Results 1 to 5 of 5

Math Help - Poker full house problem

  1. #1
    Member
    Joined
    Apr 2008
    Posts
    132

    Poker full house problem

    In poker a Full House is a hand consisting of 3 cards of one kind and 2 cards of another kind. For example, 'eights and fives' is a hand consisting of 3 eights and 2 fives. hat is the probability of being dealt a full house from a standard deck of 52 cards if the dealing is done fairly?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    The number of possible fullhouses is 13*12*6*4. See why?.

    The total number of 5 card hands is C(52,5).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Apr 2008
    Posts
    132

    not sure?

    thanks,i dont exactly see wher the 13 12 6 and 4 come from?could you give me any help?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,686
    Thanks
    617
    Hello, matty888!

    Here ya go . . .


    What is the probability of being dealt a Full House?
    We want a Triple and a Pair.

    There are 13 choices of value for the Triple.
    There are: . {4\choose3} = 4 ways to get the Triple.

    There 12 choices for the value of the Pair.
    There are: . {4\choose2} = 6 ways to get the Pair.

    Hence, there are: . 13 \times 4 \times 12 \times 6 \:=\:3744 Full Houses.


    There are: . {52\choose5} = 2,598,960 possible five-card hands.


    Therefore: . P(\text{Full House}) \;=\;\frac{3,\!744}{2,\!598,\!960} \;=\;\frac{6}{4165}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    13 is the # of suits.

    6 is the number of combinations possible for the pairs, i.e. 2 cards from 4 choices.

    4 is the # of each card type that would get 3 of a kind.

    12 is the card types left after one is used up for the 3 of a kind.

    So, we have 13*6*4*12=3744
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How many 5-card hands make full house
    Posted in the Statistics Forum
    Replies: 1
    Last Post: June 4th 2009, 12:58 PM
  2. Full-rank problem.
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: March 23rd 2009, 03:21 AM
  3. the House Problem
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: March 7th 2009, 05:20 PM
  4. full-rank matrix problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 13th 2008, 10:59 AM
  5. Poker Problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 10th 2008, 01:53 AM

Search Tags


/mathhelpforum @mathhelpforum