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  1. jay_shark jay_shark is offline

    NY
    United States

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    Feb 5th, 2007 (2/5/2007 5:49pm)

    You're utg with pocket tens at a 10 handed table . What's the probability that one of the remaining players has ace king or jacks and up ?
  2. marinersjj marinersjj is offline

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    Feb 5th, 2007 (2/5/2007 5:52pm) in reply to jay_shark

    7.8 percent.

    tens are money.

    postflop play is overrated.

    end thread.
  3. jay_shark jay_shark is offline

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    Feb 5th, 2007 (2/5/2007 5:58pm) in reply to marinersjj

    Hey Marine , if I give you $100 would that change your mind ?
    Thread Starter
  4. marinersjj marinersjj is offline

    Cambridge, MA
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    Feb 5th, 2007 (2/5/2007 6:03pm) in reply to jay_shark

    haha maybe a little bit...
    argh... ok, i'll get to work on this and get back to you.

    and although i've got nothing but love for the marines, the name is "mariners"... like the baseball team.
  5. rock3656 rock3656 is offline

    San Diego, CA
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    Feb 5th, 2007 (2/5/2007 6:24pm) in reply to marinersjj

    18-20% or so. http://www.expertinsight.com/index.p...air_principle/
     21
  6. The Sour Diesel The Sour Diesel is offline

    Santa Cruz, CA
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    Feb 5th, 2007 (2/5/2007 6:29pm) in reply to jay_shark

    1326 starting hands in holdem...

    so if ur dealt TT UTG that means there will be 9 chances for youropponent to have JJ QQ KK AA AK AKsooooooooooooooooooooted. There are 6 different ways u can haveAA= {A, A}, {A, A}, {A, A}, {A, A}, {A, A}, and {A, A}. the same applies for every other overpair.

    so its JJ,QQ,KK,AA=(4 hands)(6 possibilities)= 24+16 (12unsuited+ 4 suited) ways u can have AK=40

    (40 possible holdings) x (9 players)= (360)/(1325 remaining starting hands)= 27.169%

    -justin

    edit:
    also forgot to take into account nonsuited AK. therefore 4p3=12 ways to get AK.

    re-edit: + 4 ways suited AK= 16
  7. jay_shark jay_shark is offline

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    Feb 5th, 2007 (2/5/2007 6:32pm) in reply to rock3656

    Gordon principle is an approximation and for extreme cases , it may be significantly off .
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  8. rock3656 rock3656 is offline

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    Feb 5th, 2007 (2/5/2007 6:35pm) in reply to The Sour Diesel

    Lol looks like i was pretty close jay
     21
  9. The Sour Diesel The Sour Diesel is offline

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    Feb 5th, 2007 (2/5/2007 6:55pm) in reply to The Sour Diesel

    ok now that im done with my editing

    ...SHIP IT
  10. jay_shark jay_shark is offline

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    Feb 5th, 2007 (2/5/2007 7:05pm) in reply to The Sour Diesel

    Fairly close , but not quite .
    Thread Starter
  11. The Sour Diesel The Sour Diesel is offline

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    Feb 5th, 2007 (2/5/2007 7:25pm) in reply to jay_shark

    enlighten me then. i dont see wuts wrong. maybe the AK calculation.
  12. Goal4Ziggy Goal4Ziggy is offline

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    Feb 5th, 2007 (2/5/2007 7:30pm) in reply to jay_shark

    Ok, I'll take a shot.

    There are 50 unknown cards.

    Chances of AK= 100 * (4/50) * (4/49) = .6530612245 %
    Chances of a pair JJ or better = 100 * (4/50) * (3/49) * 4 = 1.959183673

    Add these together = 2.612244898

    Multiply by 9 because there are 9 chances for this to happen = 23.51020408 % or 3.25347222 to 1

    maybe?
  13. Moosebabies Moosebabies is offline it's all about timing

    Fort Worth, TX
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    Feb 5th, 2007 (2/5/2007 7:31pm) in reply to jay_shark

    what's wild?
  14. jay_shark jay_shark is offline

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    Feb 5th, 2007 (2/5/2007 7:37pm) in reply to The Sour Diesel

    If you hold pocket tens then there are 50c2=1225 card combos for any player .

    The probability is NOT 40/1225*9 . The reason is because two or more players may hold the same cards and so the events are not mutually exclusive .

