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Math Help - Card probability question

  1. #1
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    Card probability question

    I am having trouble with a exam review question. The question is

    How many 5 card poker hands are there? what is the probability of a hand containing 3 aces and 2 queens?

    For the first part there are 52 \choose{5} possible hands

    For the second part I am getting something like this

    13 \choose{1} 4 \choose{3} 12 \choose{1} 4 \choose{2} / 52 \choose{5}
    Is this correct? Thanks for any help that anyone can provide.
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  2. #2
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    Quote Originally Posted by fishguts View Post
    I am having trouble with a exam review question. The question is

    How many 5 card poker hands are there? what is the probability of a hand containing 3 aces and 2 queens?

    For the first part there are 52 \choose{5} possible hands

    For the second part I am getting something like this

    13 \choose{1} 4 \choose{3} 12 \choose{1} 4 \choose{2} / 52 \choose{5}
    Is this correct? Thanks for any help that anyone can provide.
    Hi fishguts! (innovative name by the way).

    You seem to be going overboard on the numerator.
    What exactly are you thinking of by multiplying your 3 aces by 13C1
    and your 2 queens by 12C1 ?

    Why do you feel that your numerator is not just 4C3 multiplied by 4C2,
    which is "any pair of queens can be matched with any triple of aces"
    ?
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    Hi fishguts! (innovative name by the way).

    You seem to be going overboard on the numerator.
    What exactly are you thinking of by multiplying your 3 aces by 13C1
    and your 2 queens by 12C1 ?

    Why do you feel that your numerator is not just 4C3 multiplied by 4C2,
    which is "any pair of queens can be matched with any triple of aces"
    ?
    Well I am multiplying it by 13C1 and your 2 queens by 12C1 because for aces we have 13 cards of value to choose from, and after we chose the aces we then multiply the queens by 12C1 because thats how many cards of value are left once aces are chosen
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  4. #4
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    Quote Originally Posted by fishguts View Post
    Well I am multiplying it by 13C1 and your 2 queens by 12C1 because for aces we have 13 cards of value to choose from, and after we chose the aces we then multiply the queens by 12C1 because thats how many cards of value are left once aces are chosen
    The numerator is just \dbinom{4}{3}\dbinom{4}{2}.
    How many ways can one choose three kings?
    How many ways can one choose two queens?
    Last edited by Plato; February 21st 2011 at 08:01 AM.
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  5. #5
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    so If the question was worded what is the probability of choosing 3 cards of the same value and 2 cards of the same value would my first approach apply to that problem?
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  6. #6
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    Quote Originally Posted by fishguts View Post
    so If the question were worded what is the probability of choosing 3 cards of the same value and 2 cards of a dfferent value (both the same value) would my first approach apply to that problem?
    See my corrections. Your method is correct
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  7. #7
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    Quote Originally Posted by Plato View Post
    See my corrections. Your method is correct
    Thanks! Also I should put on my contacts before writing back to your replies sorry for the errors.
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