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Math Help - poker odds

  1. #1
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    poker odds

    I am a (semi-)proffecional poker player with a penchant for practical maths. When i calculate the odds of certain things i often get discrepancies between what various website quote and what i get. some help would be apreciated. In texas hold 'em you get delt 2 cards and another 5 on the board. I argue this:
    if i am dealt A,K (in the hand), what is the chance of hitting either an A or a King in the next 5 cards to hit the board.
    Texas Holdem Odds - Probabilities for Poker Hands
    quotes it as being 50% (just under half way down the page). However, i calculate this:
    3 aces left in deck + 3 kings left in deck = 6 cards the satisfy my outcome. assuming a normal 52-card deck (so there are 50 cards left in the deck i get):
    (6/50)+(6/49)+(6/48)+(6/47)+(6/46)=62.5%.
    which is a huge difference.

    please tell that website has it wrong?! If not help me work out this VERY SIMPLE calculation

    thanks,
    Morlaf
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  2. #2
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    Quote Originally Posted by Morlaf View Post
    I am a (semi-)proffecional poker player with a penchant for practical maths. When i calculate the odds of certain things i often get discrepancies between what various website quote and what i get. some help would be apreciated. In texas hold 'em you get delt 2 cards and another 5 on the board. I argue this:
    if i am dealt A,K (in the hand), what is the chance of hitting either an A or a King in the next 5 cards to hit the board.
    Texas Holdem Odds - Probabilities for Poker Hands
    quotes it as being 50% (just under half way down the page). However, i calculate this:
    3 aces left in deck + 3 kings left in deck = 6 cards the satisfy my outcome. assuming a normal 52-card deck (so there are 50 cards left in the deck i get):
    (6/50)+(6/49)+(6/48)+(6/47)+(6/46)=62.5%.
    which is a huge difference.

    please tell that website has it wrong?! If not help me work out this VERY SIMPLE calculation

    thanks,
    Morlaf
    Hi Morlaf,

    The easiest way to compute the odds in this situation is to compute the probability that you will *not* see an ace or king in the next 5 cards. There are 50 cards left in the deck, of which 44 are neither aces nor kings. There are \binom{50}{5} ways to draw 5 cards, all of which are equally likely. Of these, \binom{44}{5} have no aces or kings. So the probability of not drawing an ace or king is

    \frac{\binom{44}{5}}{\binom{50}{5}} \approx 0.5126,

    and the probability of seeing at least one ace or king is approximately

    1 - 0.5126 = 0.4874.
    Last edited by awkward; February 8th 2010 at 08:02 AM. Reason: Can't do simple arithmetic!
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  3. #3
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    thanks

    elegant and cunning...... just what i expect from you mathematicians... thanks!!!

    :O)
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