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Discrete probability

  1. May 9, 2005 #1
    hi! i have a homework problem here that i'm stuck on:

    What is the probability that a hand of 13 cards contain no pairs?


    i know a 13-card hand is C(52, 13) but i have no idea how to represent the probability of that 13-card hand containing no pairs. can anyone give me a hint?
     
  2. jcsd
  3. May 9, 2005 #2

    shmoe

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    No pairs means at most one of each: Ace, two, ..., ten, jack, queen, king. So how many of each must you have in this case?
     
  4. May 9, 2005 #3

    OlderDan

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    There are probably different approaches, but if you think in terms of the probability of drawing a card that does not match any previously drawn card on every draw I think that will get you there. The first card cannot match any previously drawn card so the probability is 1. The next card has 48/51 probability of not matching the first. The next has 44/50 probability of not matching either of the first two, etc.

    The problem can also be done using a multi-type (or multivariate) hypergeometric distribution, which is, I think, the way you would want to go if you were looking for more than one of each type card instead of just one.
     
  5. May 9, 2005 #4
    that's 13 different cards and i choose 7 of them: C(13,7)?

    so [C(13,7) / C(52, 7)] is the probability?
     
  6. May 10, 2005 #5

    OlderDan

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    I don't see a 7 in the problem at all. What is 7?
     
  7. May 10, 2005 #6
    oops, i meant: [C(13,13) / C(52, 13)]
     
  8. May 10, 2005 #7

    OlderDan

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    That is not it. What has you thinking C(13,13)?
     
  9. May 10, 2005 #8
    umm...13 cards in one hand, so i choose 13 out of the 13 available cards from a suit?

    i'm just trying to make some sense out of the examples from my book... i'm so confused :(
     
  10. May 10, 2005 #9

    OlderDan

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    You have 13 different kinds of cards, with 4 of each kind. You want one card of each kind. How many combinations are there of 4 cards taken 1 at a time? How many combinations are there of four things of one kind taken one at a time, combined with 4 things of a second kind taken one at a time, combined with 4 things of a third kind taken one at a time ...... combined with 4 things of a thirteenth kind taken one at a time?
     
  11. May 10, 2005 #10
    4 things of 1st kind combined with 4 things of a 2nd kind combined with...4 things of a 13th kind = 4^13?

    so [4^13 / C(52, 13)] is the probability?
     
  12. May 10, 2005 #11

    OlderDan

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    Yes

    And if you want to be a bit more formal about it

    C(4, 1)^13/ C(52, 13)

    Note that 4*13 = 52 and 1*13 = 13

    This is a multivariate extension of the hypergeometric distribution for two kinds of things. If you wanted to know the probability of gettng exactly 3 aces in a hand of 7, you would identify aces as one kind, and everything else as another kind and you would have

    P = C(4, 3)*C(48, 4)/C(52, 7)
     
  13. May 10, 2005 #12
    awesome! thanks OlderDan! :smile: :biggrin:
     
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