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Conditional Probability with combinatorics

  1. Aug 22, 2009 #1
    If a hand of 5 cards are dealt from a 52 card pack (order doesn't matter), what is the probability that the hand will contain the ace of spades GIVEN that there is at least one ace?

    Thanks.
     
  2. jcsd
  3. Aug 22, 2009 #2
    I am not certain of my sol'n , but this is what I think :


    Let E represent the event that "In a hand of 5 cards from a 52 card pack (order doesn't matter), that the hand will contain the ace of spades GIVEN that there is at least one ace in the hand"

    Let P(E) represent the probability that event E happens

    Now since I donno how to use the Latex formatting , I am goin to use C(n,k) to represent the no of ways to select k things from a collection of n things.

    i.e. C(n,k) = n! / { ( n-k)! * k! }

    Now ,

    P(E) = P(E given there is exactly one ace ) * P( there is exactly one ace given there is at least one ace)
    +
    P(E given there are exactly 2 aces ) * P( there are exactly 2 aces given there is at least one ace)
    +
    P(E given there are exactly 3 aces ) * P( there are exactly 3 aces given there is at least one ace)
    +
    P(E given there are exactly 4 aces ) * P( there are exactly 4 aces given there is at least one ace)


    Now ,
    P(E given there is exactly one ace ) = 1/4
    P( there is exactly one ace given there is at least one ace) = C(3,0) * C(48,4) / C(51,4)




    Similarly ,

    P(E given there are exactly 2 aces ) = 2/4
    P( there is exactly one ace given there is at least one ace) = c(3,1) * C(48,3) / C(51,4)

    and so on ,

    Thus :

    P(E) = (1/4) * { C(3,0) * C(48,4) / C(51,4) } + (2/4) * { C(3,1) * C(48,3) / C(51,4) }

    + (3/4) * { C(3,2) * C(48,1) / C(51,4) } + (4/4) * { C(3,3) * C(48,0) / C(51,4) }
     
  4. Aug 22, 2009 #3
    srijithu, your approach appears to be reasonable (although dmpiq said it was a 5 card hand not a 4 card hand). However, the problem isn't that complicated.

    Let A = the event that there is at least once ace, and B = the event that there is the ace of spaces.
    P(B|A) = P(A and B) / P(A)
    Now, P(A and B) = P(B) since there is no way to get the ace of spaces without getting at least one ace. So we really just want P(B) / P(A).
    P(B) = C(51,4) / C(52,5)
    P(A) = 1 - P(zero aces) = 1 - C(48,5)/C(52,5)
    P(B|A) = C(51,4)/(C(52,5)*(1-C(48,5) / C(52,5)))
     
    Last edited: Aug 23, 2009
  5. Aug 23, 2009 #4
    Thanks for the replies, but isn't Pr(A and B) = Pr(B) if A is the event 'exactly one ace'? Here, it is at least one ace, so there can be 1 - 4 aces, but the ace of spades always has to be in the hand. Thus, the possible combinations can be:

    Ace of Spades x another 4 cards
    Ace of Spades x Ace of Hearts x Another 3 cards
    Ace of Spades x Ace of Hearts x Ace of Diamonds x Another 2 cards
    Ace of Spades x Ace of Hearts x Ace of Diamonds x Ace of of Clubs x Another card

    Is this algorithm/method correct? Thanks.
     
  6. Aug 23, 2009 #5
    To reiterate:
    A = at least once ace
    B = the ace of spades
    P(A and B) = P(B). Why? Well, suppose your hand has the ace of spades. Then, it has the ace of spades and has at least one ace. Conversely, suppose your hand has the ace of spades and has at least one ace. Then, it has the ace of spades. So the event "B" happens if and only if the event "A and B" happens, so "B" and "A and B" are the same event, so P(A and B) = P(B).

    Ah, I just realized I had a brain fart in my last post (not related to the above conceptual issue)... fixed
     
    Last edited: Aug 23, 2009
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