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Roulette RE: Mathematical Averages by juneau

Discussion in 'Roulette Forum' started by BETJACK, Oct 12, 2018.

  1. BETJACK

    BETJACK Active Member

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    Mathematical Averages by juneau

    ....If the odds in a given situation are one-in-four that one will pick correctly; pay attention to the fact that the expectation of consecutive misses is 4 to support the one-in-four picks.....therefore, pay attention to the fact that all one has to do is view the target but not commit at that first miss, leaves us at 3 misses & 4 to X without pairing or picking correctly successively; but when we do subsequently pick correctly according to the odds we are left with a 25% advantage in the game.....I can prove it!

    .....there is of course the eternal variance.....however, what game are you playing & what is the percentage of the game do you actually have in play?....I can make a comparison here taken from actual certified by the casino spins: in 24/7 x 31 days: my biggest variance with 2.70% @ work was 5.7 hits off the math average.....I'm reminded of other story's about sleeper numbers going -10.0 & upwards to -20.0...now, the wheel rolls along at a calculable average rate: we can track it; chart it; predict the future based upon mathematical probability averages certain to occur. I am in fact in a long term exercise that returns an advantage of +25% on your money.....whats wrong with that? I am of course a flat bettor...

    ....Nobody wants to talk to me.....I do in fact have the true Holy Grail.....anybody?...

    .....do the math.....what is your expectation of consecutive losses in a one-in-four game to pick correctly your allotted once in four picks.....

    .....you are correct that the odds do not change from one spin to the next.......explain please how then one comes up with their entitled 1/4th or 25% of the action if the misses do not count: i.e. one X per three misses over all; and, where you show at least one miss, one will show an average of four misses: Do the math....if one can ignore the first miss one is still at 3 misses and 4 to X for one's mathematical entitlement but does so without pairing or picking correctly successively according to the odds; but when does subsequenty pick correctly according to the odds one is 25% in advantage over the game....prove me wrong.....

    ....very well.....we shall take this step by step.....that you will be agreeable ALL ALONG THE WAY is assured.....I will be offering irrefutable guarantees....stay tuned.....

    ....there's no martingale here.....I'm talking about predicting the future with mathematical probability ...... I notice that nobody has come up with an explanation: re my earlier post Thursday..... do your simulations, 1/4, what is the average number of misses showing
    between the last X & the next X ?.....and how would you deal with it: I've already explained how I deal with it...

    ....O.K. with the odds at one-in-four: in 104 picks, if one picks every time one will expect to show an average of 26 X's or correct picks & 78 0's or misses in those 104 picks......in a fair game where one is given +3 to one for every correct pick one would of course break even...i.e. 3 0's or misses for every X posted....people say they see no patterns......I see mathematical patterns galore...I see that with at least one miss between the last and next X an averages of four misses.....i.e.: X0000X...so, simply view but not commit at that first miss after the last X and one is left with: 000X..... one breaks even with the odds of 3 misses for every X.....however, you will subsequently "pair" or pick correctly according to the odds at your 17th +3...34th +3...51st...+3....etc....picks....keep counting...it goes on forever....i.e: 000X-even...000X-even...000X-even...000XX+3
    ....now, I,m picking my average: 000X, and then "pairing" according to the odds after my 4th X....eveything I've done is just to pick according to the odds:....right or wrong?....

    ....someone has asked what happens when my selection wins when I don't bet....say what: I always bet my selection.....let's try it this way: do your math i.e. you will find that it is calculable that in a 1/4 game your average consecutive misses is FOUR... LABEL YOUR MISSES READING LEFT TO RIGHT: A1, B2, C3, & D4.....PICK AT B2.....WHEN A1 shows you merely start over..... you are always picking at 1/4...


