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Roulette Even chance percentage win rate.

Discussion in 'Roulette Forum' started by Blacksmith, Sep 13, 2020.

  1. Blacksmith

    Blacksmith Member

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    What would be a good percentage for winning bets on even chance and consider to have an edge?

    Currently my winning bet selection average between 60% and 70% with three losses being the most in a row but never lost session in about 1000spins.

    Kind regards
     
    Last edited: Sep 13, 2020
  2. gizmotron

    gizmotron Well-Known Member Founding Member

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    I would want to know how you manage to never get four losses in a row. Because if that is true I would up your winning percentage to more than 1,000%. You could use a negative progression like the Martingale.

    Have you bothered to explain your method or even drop hints at it?
     
  3. Nathan Detroit

    Nathan Detroit Well-Known Member Founding Member

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    Anyone who claims winning more than 60% is either a politician or a liar. ( John Patrick)
     
  4. Blacksmith

    Blacksmith Member

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    Thanks gizmotron .

    Its a bet selection in cycles and not a static bet and focused on events.I don't want to speak in riddles or play guess my method if the time is right and have something good It will be better to share it not in public.

    Thanks for advice on negative progression.

    Nathan Detroit early days for testing this method.Time will tel.

    Kind regards
     
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  5. Nathan Detroit

    Nathan Detroit Well-Known Member Founding Member

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    We have never tested a method . Neither Baccarat or Roulette as long as we knew the appropriate M.M. to go with it .



    ND
     
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  6. Blacksmith

    Blacksmith Member

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    Thanks Nathan Detroit it was done flat betting currently.

    Kind regards
     
  7. Nathan Detroit

    Nathan Detroit Well-Known Member Founding Member

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    I know of only one published system for flat stakes that makes sense .Otherwise it is the up and pull or a shorter version thereof .
     
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  8. Rona

    Rona Active Member

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    "three losses being the most in a row"

    I have a question for you OP.

    What on earth, do you think, prevents the wheel to produce a losing outcome for a 4th time?

    This is nonsense. In a 50-50 bet, there can easily be well over 10 losses, no matter what or when you bet. It is still 50-50.
     
    Last edited: Sep 13, 2020
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  9. Blacksmith

    Blacksmith Member

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    Thanks Rona nothing at all.

    Have you explore the Pigeonhole Principle on non random facts in several cycles ?

    Kind regards
     
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  10. Garfield

    Garfield Active Member

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    Everyone somehow will face the 4+ LIAR, but no one could tell precisely when....
    Maybe in OP case, another X years before that happens?

    Just because the math or theory said so, it MIGHT/NOT happen right away to a player....
     
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  11. Blacksmith

    Blacksmith Member

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    Thanks Garfield for your input.

    Using even chance for bet selection .What must happen after three spins.One will repeat?

    Pigeonhole theory says I can't lose no matter the outcome.

    Kind regards.
     
  12. Blacksmith

    Blacksmith Member

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    But anything is possible .

    Kind regards
     
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  13. Herby

    Herby Member

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    Hi Blacksmith,

    What defines a session, how many spins, ...

    How many bets do you have in 1000 spins ?

    Regards
     
  14. Blacksmith

    Blacksmith Member

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    Hi Herby

    Almost every spin .

    This is a interesting read from a other forum hope it can assist .


    "Here's a very abstract idea from the great master:
    Dyksexlic! wrote: Disclaimer - "The following thread contains mathematical ideas which you may not agree with. If this is likely to cause you anxiety, please refrain from reading any further. If you continue, you implicitly acknowledge that Dyksexlic is not responsible for any harm caused to your cerebral cortex."


    How to prove ANYTHING with the Pigeonhole Principle:

    There are FOUR steps involved.

    1) Decide what the 'pigeons' are. They will be the things that you’d like several of to have some 'special property'.

    2) Set up the 'pigeonholes'. You want to do this so that when you get two 'pigeons' in the same 'pigeonhole', they have the property you want. To use the Pigeonhole Principle, it is necessary to set things up so that there are fewer 'pigeonholes' than 'pigeons'.

    (Sometimes, the way to do this relies on some 'astute observation'. )

    3) Give a rule for assigning the 'pigeons' to the 'pigeonholes'. It is important to note that the conclusion of the Pigeonhole Principle holds for any assignment of 'pigeons' to 'pigeonholes', so it holds for any assignment you describe. Pick the rule so that when enough 'pigeons' occupy the same 'pigeonhole', that collection has the property you want.

