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Roulette Pigeonhole Principle - list of theorems/generalizations

Discussion in 'Roulette Forum' started by BETJACK, Oct 23, 2018.

  1. BETJACK

    BETJACK Active Member

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    Pigeonhole Principle (PP)
    If n+1 objects are placed in n boxes, then one of the boxes must contain more than 1 object.

    In 4 spins 1 dozen has to repeat (excluding zero). Dozen have 3 boxes/pigeonholes. And in 4 spins we have 4 objects/pigeons.

    (PPb)
    If n-1 objects are placed in n boxes, then one of the boxes must be empty.

    This could be the basis for the HG! If you can come up with a rule that puts less pigeons into more pigeonholes then we are guaranteed to never encounter a deadlock - a repeat will always happen before all uniques have shown.

    Extended Pigeonhole Principle (EPP)
    If nk+1 objects are placed in n boxes, then one of the boxes must contain at least k+1 objects.

    I think this is related to multiple repeats? There will be 4 repeats (5 hits) of a dozen in 13 spins. Dozens have 3 boxes/pigeonholes (n). K = 4 = number of repeats. K +1 = 5, so 5 dozens/pigeons are placed in 1 pigeonhole:
    D1
    D1 D2 D3
    D1 D2 D3
    D1 D2 D3
    D1 D2 D3

    (EPPb)
    If nk-1 objects are placed in n boxes, then one of the boxes must contain at most k-1 objects.

    Related to the above, but need to think about this.

    Generalized Pigeonhole Principle (GPP)
    If nk+s or more objects are placed in n boxes, then for each 0 ≤ m ≤ n there exist m boxes with a total of at least mk + min(s,m) objects

    (GPPb)
    If nk+s or fewer objects are placed in n boxes, then for each 0 ≤ m ≤ n there exist m boxes with a total of at most mk + max(0,s+m-n) objects

    I guess these last ones are based on the maximum pigeons being more than the average "For a non-empty, finite bag of numbers, the maximum value is at least the average value"?

    Let's say we got 3 dozens and 3 pigeonholes to put them in. The maximum number of dozens/pigeons is 3, so the average is 2. The maximum is always greater than the average + 1?

    from FALKOR2K15
     
  2. BETJACK

    BETJACK Active Member

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    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.


    thumbsup.gif

    from DYKSEXLIC!
     

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  3. BETJACK

    BETJACK Active Member

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    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.

    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..!
    cool.gif

    from DYKSEXLIC!
     

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  4. BETJACK

    BETJACK Active Member

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    Mathematics sounds the same in ANY language.
    The Pigeonhole Principle is a roulette players secret weapon.
    In general, it may not be so clear how to apply the Principle.
    Sometimes we need to cleverly construct the 'pigeons' and the 'holes'.
    If we do this correctly, the proof should be slick.
    Otherwise, the problem may seem forbiddingly difficult.
    When stuck, do not give up so easily!
    You learn and improve the most when you are stuck.
    Keep thinking of possible approaches, perhaps for a few hours,
    And you might be rewarded with an elegant solution.
    This is the ONLY way to learn mathematical problem solving.

    Cheers.

    from DYKSEXLIC!
     
  5. BETJACK

    BETJACK Active Member

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    Dyslexic's Repeater System--(for RNG only)

    Foreward: I didn't take him seriously until Tiago showed a graph that stated if you play repeaters for up to 12 spins you have an 80% chance of getting a hit.

    1) Start tracking numbers by playing red/black odd/even high/low equal amounts (or freespin)

    2) Track three different numbers (if a number repeats restart tracking)

    3) Play the last three numbers--if miss, play the last three and a chip on the new fourth number.

    4) Play the last four numbers-if miss, play the last four numbers and a chip on the fifth number (and so on, and so on...until a hit or until the 12th number misses.--End session)
    ----------------------------------------------------------------------------------------------------------------
    Suggestion: end session after the second number hits.

