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Math Help - Taking 7 cards

  1. #1
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    Taking 7 cards

    We draw the top 7 cards from a well-shuffled 52 card deck. What is the probability that the 7 cards contain exactly 3 aces?

    I get 9/1547 but I am horribly error prone on these. Would appreciate confirmation, but if you tell me it is wrong then that is ok too.

    Andrew
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  2. #2
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    Re: Taking 7 cards

    The total # of possible outcomes is 52 choose 7 = 52!/(45!*7!) = 133784560. The number of ways to get exactly 3 aces is (4 choose 3) * (49 choose 4) = 4!/(3!1!) * 49!/45!4! = 847504. Note that the "4 choose 3" is the number of combinations of 3 aces out of 4 aces in the deck, and the "49 choose 4" is the number of combinations of 4 additional cards out of the remaining 49 cards left in the deck after the 3 aces.

    So the probability you seek is 847504/133784560 = .006334842. This is different than 9/1547 = .005817712

    How did you arrive at 9/1547?
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  3. #3
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    Re: Taking 7 cards

    Don't you actually need to do 48 choose 4, not 49 choose 4? You are not allowed to include the ace that hasn't been chosen, because you could then end up choosing all four aces. If you use 48 choose 4, you do get 9/1547.
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  4. #4
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    Re: Taking 7 cards

    Quote Originally Posted by Luminary View Post
    Don't you actually need to do 48 choose 4, not 49 choose 4? You are not allowed to include the ace that hasn't been chosen, because you could then end up choosing all four aces. If you use 48 choose 4, you do get 9/1547.
    That is correct.
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  5. #5
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    Re: Taking 7 cards

    Thanks for the responses. It is precisely because of these little tricks with the 48 instead of 49 that I like to get a check on my answers. This time I think I avoided the errors though.

    Thanks again

    Andrew
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  6. #6
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    Re: Taking 7 cards

    Yes I stand corrected your answer 9/1547 is correct! Forgot about that sneaky last ace...
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