    Here is a very good approximation to the above problem .

    The probability the next player doesn't hold those cards is (1225-40)/1225. The probability all 9 players don't hold those cards is (1225-40)^9/1225^9.

    Now finally you take the complement which is the probability that one of the remaining players will hold those cards . 1-(1225-40)^9/1225^9= 25.8% .

    This is a very close approximation since the dependency is weak .

    For a precise answer you would need the principle of inclusion/exclusion to deal with these cases when the events are not mutually exclusive but it isn't necessary .
    Thread Starter
  15. The Sour Diesel The Sour Diesel is offline

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    Feb 5th, 2007 (2/5/2007 8:00pm) in reply to jay_shark

    good call on removing all the hands with tens in them. forgot about that.

    ". 1-(1225-40)^9/1225^9= 25.8%"
    thats not correct i dont think.
    ^9...maybe u meant *9?

    where is the ".1" in that calculation coming from?
    can u explain the compliment part of the calculation?

    i havent revisited stats since high school so i may be wrong.
  16. Prfsr_Cain Prfsr_Cain is offline

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    Feb 5th, 2007 (2/5/2007 8:14pm) in reply to The Sour Diesel

    ^9 means he's raising it to the ninth power. It is correct.

    There is no .1 in his calculation. It is a period, followed by a 1.
  17. jay_shark jay_shark is offline

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    Feb 5th, 2007 (2/5/2007 8:55pm) in reply to The Sour Diesel

    If you draw a circle on a piece of paper and color it red , then the complement of the red circle is the rest of the paper without the red circle . This is true if the universe set is the entire sheet of paper .

    Another example .

    The complement of drawing the ace of clubs from a 52 card deck consists of the remaining 51 cards in the deck .
    Thread Starter
  18. The Sour Diesel The Sour Diesel is offline

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    Feb 5th, 2007 (2/5/2007 9:22pm) in reply to jay_shark

    i understand that concept of a compliment calculation. i was just wondering how its related to anything in poker that has exponential behavior. looks like u have a formula that u used (or how else would u know to take that expression to the 9th power...i mean i understand y its 9...but not y its an exponent). maybe ud care to share?
  19. Decatur217 Decatur217 is offline OTs Iron Man

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    Feb 5th, 2007 (2/5/2007 9:25pm) in reply to Moosebabies

    One-eyed jacks.
  20. Maits10 Maits10 is offline

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    Feb 5th, 2007 (2/5/2007 10:01pm) in reply to Decatur217

    Obv brag post by jay shark that he knows more about useless math than most.

    Hey jay, do you actually <span>play </span>poker?
  21. jay_shark jay_shark is offline

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    Feb 5th, 2007 (2/5/2007 10:05pm) in reply to Maits10

    Maits , how is this useless math ?

    Any serious poker player has asked this question at some point or another but some just don't have the ability to answer it . I was being generous in sharing to others how this kind of calculation works .

    I think you secretly admire me .
    Thread Starter
  22. mathclub mathclub is offline

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    Feb 5th, 2007 (2/5/2007 10:36pm) in reply to jay_shark

    maths is boring, don't be a geek :)
    2
  23. khammer6 khammer6 is offline

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    Feb 5th, 2007 (2/5/2007 10:43pm) in reply to jay_shark

    The probability that one of the players will have 10's or better = 18.7% + the probability one player will have AK (10.8%) = 29.5% one of those hands are out there.
  24. The Sour Diesel The Sour Diesel is offline

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    Feb 5th, 2007 (2/5/2007 10:45pm) in reply to khammer6

    no. jay_shark is right about the contribution of mutual exclusiveness.
  25. The Sour Diesel The Sour Diesel is offline

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    Feb 5th, 2007 (2/5/2007 10:48pm) in reply to mathclub

    math club?
  26. rock3656 rock3656 is offline

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    Feb 5th, 2007 (2/5/2007 10:51pm) in reply to The Sour Diesel

    lol, then why hide the answer from us instead of jus flat out sayin it(people love feeling superior.)
     21
  27. timex timex is offline

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    Feb 6th, 2007 (2/6/2007 12:07am) in reply to jay_shark

    I never get jay_shark posts;

    Why ask stupid questions to a forum of stupid people?
    605 11
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