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Math Help - Card Problems

  1. #1
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    Card Problems

    1) 5 cards are dealt from a standard 52-card deck. What is the probability that the draws will wield:
    a) 3 aces and 2 kings?
    b) 3 of a kind and 2 of a kind (a full house)
    c) 1 ace, 1 two, 1 three, 1 four, 1 five (a straight)
    d) any straight ( 5 ordered cards)

    for a) I have that it is \frac{24}{2598960}. since it's \frac{\begin{pmatrix} 4 \\ 2 \end{pmatrix} \begin{pmatrix} 4 \\ 3 \end{pmatrix}}{\begin{pmatrix} 52 \\ 5 \end{pmatrix}}

    for b) all I know is that it's going to be \frac{3774}{2598960}

    for c) is it P^{52}_{5} = \frac{52!}{48!} for the denominator and  \binom{4}{1}^5 for the numerator which equals \frac{1024}{\frac{52!}{48!}}

    and d) I think that the denominator is the same as c) but the numerator will be 9 times the original, since there are 9 different choices which equals \frac{(9)(1024)}{\frac{52!}{48!}}

    are these right?
    Last edited by lllll; February 29th 2008 at 11:13 PM.
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  2. #2
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    Hello, lllll!

    You did pretty good . . .

    There are: . {52\choose5} \:=\:2,598,960 five card Poker hands.
    . . This denominator is used throughout the problem.


    By the way, use: {52\choose5} or \binom{52}{5}


    1) Five cards are dealt from a standard 52-card deck.
    What is the probability that the draws will yield:

    a) 3 Aces and 2 Kings?
    b) 3 of a kind and 2 of a kind (a "Full House")
    c) 1 Ace, 1 Two, 1 Three, 1 Four, 1 Five (a "Straight")
    d) any Straight ( 5 ordered cards)

    for a) I have that it is \frac{24}{2598960}, since it is \frac{ {4\choose2}{4\choose3}}{{52\choose5}} . . . . Right!

    (b) There are 13 choices for the value of the Triple.
    There are 12 choices for the value of the Pair.
    Then there are {4\choose3} to get the Triple
    . . and {4\choose2} ways to get the Pair.

    The number of Full Houses is: . 13\cdot12\cdot 4\cdot 6\;=\;3744



    for c) is it P^{52}_{5} = \frac{52!}{48!} for the denominator . . . . no
    and  \binom{4}{1}^5 for the numerator . . . . yes
    The denominator is still: . {52\choose5}



    d) I think that the denominator is the same as c)
    but the numerator will be 9 times the original . . . . not quite

    There are ten Straights: .from A-2-3-4-5 to 10-J-Q-K-A

    The probability is: . \frac{(10)(1024)}{{52\choose5}}

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