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!

Combinatorial Probability Question - Poker hand of exactly 3 suits

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

    A five card hand is dealt from a standard 52-card deck. What is the probability that there will be exactly three suits in the hand?

    2. Relevant equations



    3. The attempt at a solution
    I started with (4 C 3)(39 C 5), but this gives me all possible hands with three or less suits. How do I get rid of the the possibility of 2 or 1 suits? I tried (4 C 3)(39 C 5) - (4 C 2)(26 C5), but I am not certain that gets rid of all possible overlaps between the sets.
     
  2. jcsd
  3. Sep 17, 2012 #2
    so you do mean for instance club-club-heart-heart-spade?

    in this case I would presume these are the options:
    1. first choose 3 out of 4 suits. 3 out of 4 gives 4 possibilities.
    2. for instance, say we have chosen club,heart and spade.
    then there are 6 possibilities: cl-cl-cl-heart spade
    cl-cl-heart-heart-spade
    cl-heart-heart-heart-spade
    cl-cl-heart-spade-spade
    cl-heart-heart-spade-spade
    cl-heart-spade-spade-spade

    we have thus 4*6 possibilites= 24.
    now divide this by the total amount of possibilites. I believe this is 4^5.
     
  4. Sep 17, 2012 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I think it is easier to look at what is not in the hand. Say we have no hearts. There must be at least one each of spades, diamonds and clubs; you can represent this as (i) first choose exactly one each of spades, diamonds and clubs; (ii) then, from the remaining deck of 49 cards (13 hearts and 36 others) we must choose two more cards but no hearts. The probability you want is 4 times what you get here, because the 'absent' suit can be selected in 4 ways, and the 4 resulting events are mutually exclusive with the same probability.

    RGV
     
    Last edited: Sep 17, 2012
  5. Sep 17, 2012 #4
    better listen to ray, my analysis is a bit wrong
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Combinatorial Probability Question - Poker hand of exactly 3 suits
Loading...