    ....dear sir: with my method of play the wheel itself tells me what numbers to bet....since only 8.9 different numbers will show in 10 spins of the wheel I can only win 24% of the time X four = 96%....not enough to break even.....however, my 4th pick at B2 = 120%, because my 4th pick has the calculable average of the necessary four consecutive misses preceding it...you only pick ONCE AFTER AN X...then wait for another one...i.e. it looks like this: X now pick once (0).....X now pick once (0).....X now pick once (0).....X now pick once....now there is your calculable +3 because we now have your necessary expected consecutive 4 losses in a 1/4th game......and your 4th pick now has to be on average an X....one thing about mathematical averages, they are extremely dependable....I can do +20% in roulette...I can get rid of the house 25% advantage in Keno in a one-in-four bet....I can do +16% when dealing with the hard ways or pairing the die in Craps....or how about "the big wheel", an exciting idea....no cards in this project....I don't know how much more I can explain this, you might try listing what I've said here in a vertical posting & it should become perfectly clear.....I pick 10 single consecutive times, one number at a time in roulette...( my 4th pick @ 24% = 120% because I have the necessary predictable 4 missed average just prior to my pick....Do the math: G/N at 25% is a four miss average consecutive misses supporting our pick, our pick coming in at five presumed attempts....(only four for us)

    ....well, I can see that nobody here can prove me wrong when I say that G/N (1/4) = a predictable consecutive average miss of 4!....if nobody here can understand that predictable is a beautiful word when dealing with games of chance.....I'm out of here.....adios.....
     
  2. BETJACK

    BETJACK Active Member

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    .....can't anybody on this forum understand basic mathematics: the odds are 1-4....you will average 1 in 4, 2 in 8, 3 in 12, 4 in 16 we are even; now pair that 4th X at the 17th pick on average....prove the math is wrong....
     
  3. Sharptracker

    Sharptracker Well-Known Member

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    Alias The rebel of mathematics...
     
    Last edited: Oct 26, 2018
  4. BETJACK

    BETJACK Active Member

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    this guy knows something but the performance of the explanation is something lame
    or beyond our understanding of things
     
  5. juneau

    juneau New Member

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    .....why is it that nobody proves what I have illustrated to be wrong? anyway: one looking at a lateral posting of X's & O,s indicating "hits" & "misses" resulting from random decisions in some peoples opinion looks like "chaos", no predictable patterns can possibly be found....although we cannot predict exactly the point when an average will occur, we sure as hell can predict the rate it will do so.....i.e. 8.9 different #s on a 37# wheel will as a matter of a calculable mathematical fact appear on average in 10 consecutive spins of the wheel: i.e. 1.1 "repeaters".....we are not pursuing repeaters, our interest is that it is an irrefutable fact that on our 37# wheel we have an irrefutable guarantee that 24% of the available #s will appear on average in any 10 consecutive decisions posted......that's all that appears, that's all anybody gets. It is necessary that I explain the math first or you will not understand in the second phase of this project: do your simulations at the easier to use 25% or 1-4, & see how close you get, or, get real with your algebra and lemmas 1,2,3..etc. and prove me right that in a 1-in-4 game the average expectation of consecutive losses = 4, and, guess what? this particularly new predictable mathematical statistic is "hugely" vulnerable to exploitation.....
     
  6. Sharptracker

    Sharptracker Well-Known Member

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    I play black, i should hit at a 1/2 rate...

    OK... i'm coming and i got more than 20 red... ouch, it took everything i ve collected that was on average for such a long time...

    You cannot predict how variance will be with you otherwise it would be too easy, we would just wait a sleeper to coming back to normal or hot...

    Don't forget that an average stay an average and variance is the derivation of that average which can be very volatile at roulette (positive or negative)

    But anyway i know you won't care mate and you'll need Dame Roulette take off you pant to believe it...
     
    BlueAngel likes this.
  7. BETJACK

    BETJACK Active Member

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    He plays numbers...
    and you are right
    it is not that simple
    as you make it to be...
    but the problem is that I still
    can not understand what @juneau is talking about but it is not your example...

    i need to think MOOORE about this.
     

  8. Sharptracker

    Sharptracker Well-Known Member

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    he's speaking about mathematical average, hit rate... all bets are concerned...
     
