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Probability of being dealt one pair from a 52-card deck

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Since no one wants to answer my heat conductor problem (*hint hint*), here's a more interesting one.

    If it is assumed that all (52 choose 5) poker hands are equally likely, what is the probability of being deal one pair? (This occurs when the cards have denominations a, a, b, c, d, where a, b, c, and d are all distinct.)

    2. Relevant equations

    Basic rules of probability

    3. The attempt at a solution

    First, let me harken back to an example in the book and then employ the same reasoning.

    The book then goes on to say that we can think of this as a

    This gives the result (10 choose 5)*25 / (20 choose 5).

    Now, it seems like I can use an analogous 5-stage experiment.

    Stage 1:

    We'll have 4 different denominations represented. The number of ways to choose these is (13 choose 4).

    Stage 2:

    Then, from 1 of the chosen denominations, we'll have (4 choose 2) different suits to complete our pair.

    Stages 3-5:

    Then we'll choose 4 different suits from each of the remaining 3 denominations, which allows for 43 possibilities.

    Since the total number of ways to be dealt 5 cards from a deck of 52 is (52 choose 5), the probability of getting a single pair is therefore

    (13 choose 4) * (4 choose 2) * 43 / (52 choose 5).

    However, this isn't the answer in the back of the book. :frown: :frown: :frown: :frown: :frown: :frown: :frown: :frown: :frown: :frown: :frown: :frown: :frown: :frown: :frown:

    Explain my fallacious reasoning.
     
  2. jcsd
  3. Apr 10, 2012 #2

    Ray Vickson

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    It would have been helpful to tell us what the book's answer was. However, I think your reasoning by analogy starts off incorrectly. You seem to break up the cards into 4 groups of 13, but I think the way the book's reasoning went, they would have broken the cards into 13 groups of 4, then ask whether they get two from the same group and 3 from three other distinct groups.

    RGV
     
  4. Apr 10, 2012 #3
    The book's answer is [13*(4 choose 2)*12*(4 choose 1)*11*(4 choose 1)*10*(4 choose 1) / 3!] / (52 choose 5)
     
  5. Apr 10, 2012 #4

    vela

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    Your count for the number of hands is
    $$\binom{13}{4} \binom{4}{2} 4^3 = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\binom{4}{2}4^3$$ while the book's is
    $$\binom{13}{1} \binom{4}{2} \binom{12}{3}4^3 = \frac{13}{1}\times \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \binom{4}{2} 4^3$$ The only difference between your answer and the book's answer is a factor of 4. Can you see why the factor of 4 needs to be there?
     
  6. Apr 10, 2012 #5

    Because after choosing 4 denominations from 13, I then had 4 choices to which I could assign the pair. Correct?
     
  7. Apr 10, 2012 #6

    vela

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    Right.
     
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