Results 1 to 4 of 4

Math Help - Probability (Bridge Hands)

  1. #1
    Newbie
    Joined
    Sep 2007
    Posts
    3

    Smile Probability (Bridge Hands)

    Hello this is my first posting so I hope I'm doing everything right!

    Okay so here is the question
    ----------------------------
    What is the probability that a Bridge Hande (13 cards) contains:
    a) 4 spades, 5 hearts, 3 diamonds, and 1 club

    b) no spades?

    c) a void? (at least one suit that is not represented)
    -----------------------------

    Now I think I answered the first one correctly, here it goes:
    -----------Answers-----------
    a) (13C4 * 13C5 * 13C3 * 13C1)/(52C13)
    = 0.00538779

    b) now im not sure for this one but (is it correct?)...
    (39C13)/(52C13) = 0.01279094804

    c) I thought that the answer to this would be the same as b) since one suit was not represented in b)
    -------------------------------

    Well thanks in advance to anyone who responds with some input, I'm not great at this stats stuff, and our prof is not giving great notes :S.

    Edit: woops sorry, I didn't see that I was in the Elementary and middle school section, I will try to repost in the correct section if someone could delete this posting
    Last edited by shadarap; September 13th 2007 at 06:13 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by shadarap View Post
    Hello this is my first posting so I hope I'm doing everything right!

    Okay so here is the question
    ----------------------------
    What is the probability that a Bridge Hande (13 cards) contains:
    a) 4 spades, 5 hearts, 3 diamonds, and 1 club

    b) no spades?

    c) a void? (at least one suit that is not represented)
    -----------------------------

    Now I think I answered the first one correctly, here it goes:
    -----------Answers-----------
    a) (13C4 * 13C5 * 13C3 * 13C1)/(52C13)
    = 0.00538779

    b) now im not sure for this one but (is it correct?)...
    (39C13)/(52C13) = 0.01279094804

    c) I thought that the answer to this would be the same as b) since one suit was not represented in b)
    -------------------------------

    Well thanks in advance to anyone who responds with some input, I'm not great at this stats stuff, and our prof is not giving great notes :S.

    Edit: woops sorry, I didn't see that I was in the Elementary and middle school section, I will try to repost in the correct section if someone could delete this posting
    Next time, just leave it. the moderators can move it to the correct forum
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,811
    Thanks
    701
    Hello, shadarap!

    Welcome aboard!


    What is the probability that a Bridge Hande (13 cards) contains:

    a) 4 spades, 5 hearts, 3 diamonds, and 1 club?

    b) no spades?

    c) a void? (at least one suit that is not represented)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Now I think I answered the first one correctly, here it goes:


    a)\;\;\frac{(_{13}C_4)(_{13}C_5)(_{13}C_3)(_{13}C_  1)}{_{52}C_{13}} \;=\; 0.00538779 . . . . correct!


    b) now im not sure for this one ... but is it correct? . . . . yes!
    . . \frac{_{39}C_{13}}  {_{52}C_{13}} \:= \:0.01279094804


    c) I thought that the answer to this would be the same as b)
    since one suit was not represented in b)

    In part (b), we have the probability that one particular suit is void.
    . . And there are four suits that could be void.

    Note that it says: at least one suit is void.
    . . This means: one suit is void, or two suits are void, or three suits are void.


    We must consider the three cases:

    One suit void
    There are 4 choices for the suit to be void.
    Then there are: _{39}C_{13} ways to select 13 cards from the other 39 cards.
    . . P(\text{1 suit void}) \;=\;\frac{4\cdot(_{39}C_{13})}{_{52}C_{13}}

    Two suits void
    There are _4C_2 = 6 ways to choose the two suits to be void.
    Then there are: _{26}C_{13} ways to choose the 13 cards from the other 26 cards.
    . . P(\text{2 suits void}) \;=\;\frac{6\cdot(_{26}C_{13})}{_{52}C_{13})}

    Three suits void
    There are _4C_3 \,=\,4 choices for the 3 suits to be void.
    If three suits are void, all 13 cards are from the one remaining suit.
    There is one way to select 13 cards of one suit.
    . . P(\text{3 suits void}) \;=\;\frac{4\cdot1}{_{52}C_{13}}


    The answer is the sum of these three probabilities.

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2007
    Posts
    3
    Thank you so much for adding that reply, the detail really helped me understand what was meant. It is greatly appreciated, now hopefully I can help someone else out some day to return the favour

    ~Randy
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Bridge hands
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: May 16th 2010, 11:56 AM
  2. Probability in a game of Bridge
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: March 17th 2009, 09:08 PM
  3. Bridge Probability
    Posted in the Statistics Forum
    Replies: 3
    Last Post: January 31st 2009, 07:48 AM
  4. Probability of Poker Hands
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: September 17th 2007, 06:54 PM
  5. Probability (with Bridge Hands)
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: September 17th 2007, 03:47 AM

Search Tags


/mathhelpforum @mathhelpforum