr).
    Here we got an even higher amount, a 5.56 level.

    Never forgetting that the winning or losing sequences must be always put in relationship with the ideal zero point.
    Hence a -18 overall losing sequence following a +12 situation isn't a sort of statistics disaster: here we're just 6 step far from the zero point.

    In this example, starting the L progression at the zero point would be a far better idea than starting it at the +12 level (6 losses vs 18 losses).
    That's why we have to adjust the progression even on the W side, this time lowering the bets.

    as.     

     

       
    Winners are simply willing to do what losers won't

    Offline AsymBacGuy

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    Re: A progression that can't lose
    « Reply #9 on: May 11, 2016, 10:29:28 PM »
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  • Ok I now understand you are going for two consecutive wins after the losses.  Then I don't understand the last column 'gain after the equilibrium will be reached'.  you are just trying to find two wins after a possibly huge series of losses?

    Nope, but I understand as I haven't clarified well the topic.

    The first column is the most important. It tells us how many losses we got.
    That is HOW MANY BETS I'M BEHIND TO GET THE EQUILIBRIUM (W=L)

    If I'm losing 7 bets (consecutively or not, what it counts is the overall total) I'll have to stay at this level for 7 bets as I'll get the equilibrium after those 7 bets. Giving me a profit of $280.

    In a word, we have to take into account the L number and bet what that level dictates up to the equilibrium point.

    If we're at level #7 and we win the first hand we keep playing this level up to the equilibrium. If after a win at a given level we lose, we don't go to #8 level as the number of losses hasn't change (always 7). Only if we'll have a first loss at a given level we'll go to #8 and so on.

    Obviously final levels can get a profit even though we won't reach the equilibrium point.

    But under normal circumstances, it would be a mistake to stop the searching of the eq. point as we want to get a profit PROPORTIONALLY placed to the risk involved.


    as. 
     

     


     
    Winners are simply willing to do what losers won't

    Offline 21 Aces

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    Re: A progression that can't lose
    « Reply #10 on: May 11, 2016, 10:50:24 PM »
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  • Now I understand.  This is somewhat how I win back.  If I am building a net loss on mini and midi, I have frequently gone into rally mode and decisively won back on midi.  I have never gotten that deep, and my best sessions never really went too rough at any point.

    If you are having difficulty it is because it is most likely a bad combination of you and difficult progressions.  All it takes is some progressions that are straight forward and clear to you and you can roll.
    Life is something you dominate if you're any good. - Tom Buchanan

    Online alrelax

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    Re: A progression that can't lose
    « Reply #11 on: May 11, 2016, 10:53:18 PM »
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  • As, The question is, what do you do when there is a black panther running through the casino followed by black horses and their henchman followed by Vanilla Ice followed by dogs and cats on skateboards???  1-3-2-6 and pump it up to $1,000 on the first?????
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    Offline Albalaha

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    Re: A progression that can't lose
    « Reply #12 on: May 11, 2016, 10:59:33 PM »
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  • Asym,
                 You started with a wrong direction. There is no equilibrium in a game of house edge as every bet is subjected to that and in long run, all bets will go far from equilibrium in terms of "extra losses". Variance can take them even more far. Even in a game without any house edge, a bet might not get equilibrium even after a billion trials.
                   D'alembert is a classic comedy of errors and based on ideas that do not work in real life. It has no mathematical basis to make it a winner.
    Email: earnsumit@gmail.com - VIsit my blog: http://albalaha.lefora.com

    Offline 21 Aces

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    Re: A progression that can't lose
    « Reply #13 on: May 11, 2016, 11:03:24 PM »
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  • Asym,
                 You started with a wrong direction. There is no equilibrium in a game of house edge as every bet is subjected to that and in long run, all bets will go far from equilibrium in terms of "extra losses". Variance can take them even more far. Even in a game without any house edge, a bet might not get equilibrium even after a billion trials.
                   D'alembert is a classic comedy of errors and based on ideas that do not work in real life. It has no mathematical basis to make it a winner.

    This reality only exists in the chambers of The Dark Wizard.  Please exercise extreme caution or he may summon his Black Riders to come and get you!
    Life is something you dominate if you're any good. - Tom Buchanan

    Offline roversi13

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    Re: A progression that can't lose
    « Reply #14 on: May 12, 2016, 01:56:27 AM »
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  • 31 B or P in a row is "almost" impossible,but a gap W/L of 31 is frequent.
    Mathematically, in 992 hands (31.31 + 31) a gap W/L of 31 has more than 50% probability to occur ,without never reaching the equilibrium W/L in the meantime.