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TurboGenius Law of Thirds.....

Discussion in 'TurboGenius's Forum' started by TurboGenius, Nov 8, 2020.

  1. Smitridel

    Smitridel Active Member

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    Thanks for the example.
    But it certainly isnt what this thread is about.
    A drawdown of this magnitude can also be caused by any other method even by picking the first four and sticking with them until in profit.
    Pick any repeater method you want.
    Since you added progression into the mix anything is possible given enough spins..

    Unless I cant see something hidden in there, I really dont see how this contributes to your percentages examples since you're counting on the 24% percentage of the first of the first to get to profit..
     
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  2. Smitridel

    Smitridel Active Member

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    For what its worth here's my take on the whole repeater theory.

    Repeaters are a mental construct.
    They dont exist unless you wish them to be (because one's repeaters may be another's starting numbers and bla bla etc)

    There are indeed many things abstract enough in life such as the construct of Time that are observer-dependant, but that is another conversation for another forum maybe..

    All in all, this law of thirds can be applied anywhere inside a frame of thirds in roulette.
    That said, you CAN just limit your bets and your numbers on smaller subsets that should be in accordance to your bankroll.

    You dont need to cover all the table like TG has done in the link he so politely offered.

    Most online tables are a bit (or a lot) biased towards the house, so any attempt to even prove it may be futile.
    Now the B&M ones are a bit ambiguous if you ask me..MrJ can shed some light on it.

    The theoretical aspect of it is interesting but we wont really know whether we made it, or we were just lucky (personally I have a loooooong list zipped in file of successes in testing AND in Real play but I've also had my share of failures).
     
    Last edited: Nov 17, 2020
  3. Gigi666

    Gigi666 Active Member

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    As repeats go I don't think I've seen anything like this outside of RS
    upload_2020-11-18_17-20-51.png
     
  4. Gigi666

    Gigi666 Active Member

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    Can we focus on getting more players to get to a "working method" rather than the back and forth, someone is this or that?

    Turbo would you care to comment why in your example on RS you play on 1st number showing in a street even though it has low chance of repeating before any of the other 2 show?
     
  5. Luckyfella

    Luckyfella Well-Known Member

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    LOTT is actually the frequency distribution of roulette spins. When we apply math based betselection the positive edge bet can be clearly seen in this session I played. 100k units profit in 1 continuous session.
     

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  6. Gigi666

    Gigi666 Active Member

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    100k units would suggest your unit =1$? in 203 spins? Care to expand on this?
     
  7. Luckyfella

    Luckyfella Well-Known Member

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    I bet inside numbers using 100units betsize

    Took me 2hrs to play this session. Since it's the weekend I put in this once off effort.

    I posted the long term graph on my thread on Steve's forum.
     
    Last edited: Nov 22, 2020

  8. Gigi666

    Gigi666 Active Member

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    Thanks Lucky, but RF is bit messy, not sure what's your nickname there. Do you expand on your bet selection (I assume you wait 37 spins to see how the numbers split into 3 groups).?
     
  9. thereddiamanthe

    thereddiamanthe Well-Known Member

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    Turbo, when you were doing this count → table arrays
    did you use the 3/3 DZ as 1/3?

    As otherwise there is a discrepancy between the arrah & actual betting on every spin.


    Also, very important to me, how would you quantify the presented principle relevant to the parachute betting?
     
  10. gianfranco

    gianfranco New Member

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    hi turbo, did you write the dgt file for rx. i wuld like to test it.
     
  11. thereddiamanthe

    thereddiamanthe Well-Known Member

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    Lucky, if you mind ..
    this being the most important thread .. thus including the most important principle(s) .. I'd like to start hereof.

    I've attempted various betting models on dzs, but so far didn't encompass that probability with them to result in flat bet positive expectation.

    One of the models --
    based on 50% dozen binomial distribution probability = 1 hit in 3 spins.
    Seems to fit/match the 'other two yet unappeared dz's ≈74% overall advantage.
    Bet 2u on the first spin (congested time), bet 1u on the 2nd = 1 hit in 3 positions.