  9. juneau

    juneau New Member

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    .....the soul of the 37# wheel: "once your average # shows (any #)...it is calculable it will re-show by the 10th consecutive spin 24% of the time....so all #'s will do so once they appear....so: 24% of all the available #'s on the 37 # wheel will show on average by the 10th consecutive spin....convince yourself that the foregoing is an irrefutable mathematical fact & I'll show you how to take advantage of this fact to the tune of a +20% player advantage over the house in roulette.....variance is a fact in any mathematical exercise, we still however will come to realize our 24% math entitlement.....we flat bet....10 consecutive times & are correct 24% of the time.....take note on another angle: people say that if you miss a pick @ 1-4 the odds don't change....which is true, however one will get "credit" for that miss on our scoreboard of 3 average misses for every X posted, or, said otherwise: where you show at least one miss between the last X & next X your average will be 4 misses.....again: simply view the target but not commit at that first miss you are still at 1-4 to break even with 3 000's for every X, your mathematical entitlement...& "pair" your attempts according to the odds...i.e. at the 17th pick, 34th pick, 51st pick, etc....all +3 units when you do.....
     
  10. BETJACK

    BETJACK Active Member

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    the Quote is what I understand from what you say.
    Ok You're right ... 24% of all numbers.
    It is not important that one or two are repeated
    so what's up then ...
    more numbers come out.
    more 50% of all numbers up to 24 turns?
    what do we do with repeaters ...?

    this 1-4 .... the spins divide it into 4x9 = 36?!
    or 1-4 ... 10 spins +10spins...+10 spins +10 spins
    What about the groups ?
    why A1 B2 C3 D4?
     
  11. BETJACK

    BETJACK Active Member

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    Do you understand what @juneau ... talking about?
    how to make this 1-4 ?
    how to make the groups?
     
  12. John Blerg

    John Blerg Well-Known Member 👹 Troll 👹

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    You don't get tired of posting all this elaborate technical mathematical statistical average stuff why don't you go implement and play it and make money to the max??????????????

    Cause it cannot hold true to theory in the real live casino!!!!!!!!!!!!!!!!!!
     
  13. BlueAngel

    BlueAngel Active Member

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    In short what he says is to bet what is missing, old time fallacy, specifically when a 9 wheel sector misses for 4 spins flat bet it till you are in profit or broke.
    This is what he suggests in simple words, I don't blame him, he is French.
     
  14. Dr. Sir Anyone Anyone

    Dr. Sir Anyone Anyone Well-Known Member Lineage to Founders

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    This is stupid.
     
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  15. Sharptracker

    Sharptracker Well-Known Member

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    @BA French from Alaska??? By the way why do you surestimate french like this? :)

    @Dr Sir Not more not less than any other EV-system actually...The stupid thing is actually to calculate a virtual edge according to the actual results.
     
  16. Dr. Sir Anyone Anyone

    Dr. Sir Anyone Anyone Well-Known Member Lineage to Founders

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    Unfortunately for Juneau the law of the third Indicates that there are one or two too many pockets in the wheel for it to win. He would need a 35 pocket wheel in order to win.

    I doubt he comprehends it.
     
    Last edited: Oct 28, 2018
  17. John Blerg

    John Blerg Well-Known Member 👹 Troll 👹

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    Math wizes for school book math! Lol!!!!!! Be a professor in teaching, 100,000,00 in salary and the benefits!!!!!!!!!!
     
    Last edited: Oct 28, 2018
  18. juneau

    juneau New Member

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    .....this method of play has nothing to do with cards.....O.K. BETJACK, your'e the only one who hasn't dismissed my claims outright or called me stupid.....Listen up: first a simple explanation of A1,B2,C3,D4: calculating at the easier 25% (1/4): in 400 attempts A1 will show an average of 100 times; the 300 times it doesn't it moves on to B2 & it shows it's 75 times (25%)....if again a no-show on to C3 showing it's 25% of 225 attempts....if a no show then on to D4 etc....when either A,B,C,D, do appear according to the odds we start over at A1....now, take this regime and apply it to: in 3,700 spins our average # is calculated to show 100 times (our 2.70%) now deduct your 100 "hits" from 3,700 you have left 3,600 x 2.70 at B, & on & on: after any "hit" at any letter we start over at A1.....
     
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  19. Dr. Sir Anyone Anyone

    Dr. Sir Anyone Anyone Well-Known Member Lineage to Founders

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    Juneau,

    The house payout is still short of your odds of winning. It won't work.
     
  20. John Blerg

    John Blerg Well-Known Member 👹 Troll 👹

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    Numbers played in a real casino will make u lose, really and simply!!!!!!! N
     

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