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Math Help - Bridge Odds: How to calculate distribution of aces?

  1. #1
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    Bridge Odds: How to calculate distribution of aces?

    Total number of ways to distribute Aces is: {7 \choose 4}=35

    4 aces in one player's hands:
    (4,0,0,0) = 4 ways

    3 aces in 1 player's hands:
    (3,1,0,0) = 12 ways

    2 aces in 1 player's hands:
    (2,2,0,0) = 6 ways
    or
    (2,1,1,0) = 12 ways

    1 aces in each player's hand:
    (1,1,1,1) = 1 way

    Total is 35 ways.

    But can anybody tell me how to calculate the probability of each scenario happening? Say, (2,1,1,0).

    Thanks
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  2. #2
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    Hello, chopet!

    We need some serious clarification . . .

    You said "bridge hands", so I assume 13 cards are dealt to each of four players.

    Next, are the four players distinguishable? .Can we call them A, B, C, and D?

    If so, there are: . {\color{blue}{52\choose13,13,13,13}} possible ways to deal the cards.

    And the problem is far more complicated than you think . . .



    Four Aces in one player's hand

    There are 4 choices of players to get the four Aces.
    There is 1 way for him get the four Aces.
    For his other 9 cards, there are: {\color{red}{48\choose9}} choices.
    Then the remaining cards are given to the other three players in: . {\color{red}{39\choose13,13,13}} ways.

    Hence, there are: . {\color{blue}4\cdot{48\choose9}\cdot{39\choose13,1  3,13}} ways for one player to get the four Aces.



    Three Aces in one player's hand, one Ace in another's

    There are 4 choices of players to get the three Aces.
    There are: {\color{red}{4\choose3}} ways for him to get three Aces.
    For his other 10 cards, there are: {\color{red}{48\choose10}} ways.

    There are 3 choices of players to get the one Ace.
    There is one way for him to get the one remaining Ace.
    For his other 12 cards, there are: {\color{red}{38\choose12}} ways.

    The remaining 26 cards are distributed to the other two players.
    There are: {\color{red}{26\choose13,13}} ways.

    Hence, there are: . {\color{blue}4\cdot{4\choose3}\cdot{48\choose10}\c  dot3\cdot{38\choose12}\cdot{26\choose13,13}} ways


    Get the idea?



    Edit: corrected an error . . . sorry for the blunder.
    .
    Last edited by Soroban; November 16th 2007 at 08:33 PM.
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  3. #3
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    Quote Originally Posted by Soroban View Post

    Hence, there are: . {\color{blue}4\cdot{4\choose3}\cdot{48\choose10}\c  dot3\cdot{48\choose12}\cdot{26\choose13,13}} ways
    Hi, thanks!
    but for the guy with only ace. should it be:
    . {\color{blue}4\cdot{4\choose3}\cdot{48\choose10}\c  dot3\cdot\color{red}{39\choose12}\cdot\color{blue}  {26\choose13,13}}
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  4. #4
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    Hello again, chopet!

    Thank you for pointing out my error . . .

    After the first player gets his 13 cards (including 3 Aces),
    . . there are 3 choices for the next player.

    There is one way for him to get the remaining Ace.
    His other 12 cards come from the remaining 38 cards.
    So there are: {\color{red}{38\choose12}} ways.

    There are: . {\color{blue}3\cdot{38\choose12}} ways for the second player.


    Three Aces in one player's hand, one Ace in another's

    . . {\color{blue}4{4\choose3}{39\choose12}\cdot 3{38\choose12}\cdot{26\choose13,13}}

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  5. #5
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    Hi,

    I was looking at this problem, and just want to clarify one more point. Say for the case (2,1,1,0), where one player has 2 aces, and 2 players have 1 ace each:

    we have 4 ways to choose Player 1, and {4 \choose2}=6 ways to choose his 2 aces, followed by {3 \choose2}=3 ways to choose Player 2 and 3, and {2 \choose1}=2 way to choose their 1 ace:

    hence giving us:
    4*6*3*2=144

    Is that what you mean?
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  6. #6
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    bridge hands

    here is my crack at this. there are a total of (52/13) hands equal to 52 factorial divided by the product 0f 13 factorial and 39 factorial.
    the hand containing 4 aces is filled from the remaining 48 cards taken 9 at atime.
    by eliminating the factorials the equation becomes

    13x12x11x10 dividedby 52x51x50x49 equals .0026 the probability
    0f a designated player getting the 4 aces . probability of one of the four getting them is 4times greater or ,01. four aces would appear once in every 100 deals
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