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Combinatorics question

  1. May 5, 2009 #1
    This question is out of the book, graduate problems in physics. problem 4 mathematical physics.

    In dealing 52 cards among two teams (containing two partners), what is the probability a particular team will obtain one complete suit between them?

    To determine the answer we need to find the number of ways one team can be dealt one particular suit regardless of whether they have another complete suit. and the answer is (39!)(26!)/(13!). but I don't understand how they got this part to help solve the answer. Can you help me?

    I know there are 13 cards in a suit and (52-13)=39 which is the number of cards left after one team has a suit. I don't understand what the 26 represents. But I do know that with combinatorics, the number in the numerator is the total probability of cards and the number in the deminator is the number by which you over count the total probability of cards.

    Thanks
     
  2. jcsd
  3. May 6, 2009 #2
    Welcome to PF, Physicbum.

    First let me make clear that this formula is just one step in the overall solution, not the final answer.

    Also, in their solution, they are treating the size of sample space as 52! which is the number of ways to distribute 52 cards onto an ordered sequence of 52 possible positions.

    Let's count the number of ways that our team can be dealt the entire suit of hearts regardless of whether we have another complete suit. Our team owns 26 of the possible card positions. Let's count the number of ways to distribute the 13 heart cards into 13 positions chosen from an ordered sequence of 26 available positions.

    Let the cards choose the position they are going to go to.

    The ace of hearts has 26 positions to choose from.
    Then the 2 of hearts has 25.
    etc.
    Then the K of hearts has 14.

    This gives 26*25*24*...*14.

    But 26*25*24*...*14 = 26!/13!. That's where the 26! and 13! come from.

    Then the 39! is the number of ways to distribute the remaining 39 cards onto the remaining 39 positions.
     
  4. May 10, 2009 #3
    Thanks for helping me out on the 26!*39! / 13! I understand now that the 39! and 26! / 13! are separate probabilities that occurring at the same time.Thus their probabilities need to be multiplied. Thanks for your help.
     
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