    First dz known → bet other 2DZ flatbet 1u/each, hit =profit, restart
    First dz known → bet other 2DZ flatbet 1u/each, no-hit → bet the remaining DZ, hit =cycle break-even restart
    First dz known → bet other 2DZ flatbet 1u/each, no-hit → bet the remaining DZ, no-hit =cost 3u/pop

    Hmmm .. ofc there's a complication
    • 1st dz repeats or Zero appears → either bet both again or bet which? one of them.

    For latter variation I've used Monty Hall, betting penultimate on the marquee, avoiding the quite dz -- but that goes against the initial probability & breaks the core idea


    Another model;
    betting only 1DZ two times in 3-spin cycle.
    1st dz known → two remain, of the two bet the one according to Monty, penultimate on the marquee =hit (+2)

    no-hit
    hereon two possibilities for the second DZ bet on the 3rd spin of the cycle
    first off, first appeared dz is off the table due to the other two dz ≈74% hit appearance probability
    • bet the remaining dz
    • bet for the repeat of the second appeared (according to Monty Hall, avoid quite dz)


    All of these options I've tried flat-bet with none of them showing positive expectation already in 200 spin permanence.


    Question:
    Is the core idea erroneous if compared as a benchmark to what you what you had in mind when responding the now quoted -- is the solution within described, but requires some restructural refinement or an addition -- or completely off the mark?
     
    Last edited: Oct 29, 2021
  12. Luckyfella

    Luckyfella Well-Known Member

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    Sir Anyone is correct that Monty Hall model cannot be used in roulette. He can't provide the explanation why not.

    There are 3 events,
    1. Player selects the door,
    2. Host reveals the goat, and
    3. Player decides to stay or switch.

    The 2nd event is dependent since the host will not reveal the car.

    This is not the case with roulette outcomes.

    In the show the player is given one chance to decide stay or switch. It's the same with roulette. One chance.
     
  13. thereddiamanthe

    thereddiamanthe Well-Known Member

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    OK, I agree with the above. When I am referring to Monty Hall I mean in a form adapted to roulette.
    Exactly, there are two states in dzs. All three alternating or two repeating with one going quite.
    This two states sometimes flow directly into each other, more likely they solidify for a while.
    A break out of this state is or its ending is clearly signalled by a repeat of dz or 3rd dz appearing respectively.

    The latter two dz repeating state happens quite often & on various lengths, with a clear end & break off out of solidity.
    Meanwhile in this state, formed as an isolated & continuous event, the probability of any dz appearing is ≈50/50 including the Zero. Of course there's a chance of solidity breaking apart ..

    This is what Turbo meant with the "Yet if I walk up during the "game" when there are only 2 doors left to open - MY chances are 50/50."

    Nonetheless, the pure Monty Hall definitely dies not apply to roulette; only a temporary state.



    But the core of my question relates to the 50/50 remodelling.
    What I have trouble with .. is to encompass ≈74% throughout the cycle, so the 2nd&3rd spin or 4th+ with Zero .. & still keep the bet (sequence) structured as 50/50.

    Inducing Monty Hall adaptation has been the consequence of this & perhaps, as you are probably implying a step in the wrong direction.


    To encompass is the keyword.

    ?
     
  14. TwoUp

    TwoUp Well-Known Member

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    We don't even need statistical evidence, it takes no belief as the proof comes directly from the probability.

    Hitting a dozen is a 12/37 event, therefore on average we just need flip the probability to calculate the spins required to see the event. So a dozen will take on average 37/12 = 3.07 spins to appear.

    So to work out the probability of seeing all three dozens we need to first work out the probability of the first dozen appearing, the probability of hitting the next dozen and the probability of hitting the final dozen.

