1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
    52!/(13!)^2

    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

    User Avatar
    Science Advisor
    Homework Helper

    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?

    RGV
     
  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

    User Avatar
    Science Advisor
    Homework Helper

    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.

    RGV
     
    Last edited: Apr 30, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Probability of each of 4 players getting an Ace
  1. Probability 4 (Replies: 6)

Loading...