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Poker 10 Decks, 10 cards, independent 4 Aces

  1. Nov 11, 2006 #1
    Silly question of very little importance :smile:

    My dad just asked me the following question: You have 10 standard independent decks (52 cards), you draw 1 card from each deck 1 at a time, what is the probability (count) of getting 4 aces (not necessarily distinct suits) out of the 10 cards you draw.
    -------

    I am thinking that the number of ways to get such a "hand" of 10 cards is the following:
    [tex]\binom{10}{4}\left(\frac{4}{52}\right)^4\left(\frac{48}{52}\right)^6[/tex]

    Reasoning being that you want 4 aces and there are 4 per deck, hence the (4/52)^4 and for the other 6 cards you want you have (48/52)^6, then 10 choose 4 for all the different ways you can get the 4 aces out of the 10 cards.

    And if we wanted the aces to have distinct suits we would have 4! instead of 4^4

    Does this look correct? Thanks.
     
  2. jcsd
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