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Deck of Cards Probability Question

  1. Jan 9, 2010 #1
    1. The problem statement, all variables and given/known data

    What is the probability of gettting four-of-a-kind in a thirteen card hand dealt from a standard fifty two card deck?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 9, 2010 #2
    Are you asking for the probably at least one four of a kind? Or is having multiple four of a kinds in the hand illegal?
     
  4. Jan 10, 2010 #3
    I am asking for at least one four of a kind.
     
  5. Jan 10, 2010 #4

    vela

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    I got about 3.43%. What have you tried so far?
     
  6. Jan 10, 2010 #5
    I got this answer as well.
     
  7. Jan 10, 2010 #6
    I got .0753 = ((13)(52C9))/(52C13) although i suspect that there are some repetitious combinations ie. 4,4,4,4,5,6,7,7,7,7,5,6,2 and 7,7,7,7,5,6,4,4,4,4,5,6,2.

    Could you explain how you arrived at your answer?
     
  8. Jan 10, 2010 #7

    vela

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    Your mistake is in the numerator. After you draw the four of a kind, how many cards are left from which to draw the remaining 9 cards?
     
  9. Jan 10, 2010 #8
    ahh, i see
    Thanks!
    I'm still not sure how to eliminate the repetitious combinations though...
     
  10. Jan 10, 2010 #9

    vela

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    You haven't eliminated repetitions like getting two or three sets of four of a kind, but that's okay because you said you wanted the probability of at least one four of a kind, not exactly one four of a kind.
     
  11. Jan 10, 2010 #10
    "You haven't eliminated repetitions like getting two or three sets of four of a kind"

    In calculating the combinations for any particular four-of-a-kind, for example for the combinations of the four-of-a-kinds for cards 4 and 7, there will be one combination that is counted twice: 4,4,4,4,5,6,7,7,7,7,5,6,2 and 7,7,7,7,5,6,4,4,4,4,5,6,2.
     
  12. Jan 10, 2010 #11

    vela

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    Yeah, you're right. I'll have to ponder this a bit.
     
  13. Jan 10, 2010 #12

    vela

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    This might help:

    [tex]\{\texttt{hands with three four-of-a-kinds}\} \subset \{\texttt{hands with at least two four-of-a-kinds}\}[/tex]
    [tex]\subset \{\texttt{hands with at least one four-of-a-kind}\}[/tex]
     
    Last edited: Jan 10, 2010
  14. Jan 10, 2010 #13

    LCKurtz

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    I agree, but remember probabilities are numbers between 0 and 1, not percentages.

    So you mean .034 approximately.
     
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