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Math Help - Card draw probability

  1. #1
    Newbie
    Joined
    Mar 2010
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    Card draw probability

    Hallo,

    I'm working on card drawing probabilities and I'm not quiet getting the results I was hoping for.
    I know that the odds of drawing at least one copy of a desired card in the
    first 4 cards from a deck of 12 cards containing x copies of a desired card is:

    x=1 copy: 0,3333
    x=2 copies: 0,5758
    x=3 copies: 0,7455
    x=4 copies: 0,8586
    x=5 copies: 0,923
    x=6 copies: 0,9697

    My formula for this is:
    Y=1-[(12-x)/12*(11-x)/11*(10-x)/10*(9-x)/9], which yields the above results.
    The part in parantheses is the probability of not drawing on of the desired
    cards for 4 draws from a deck of 12 which then gets subtracted from 1,
    yielding the reverse probability (the prob. of drawing at least 1 copy in that draw).

    Now i want to know, what are the odds of drawing at least 2 copies of a certain card
    in the first 4 drawn cards. My formula for this would be:

    y=(1-[(12-x)/12*(11-x)/11*(10-x)/10*(9-x)/9])*(1-[(12-x)/11*(12-x)/10*(12-x)/9])

    I take the probability of drawing at least 1 copy, then i multiply with the probability of drawing at least 1 copy when drawing 3 cards from a deck of 11, containing x-1 copies of a desired card. I'm not sure if this is the right way of calculating these odds, but the results would be:

    x=1 copy : 0
    x=2 copies: 0,1570
    x=3 copies: 0,3660
    x=4 copies: 0,5672
    x=5 copies: 0,7322
    x=6 copies: 0,8522

    Going forth with this line of thought, the odds of drawing at least 3 copies would then be:

    y=(1-[(12-x)/12*(11-x)/11*(10-x)/10*(9-x)/9])
    *(1-[(12-x)/11*(12-x)/10*(12-x)/9])*(1-[(12-x)/10*(12-x)/9])
    yielding

    x=1 copy : 0
    x=2 copies: 0
    x=3 copies: 0,0732
    x=4 copies: 0,2143
    x=5 copies: 0,3905
    x=6 copies: 0,5681


    and for drawing at least 4 copies:
    y=(1-[(12-x)/12*(11-x)/11*(10-x)/10*(9-x)/9])
    *(1-[(12-x)/11*(12-x/10*(12-x)/9])
    *(1-[(12-x)/10*(12-x)/9])*(1-[(12-x)/9])
    yielding


    x=1 copy : 0
    x=2 copies: 0
    x=3 copies: 0
    x=4 copies: 0,0238
    x=5 copies: 0,0868
    x=6 copies: 0,1894

    Now, if x=4, drawing all 4 desired cards from a deck of 12 in the first 4 draws can also be calculated by
    y=4/12*3/11*2/10*1/9=0,00202, yiedling a results that is different by a factor of >10 from the result obtained
    with my formular.

    Does anyone of you math wizs know how to do this properly?
    Thanx in advance,
    chromeboy
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  2. #2
    MHF Contributor

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    To find probability of at least two we do the following.
    Find the probability of none plus the probability of exactly one.
    Subtract that sum from 1.
    That is your answer.
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