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Homework Help: Probability of each of 4 players getting an Ace

  1. Apr 28, 2012 #1
    1. The problem statement, all variables and given/known data

    A well-shuffled 52-card deck is dealt to 4 players. Find the probability that each of the players gets an Ace.

    2. Relevant equations

    Monomial coefficient.

    3. The attempt at a solution

    Each player gets 13 cards.

    Total possible partitions from a 52 card deck with 4 groups each having 13 cards is:

    Total permutations of an Ace is 4!

    Why is the answer not 4!/(52!/(13!)^2)?
  2. jcsd
  3. Apr 29, 2012 #2

    Ray Vickson

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    Think of dealing 13 cards first to player 1, then another 13 to player 2, etc. In a draw of 13 cards from a deck having 4 aces and 48 non-aces, player 1 must get exactly 1 ace. What is the probability of that? Now to player 2 we deal 13 cards from a deck having 3 aces and 36 non-aces, and player 2 must get exactly 1 ace. What is the probability of that? Then we deal 13 cards to player 3 from a deck having 2 aces and 24 non-aces, and player 3 must get exactly 1 ace. What is the probability of that? Of course, if players 1-3 each have exactly 1 ace we are done: player 4 will also get one ace. So, what is the final probability?

  4. Apr 29, 2012 #3
    Thank you for the reply. I know that the problem can be done as you suggested but I was wondering what the fallacy is in the method that I proposed.
  5. Apr 29, 2012 #4

    Ray Vickson

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    You counted things incorrectly. Suppose we want to distribute N (distinct) cards to A, B, C and D, with Mr. A getting a cards, Mr. B getting b cards, Mr. C getting c cards and Mr. D getting d cards (with a+b+c+c=N). How many different dealings there?
    [tex]\text{Answer} = \frac{N!}{a! b! c! d!}.[/tex]
    If you don't know this, just accept it for now; we will prove it later.

    So, the number of dealings = 52!/(13!)^4.

    How many dealings give each of A,B,C,D exactly one Ace (assuming 4 aces in the deck)? Well, there are 4! ways to give each player one ace, then there are
    [tex] \frac{(N-4)!}{(a-1)! (b-1)! (c-1)! (d-1)!}[/tex] dealings of (N-4) non-aces to the players, giving (a-1) non-aces to Mr. A, etc. For each of these dealings, we just give each player one of the aces, so the total number of such dealings is
    [tex] \frac{4! (N-4)!}{(a-1)! (b-1)! (c-1)! (d-1)!}.[/tex] In the example we have N=52, a=b=c=d=13, so the number of one-ace-per-player dealings is 4!*48!/(12!)^4. The required probability is
    [tex] \text{prob.} = \frac{(13!)^4}{52!} \cdot \frac{4! 48!}{(12!)^4} = \frac{2197}{20825} \doteq 0.1050,[/tex] You would get exacctly the same probability by implementing the approach in my previous post.

    Finally, why is the count as stated above? Look at it sequentially: for the case of the total number of dealings, we first determine the number of distinct hands Mr. A could have; that is C(N,a) = N!/[a! (N-a)!]. Now there are (N-a) cards from which we deal b cards to Mr. B, giving him a total of C(N-a,b) possible hands. Then Mr. C can have C(N-a-b,c) hands, and that leaves Mr. D with one hand. The total is C(N,a)*C(N-a,b)*C(N-a-b,c), which equals the previous result when we recognize that d = N-a-b-c.

    Last edited: Apr 30, 2012
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