1. Limited time only! Sign up for a free 30min personal 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 a single pair poker hand

  1. Oct 9, 2016 #1

    hotvette

    User Avatar
    Homework Helper

    1. The problem statement, all variables and given/known data
    How many one-pair 5-card poker hands are there in a standard 52-card deck?

    2. Relevant equations
    [tex]
    {}^nC_k = \frac{n!}{(n-k)!}
    [/tex]

    3. The attempt at a solution
    I've seen the following solution in several different places:
    [tex]
    n = {}^{13}C_1 \cdot {}^{4}C_2 \cdot {}^{12}C_3 \cdot {}^{4}C_1 \cdot {}^{4}C_1 \cdot {}^{4}C_1 = 1098240
    [/tex] and every term makes perfect sense. What I don't understand though, is why the following:
    [tex] n = {}^{13}C4 \cdot {}^{4}C_2 \cdot {}^{4}C_1 \cdot {}^{4}C_1 \cdot {}^4C_1
    [/tex] underestimates the answer by a factor of 4. To me 13 choose 4 makes just as much sense as 13 choose one followed by 12 choose three, but of course one is correct and the other isn't. Can someone explain?
     
    Last edited: Oct 10, 2016
  2. jcsd
  3. Oct 9, 2016 #2

    MarneMath

    User Avatar
    Education Advisor

    That's fine but you need to count the number of suites you can pick.
     
  4. Oct 9, 2016 #3

    hotvette

    User Avatar
    Homework Helper

    Hmm, doesn't the 4C2 take care of suits? After thinking about this for a while, it seems to me the factor of 4 comes from the fact that {xx yzw} can be arranged 4 ways given that the order of yzw doesn't matter. Is this the correct way to look at it?

    I think so, because I tried the same logic with 3 of a kind {xxx yz}. One solution is 13C1 x 4C3 x 12C2 x 4C1 x 4C1 = 54912 but a second way is 13C3 x 4C3 x 4C1 x 4C1 x 3C1 where the last 3C1 is the number of ways to arrange 3 values in {xxx yz} given the order of yz doesn't matter.

    I also tried it with 4 of a kind {xxxx y}. One solution is 13C1 x 4C4 x 12C1 x 4C1 = 624 but a second way is 13C2 x 4C4 x 4C1 x 2C1 where the last 2C1 is the number of says to arrange 2 values in {xxx y}.

    If my logic is correct, this is all (finally) making sense.
     
    Last edited: Oct 9, 2016
  5. Oct 10, 2016 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Pleas turn off the bold font; it makes it look like you are yelling at us, and is very distracting.

    Anyway, I do not understand where your ##{13 \choose 4} \equiv {}^{13}C_4## comes from. What is the logic behind that way of writing things?
     
  6. Oct 10, 2016 #5

    hotvette

    User Avatar
    Homework Helper

    Strange, I never turned bold on in the first place, but clearly it was bold. Fixed now. Anyway, my logic was that for a single pair, I'm ultimately choosing 4 different cards from a pool of 13, one for the pair plus the remaining three. The math works if I consider that there are 4 different ways to group the 4 cards when using that scenario (each of the 4 cards could be the pair and the order of the remaining three does not matter). The logic seems to work for all of the scenarios I tried (one pair, two pair, four of a kind, 4 of a kind).
     
  7. Oct 10, 2016 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    After you have chosen 4 different cards from the 13 ranks, you need to pick ONE of those 4 ranks for duplication across suits, so there are ##C^4_1 = 4## ways of choosing the rank to be paired, then ##C^4_2## ways of choosing the pair from the 4 suits at that rank. Then, of course, for each of the other three ranks there are ##C^4_1## ways of choosing the suit. That gives the total number as ##C^{13}_4\, C^4_1 \,C^4_2 \, C^4_1 \, C^4_1 \, C^4_1 = 1098240##. THIS is the same as the other computation you mentioned!
     
  8. Oct 12, 2016 #7

    hotvette

    User Avatar
    Homework Helper

    Thanks, I think you confirmed what I was asking.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Probability of a single pair poker hand
Loading...