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calling all math wizards

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  1. jesterwords jesterwords is online now

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    May 30th, 2007 (5/30/2007 7:56pm)

    at a full table in omaha, what is the probability of someone hitting quads when all 5 board cards are out and he's only holding one in his hand.

    its happened to me 2 times in 2 days and 3 times in 5 days... so I'm just wondering what the probability would be.

    I'm dyslexic and this type of math is way out of my league.
  2. nevertilt22 nevertilt22 is offline

    Panama

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    May 30th, 2007 (5/30/2007 8:32pm) in reply to jesterwords

    1 in every 10 hands
     
  3. jesterwords jesterwords is online now

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    May 30th, 2007 (5/30/2007 8:35pm) in reply to nevertilt22

    ty never, ever so helpful.. LMFAO
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  4. hotriceaa hotriceaa is offline

    Newark, DE
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    May 30th, 2007 (5/30/2007 8:40pm) in reply to jesterwords

    1/4324..?

    edit
    [

    1/4324 ~ .023%

    ]

    anyone correct me

    4 cards

    12 possible cards to pair

    flop:

    12/48 (pair cards / cards in deck) * 2/47 (three of a kind card possibilities / cards in deck) * 1/46 (quads / cards in deck)

    turn: 45/45

    river: 44/44

    i guess this would only be for flopping it
  5. Camz Camz is offline

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    May 30th, 2007 (5/30/2007 8:40pm) in reply to jesterwords

    What? You act like you were expecting a serious response.
  6. Prfsr_Cain Prfsr_Cain is offline

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    May 30th, 2007 (5/30/2007 8:58pm) in reply to hotriceaa

    You are very close hotriceaa. I'll be a little more specific.

    Suppose you have four unpaired cards. There are 12 possilbe cards to pair.

    (12/48)(2/47)(1/46)(45/45)(44/44)*100 = 0.0231%

    But that's just the odds of flopping it. The board has 5 cards. We need 3 to give us quads. So we have 5C3 = 10 different ways of hitting quads. (you forgot this part)

    Therefore, the probability is 0.0231*10 = 0.231%

    Now suppose you have two unpaired cards and two paired cards. So there's only 6 possible cards to pair (you said quads w/only one in hand, so I won't do it for the paired cards). As with before, there are 5C3 = 10 different ways of hitting quads.

    So P = 10(6/48)(2/47)(1/46)(45/45)(44/44)*100 = 0.115%

    Hope this helps.
  7. jesterwords jesterwords is online now

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    May 30th, 2007 (5/30/2007 9:06pm) in reply to Prfsr_Cain

    ok, so then the % of time it would happen in any one random hand would then be??? Prfsr?
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  8. hotriceaa hotriceaa is offline

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    May 30th, 2007 (5/30/2007 9:34pm) in reply to jesterwords

    .231%
  9. jesterwords jesterwords is online now

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    May 30th, 2007 (5/30/2007 9:41pm) in reply to hotriceaa

    so out of 1000 hands it would happen roughly 2.31 times... thats just sick.
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  10. freshbeaterss freshbeaterss is offline

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    May 30th, 2007 (5/30/2007 9:47pm) in reply to hotriceaa

    .8810%
  11. MCMLXXXIII MCMLXXXIII is offline

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    May 30th, 2007 (5/30/2007 10:46pm) in reply to jesterwords

    All these calculations are good if you want to know what the odds are that you yourself will make quads with no prior information on any given hand, but I don't think that answers the OPs question. If you know your 4 cards and the 5 on the board, there are 43 cards left. The odds that someone does not have the 4th card to make quads can then be calculated. For one opponent odds they do not have the 4th card are: 42/43*41/42*40/41*39/40 = 0.907
    So the odds the one opponent has the 4th card are 1-0.907 = 0.093 or 9.3%
    That is for one opponent. If you are up against 7 opponents then multiply 9.3%*7=65.1% chance someone has that 4th card. As you can see, you should not be surprised.
  12. MCMLXXXIII MCMLXXXIII is offline

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    May 30th, 2007 (5/30/2007 10:51pm) in reply to Camz

    LOL
    See my next post in this thread
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