    The first dozen any of the 36 pockets will do, the second dozen we have to hit one of the remaining 24 pockets and the final dozen leaves just 12 pockets we have to hit.

    The probability for each of these events is 36/37, 24/37 and 12/37.

    To work out the required spins we just flip the probabilities and add them up.

    To get the first dozen it's 37/36 = 1.03 spins.

    To get two dozens it's 37/36 + 37/24 = 2.57 spins.

    And to get all three dozens is 37/36 + 37/24 + 37/12 = 5.65 spins

    Turbo showed his statistical sample requiring 5.46 spins to see all three dozens which is an underestimate. The pure math says it takes 5.65 spins (on average) to see three dozens hit so it's actually more than what Turbo said and closer to 6 spins than 5.
     
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  15. thereddiamanthe

    thereddiamanthe Well-Known Member

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    this only reinforces the fact further .. can we break this math even deeper .. starting solely with dozens, as the best & simplest representation of ⅓s .. before delving further even deeper; meaning, on a spin to spin basis, by betting only one ⅓ .. to supersede this
    .

    .. so, if we look at these figures alone;

    • we see that after the 1st ⅓ has already appeared (12ⁿ, ⁿ=numbers played), the likelihood of the 2nd or 3rd ⅓ (24ⁿ) ..
    cascades from the average 1.03 to 2.57 mark -- doesn't this imply that on the 2nd spin, ie. 2.00 mark, the repeat of the 1st ⅓ (12ⁿ) has a greater likelihood > the 2nd+3rd ⅓ (24ⁿ) -- so much so, since only 12ⁿ vs 24ⁿ are played;

    1.03+(2.57-1.03)/2 = 1.03+0.77 = 1.8 !! as the broken down average between the two mark notches..

    .. also, now having two factors to regard, not only the spin count but also the numbers played per spin --
    if that's the correct wording, multi-level calculation, due to betting only a ≈⅓ of numbers??


    • given the 1st ⅓ repeated, the game is in plus .. irregardless, since we are counting the three ⅓s to show, the implication of that is that the likelihood or probability shifted a notch further, now to & on the 2nd spin .. tilting towards the 2nd or 3rd ⅓ to show

    +

    given
    .. & due to the average likelihood averaging at 2.57 & 5.65, implying that there are instances worse than that to average it so ..

    .

    it would still only make sense to play for
    a) the 1st ⅓ to repeat again


    & only after the 2nd ⅓ appears, leaving the 3rd ⅓ out of the equation (due to its avg being 5.65, so at least 2 or some change spins later) + 1st as well (due to its repeat on each further spin the likelihood for another repeat diminishing also further) + playing 12ⁿ instead, giving the wheels another 1x spin to 'work in favor' or 'produce favorable result' at a lower price, & another 2x spins to break-even (rather than giving the wheel only 1x spin opportunity to do the same, at double the price)

    b) for the 2nd appeared ⅓ to repeat again

    .

    Does the math confirm my logic?
     
    Last edited: May 13, 2022
  16. thereddiamanthe

    thereddiamanthe Well-Known Member

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    & at which spin, it would make the most sense to start betting on the 3rd ⅓ to show;
    as the second degree of the shift in/of the INFLECTION POINT?

    @TwoUp
     
    Last edited: May 13, 2022
  17. thereddiamanthe

    thereddiamanthe Well-Known Member

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    I have a few more questions, a bit more complex .. so fundamentally, I rather start & complete this first.
     
  18. thereddiamanthe

    thereddiamanthe Well-Known Member

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    alternatively;

    after the 2nd ⅓ appears, betting 24ⁿ as both appeared 1st+2nd ⅓, & up till which spin the most optimized ..
    shift of the inflection point .. to then bet either 1x ⅓ only as the 3rd ⅓?
     
  19. 6th-sense

    6th-sense Active Member

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    all presumptions its a 3rd....its not....it,ll go one way....then the other
     

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  20. 6th-sense

    6th-sense Active Member

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    up and down like a yoyo
     

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