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Homework Help: A probability question

  1. Jun 28, 2010 #1
    52 cards, 13 values (A to K) and 4 suits. what's the probability of getting 5 cards with a four-of-a-kind (same value)?
    My solution: first card 52/52, second 3/51, third 2/50. fourth 1/49, fifth 48/48
    so (52/52)*(3/51)*(2/50)*(1/49)*(48/48)~4.8e-5
    The solution provided by the book 13x48/(C 52 5)~2.4e-4
    My solution is 1/5 of the answer. Did I miss something? I think both ways of solving it make sense

    Thanks
     
  2. jcsd
  3. Jun 28, 2010 #2

    HallsofIvy

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    Science Advisor

    Your solution is the probability that the first four cards dealt are from the same suit and the fifth card from a different suit. It is easy to show that the probability of "second, third, fourth, fifth cards from the same suit but first card from a different suite", "first, third, fourth, fifth cards from the same suit, second card from a different suit", etc. are the same. That "different suit" card can be any one of the five cards dealt so the value you got is multiplied by 5.

    More formally, there are [tex]\begin{pmatrix}5 \\ 1\end{pmatrix}= 5[/tex] ways of arranging five things, four of which are the same.
     
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