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Math Help - poker hands

  1. #1
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    poker hands

    I am doing the following question.

    "5. How many poker hands are there that contain at least one of 10@, QB, AC, or 6D ?
    Show your work."

    For my solution I did

    A=52*(51C4)
    B=52*51*(50C3)
    C=52*51*50*(49C2)
    D=52*51*50*49*(48C1)

    then

    A+B+C+D=E

    Does this seem to make sense?
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  2. #2
    jar
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    Re: poker hands

    think of what the question is really asking. "at least one of or "

    and what does A + B + C + D mean.

    good luck,
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  3. #3
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    Re: poker hands

    The question asks at least 1.

    So that means how many ways are can you get one of them or two of them or three or all 4 of them.

    A=the amount of ways to get 1
    B=the amount of ways to get 2
    C=the amount of ways to get 3
    D=the amount of ways to get 4

    edit: I did mess up though. It should be

    A=52*(51C4)
    B=(13C2)(4)(4)(50C3)
    C=(13C3)(4)(4)(4)(49C2)
    D=(13C4)(4)(4)(4)(4)(48C1)
    Last edited by ehpoc; November 30th 2011 at 09:41 PM.
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    Re: poker hands

    could anyone confirm if they agree with this solution?
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  5. #5
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    Re: poker hands

    Quote Originally Posted by ehpoc View Post
    could anyone confirm if they agree with this solution?
    Well, as usual with most of your postings you steadfastly refused to tell anyone what the totally non-standard notations means.
    For example, we have no idea what "10@, QB, AC, or 6D" means.
    If you want help then explain what that weird string means.
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  6. #6
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    Re: poker hands

    10 of some specific suit, queen of some specific suit, Ace of some specific suit, 6 of some specific suit
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  7. #7
    jar
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    Re: poker hands

    Quote Originally Posted by ehpoc View Post
    A=the amount of ways to get 1
    B=the amount of ways to get 2
    C=the amount of ways to get 3
    D=the amount of ways to get 4
    Agree with the above thinking.

    Quote Originally Posted by ehpoc View Post

    A=52*(51C4)
    B=(13C2)(4)(4)(50C3)
    C=(13C3)(4)(4)(4)(49C2)
    D=(13C4)(4)(4)(4)(4)(48C1)
    Please look into what does it really give you from each of the above statements.

    As Dr. Kocay said sometime you can pull it out from the hat. Personally, I think mathematics should not be pulling out from a hat.
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  8. #8
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    Re: poker hands

    Hello, ehpoc!

    I hope I understand the problem . . .


    \text{How many poker hands are there that contain at least one of }10\heartsuit,\:Q\spadesuit,\:A\clubsuit\text{ or }6\diamondsuit\,?

    There are:. {52\choose5} \:=\<2,\!598,\!960 possible poker hands.


    How many hands contain none of the four cards?

    There are:. {48\choose5} \:=\:1,\!712,\!304 such hands.


    Answer: . 2,\!598,\!960 - 1,\!712,\!304 \:=\:886,\!656

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