    4) Apply the Pigeonhole Principle to your system and get the desired conclusion.


    [​IMG]
    Dyksexlic! wrote: The 'pigeons' are merely a mathematical metaphor.

    Consider the principle a 'conceptual visualisation tool'.

    A 100% Mathematically guaranteed roulette bet is possible with the correct application.

    This thread is ONLY concerned with the Pigeonhole Principle itself.

    For 'winning bet selections', kindly refer to Norman Bates... [Sarcastic smile [​IMG]]!

    However, consider the following example..

    QUESTION -
    Prove that in a streak of 10 LOW roulette spins {1, 2, . . . , 18}, the selection
    includes integers a and b such that a|b (that is, a divides b – there exists an integer k such
    that ak = b).

    ANSWER -
    Let the 'pigeons' be the 10 spins selected.

    Define nine 'pigeonholes' corresponding to the odd spins 1, 3, 5, 7, 9, 11, 13, 15, and 17.

    Place each spin selected into the pigeonhole coresponding to its largest odd divisor (which must be one of 1, 3, 5, . . . , 17).

    Notice that if x gets placed in the pigeonhole corresponding to the odd spin m, then x = 2km for some spin k ≥ 0.

    Since 10 spins are selected and placed in nine pigeonholes, some pigeonhole contains two spins a and b, where a < b.

    Suppose this pigeonhole corresponds to the odd spin t. Then, a = 2rt and b = 2st, where are < s, so that a2s−r = b.

    Since s−r is a positive spin, it follows that a|b.

    Welcome to the Magical Land of Mathematics..!

    8)
    If he's creating 9 odd pigeonholes 1,3,5,7,9,11,13,15,17 that are fewer than 0,2,4,6,8,10,12,14,16,18 pigeons, and has a rule that can place all the latter into the former, then what's he betting on!?

    Another concept here is playing with half the table."


    I dont hope my Pigeons fly away :)

    Kind regards
     
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  15. Herby

    Herby Member

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    I think I just caught some (10) pigeons. Don't know if they are just 10 mosquitos. But their number is 10. :bear:
    Regards
     
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  16. Nathan Detroit

    Nathan Detroit Well-Known Member Founding Member

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    Roma,


    Find the answer with Frank Barstow`s principle of diminishing probabilities regarding repeating pattern of EC.
     
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  17. Proofreaders2000

    Proofreaders2000 Well-Known Member Founding Member

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    Don't forget the zero. Even a tie affects the flow of cards in Baccarat.
     
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  18. Blacksmith

    Blacksmith Member

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    Thanks ND

    Where can I get a copy to read?

    Kind regards
     
  19. Punkcity

    Punkcity Well-Known Member

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    Many people I know say this tie reference to me, usually that means the next bet they lost etc. personally I have to say that in my method it means nothing. If I bet player and a tie comes, money back . Then I re bet the player as that was my selected criteria pre the tie and it wins , so for me I would have to say the TIE affected my result?
    No
    for me I would say it is irrelevant , we cannot be 100% assured that the player was AWAYS going to occur after that tie, now it’s affecting the flow how ?
    So after that tie it’s a banker win and I lose YET someone else at the table bet the banker for the win .
    Now the flow is affected for me but not the other player? In my opinion to think that way opens the door for degenerate gambling fallacy to creep in , at the very least the casino is targeting me or god is punching me in the wallet.
    If the method you employed says jump TO Opposite YOUR initial decision after tie and it stayed the same as your initial decision is that tie still affecting the flow of cards? See where this goes? Because we cannot be 100 gazillion % absolutely SURE the next result after zero or tie is red /black or play/bank it has to be A irrelevant , it is something but not the world ender , you have lost or won , if you’re martingale or progress beting tough luck , but flat betting it’s just another unit.
    For mine not a problem, for others maybe and if that’s the case for them perhaps they should re evaluate their process. Cheers
     
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  20. Garfield

    Garfield Active Member

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    There's might be a slight different how tie affect different games....in bac money back, in roulette and dragon tiger, you'll lose if you didn't bet on it...

    That's why some have different PoV regarding tie... In bac probably we could see what's the consistency after a tie, does it cut, or follow? Is it continuing the trend, or break it?

    Although it can't be scientifically proved, many tends to see tie as a trigger...

    IMHO
     

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