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Card Hand probabilities: Discrete Math

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data
    In the following you are given a 5-card hand from a 52 card deck.
    a) given that you have at least one ace, what is the probability you have at least 2 aces?
    b) given that you have the ace of diamonds, what is the probability that you have another ace?
    c) given that you have a red ace, what is the probability you have another ace?
    d) Suppose you select a card from your hand at random and it is an ace. What is the probability you have another ace.


    2. Relevant equations
    pr(not a) = 1 - pr(a)
    pr(A given B) = Pr(A and B) / Pr(B)


    3. The attempt at a solution

    C(X,Y) is X choose Y here.

    Pr(A): The chance of drawing AT LEAST one Ace is 1 - C(48,5)/C(52,5) =~ .6588

    Pr(B): The probability of having at least 2 aces: ?
    Pr(A and B): ?

    pr(C): That you have the ace of diamonds in a 5 card hand is 1-C(51,5)/C(52,5) =~ .096
    Pr(C and B):?

    pr(D): That you have at least one red ace in a 5 card hand: 1-C(50,5)/C(52,5) =~ .1848
    (im not sure if i want the probability of having exactly 1 red ace, or at least one)
    Pr(D and B):?


    Finding all of the "?" probabilities should allow me to calculate problems a-c, but i really dont know how to find PR(X AND Y) with these problems, nor do I know how to get the probability of having at least two aces.
     
  2. jcsd
  3. Feb 25, 2009 #2

    jambaugh

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    Probability of having at least two aces...
    Hmmm...
    You can calculate the probability of having exactly 4 aces + exactly 3 aces + exactly 2 aces.

    Pr(4) = C(4,4)C(48,1)/C(52,5)
    Pr(3) = C(4,3)C(48,2)/C(52,5)
    Pr(2) = C(4,2)C(48,3)/C(52,5)
    Add 'em up.
     
  4. Feb 25, 2009 #3
    Thanks, that makes sense.

    How do you go about getting the probability of the union though?
     
  5. Feb 25, 2009 #4

    jambaugh

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    Since you are enumerating distinct hands (no hand with exactly 2 aces is a hand with exactly one ace) you can simply add the probabilities.
    Pr(x>1)= Pr(x=2) + Pr(x=3) + Pr(x=4).
     
  6. Feb 25, 2009 #5
    what about pr(1 ace red AND at least 2 aces) one is not necessarily in the other in this case, same with 1 ace diamond
     
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