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Counting poker hands ?

  1. Jul 1, 2011 #1
    1. The problem statement, all variables and given/known data
    A poker hand conatins 5 cards chosen from a deck of 52.
    1. How many hands are possible
    a. 4 of a kind
    b.3 of a kind
    c. flush
    d. full house
    e. straight
    f. straight flush
    3. The attempt at a solution
    When i write (5,2) I mean 5 choose 2 .

    1. (52,5)
    a. I have 13 different choices from 2 .... king, ace .
    so (13,1)*(4,4)*(12,1)*(4,1) then my last card has to come from the remaining 12 and then of the 4 of those cards I have to pick 1 .
    b. (13,1)(4,3)(12,1)(4,1)(11,1)(4,1)
    When I pick my last cards I have to pick from different rows to avoid getting 4 of a kind or a full house.
    c. I have 4 different suits to pick from so I pick 1 of the four then pick 5 of the suits 13 cards. so it will be (4,1)(13,5)
    d. full house will be (13,1)(4,3)(12,2)
    e. Well there are 9 ways to get a straight with 5 cards numerically in a row.
    and their are 4 copies of an ace so i think the answer is 4^5(9)
    f. well there are 9 ways to get a straight an 4 suits so 9*36 but it would be (9,1)(4,1)
     
  2. jcsd
  3. Jul 2, 2011 #2

    vela

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    I think you have the wrong answers for b and d.

    With three-of-a-kind, you're double-counting since you count XXXab and XXXba separately.

    With the full house, the factor of (12,2) isn't right because you want the final two cards to be a pair, and you left out the factor because of the possible suits for the pair.
     
  4. Jul 2, 2011 #3
    yes your right. part d. should be (13,1)(4,3)(12,1)(4,2)
    and on b. I am off by a factor of 2 .
    for part b. (13,1)(4,3)(12,2)(4,1)(4,1)
     
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