    Bankroll: 75 units or more.
     
  6. SERGIO

    SERGIO Active Member

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    I think it's good to start this topic, I forgot to send you all this information Skype and some things, later when I get home I send what I have to all this to publish if you want.
    When you're on Skype let me know and we'll talk, regards
     
  7. BETJACK

    BETJACK Active Member

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  8. Dr. Sir Anyone Anyone

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

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    This is the dumbest thread that I've read in a long time. Did a child make up the bullshit math or did some forumtard? It reads like some kind of stupid gypsy scam.
     

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  9. Nathan Detroit

    Nathan Detroit Well-Known Member Founding Member

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    LOL



    Stu Pit was the Godfather of this topic .


    ND
     
  10. Dr. Sir Anyone Anyone

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

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    The gypsies suck at math and are generally very poorly educated, but this is stupid even for a gypsy!
     
  11. BETJACK

    BETJACK Active Member

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    I have not read anything yet.
    ... nothing of what you write ...
    ...to prove that math is wrong ...
    "here is the mistake"
    you just say math is stupid.
     
  12. Dr. Sir Anyone Anyone

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

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    The "math" cough, cough, written above is just made up nonsense created by some forumtard. It's not math. LOL!!!

    Dyslexic = forumtard.
     
    Last edited: Oct 23, 2018
  13. Nathan Detroit

    Nathan Detroit Well-Known Member Founding Member

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    Is this cat for real giving a floor show with a box at roulette table in a casino. ROFLMAO.
     
  14. Dr. Sir Anyone Anyone

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

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    tenor.gif

    I love these math terms, "the odd spin m, then x = 2km for some spin k ≥ 0." Expecially the "≥ 0" Yep, those are terms right out of statistics and calculus 101! NOT!!! LOL!!!!

    I have no doubt that Roulette Ghost is probably hypnotized by it all!
     

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    Last edited: Oct 24, 2018
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  15. Madi

    Madi Member

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    41EE2173-8445-4438-A7E6-40D65FD2F493.jpeg Sir

    Why do u think that you are not a junkie? Till now we havent seen anything reasonable from you rather than barking.

    Broken wheel only can be seen in museum or mostly in dream. Ur behaviour is like old homeless drunk. Always senseless.
     
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  16. BlueAngel

    BlueAngel Active Member

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    You are wrong!
    DYKSEXLIC = DickSexLick
    Do you UNDERSTAND now?
     
  17. BETJACK

    BETJACK Active Member

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    BlueAngel what do you think for this
    Pigeonhole Principle
     
  18. BlueAngel

    BlueAngel Active Member

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    It is valid but if you don't know how to use it then it's just another theory.
    Usually the problem is in communication, it's simpler to complicate something and complicated to simplify it...
    What I'm tying to say is that theories you find around the web have formal presentation and specialized terminology, all these fancy wording and illustrious vitrine make you forget the essence of it.
    So what's the essence of it, I like to speak with metaphors, just imagine a marathon of 37 km, contesting 37 marathon runners, at some point one of them is getting ahead, the distance he has to cover till the end is less than what any other runner has to cover...
     
  19. Dr. Sir Anyone Anyone

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

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    It's retarded. Anybody that says otherwise is full of shit.
     
    Last edited: Oct 24, 2018
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  20. Sharptracker

    Sharptracker Well-Known Member

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    Sigh to note that human chose to solve the problem by inventing all systems possible and check how it is going with a paper and a pen on 100 spins instead to understand only once, YOU CANNOT REVERSE A NEGATIVE EDGE WITH MATH OR WITH A PROGRESSION. ANY SYSTEM THAT'LL CHOOSE WILL MEET GOOD AND BAD STREAKS, BUT SOON OR LATE YOU'LL PAY THE DUE. YOU GOT TO PLAY WITH ODDS IN YOUR FAVOUR, THERE'S NO OTHER WAY.
     

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