5-Card Poker Two-Pair Probability Calculation












1












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Question: What is the probability that 5 cards dealt from a deck of 52 (without replacement) contain exactly two distinct pairs (meaning no full house)?



Solution: $$frac{binom{13}{2}binom{4}{2}binom{4}{2}binom{11}{1}binom{4}{1}}{binom{52}{5}} = 0.047539$$



Why doesn't ${13choose 1}{4choose 2}{12choose 1}{4choose 2}{11choose 1}{4choose 1}over{52choose 5}$ OR ${13choose 3}{4choose 2}{4choose 2}{4choose 1}over{52choose 5}$ work?






probability discrete-mathematics poker







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  • 1




    $begingroup$
    You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Aug 29 '15 at 23:32
















1












$begingroup$


Question: What is the probability that 5 cards dealt from a deck of 52 (without replacement) contain exactly two distinct pairs (meaning no full house)?



Solution: $$frac{binom{13}{2}binom{4}{2}binom{4}{2}binom{11}{1}binom{4}{1}}{binom{52}{5}} = 0.047539$$



Why doesn't ${13choose 1}{4choose 2}{12choose 1}{4choose 2}{11choose 1}{4choose 1}over{52choose 5}$ OR ${13choose 3}{4choose 2}{4choose 2}{4choose 1}over{52choose 5}$ work?






probability discrete-mathematics poker







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  • 1




    $begingroup$
    You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Aug 29 '15 at 23:32














1












1








1





$begingroup$


Question: What is the probability that 5 cards dealt from a deck of 52 (without replacement) contain exactly two distinct pairs (meaning no full house)?



Solution: $$frac{binom{13}{2}binom{4}{2}binom{4}{2}binom{11}{1}binom{4}{1}}{binom{52}{5}} = 0.047539$$



Why doesn't ${13choose 1}{4choose 2}{12choose 1}{4choose 2}{11choose 1}{4choose 1}over{52choose 5}$ OR ${13choose 3}{4choose 2}{4choose 2}{4choose 1}over{52choose 5}$ work?






probability discrete-mathematics poker







share|cite|improve this question











$endgroup$




Question: What is the probability that 5 cards dealt from a deck of 52 (without replacement) contain exactly two distinct pairs (meaning no full house)?



Solution: $$frac{binom{13}{2}binom{4}{2}binom{4}{2}binom{11}{1}binom{4}{1}}{binom{52}{5}} = 0.047539$$



Why doesn't ${13choose 1}{4choose 2}{12choose 1}{4choose 2}{11choose 1}{4choose 1}over{52choose 5}$ OR ${13choose 3}{4choose 2}{4choose 2}{4choose 1}over{52choose 5}$ work?






probability discrete-mathematics poker




probability discrete-mathematics poker






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edited Aug 30 '15 at 4:53









Graham Kemp

87.8k43578




87.8k43578










asked Aug 29 '15 at 23:05









Adam TachéAdam Taché

255




255








  • 1




    $begingroup$
    You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Aug 29 '15 at 23:32














  • 1




    $begingroup$
    You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Aug 29 '15 at 23:32








1




1




$begingroup$
You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Aug 29 '15 at 23:32




$begingroup$
You can obtain $binom{n}{k}$ by typing binom{n}{k} in math mode. Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Aug 29 '15 at 23:32










3 Answers
3






active

oldest

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8












$begingroup$

In your first alternative, you're overcounting. You could first choose two eights and then choose two sevens, or first choose two sevens and then choose two eights, and you'd be counting this as two different possibilities, but they yield the same hands.



In your second alternative, you're undercounting. You're just choosing three card values, but you're not saying which of them are pairs and which is the singlet, so even though $(7,7,8,8,9)$, $(7,7,8,9,9)$ and $(7,8,8,9,9)$ all yield different hands, you're counting all three as one.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:



    $$
    binom{13}{2}
    cdot binom{4}{2}
    cdot binom{4}{2}
    cdot binom{13 - 2}{1}
    cdot binom{4}{1}
    = 123,552
    $$



    Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      First, let's look the steps in the process of forming a 5-card hand pairs, say 2 eight and 2 aces:




      1. Choose two face values out of 13 possible face values for the pairs (eights and aces):
        13 choose 2 = 78

      2. Choose two cards from the smaller face value (8(clubs), 8(spades)):
        4 choose 2 = 6

      3. Choose two cards from the larger face value (A(spades),A(clubs));
        4 choose 2 = 6

      4. Choose one card from those remaining (9(diamond)):
        44 choose 1 = 44 (eliminating the 8 face cards already considered: 52-8=44)


      By the product rule, H=78*6*6*44=123552






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Welcome to MSE. Your answer adds nothing new to the already existing answers.
        $endgroup$
        – José Carlos Santos
        Dec 13 '18 at 14:26










      • $begingroup$
        Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
        $endgroup$
        – hardmath
        Dec 13 '18 at 14:28














      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes









      8












      $begingroup$

      In your first alternative, you're overcounting. You could first choose two eights and then choose two sevens, or first choose two sevens and then choose two eights, and you'd be counting this as two different possibilities, but they yield the same hands.



      In your second alternative, you're undercounting. You're just choosing three card values, but you're not saying which of them are pairs and which is the singlet, so even though $(7,7,8,8,9)$, $(7,7,8,9,9)$ and $(7,8,8,9,9)$ all yield different hands, you're counting all three as one.






      share|cite|improve this answer









      $endgroup$


















        8












        $begingroup$

        In your first alternative, you're overcounting. You could first choose two eights and then choose two sevens, or first choose two sevens and then choose two eights, and you'd be counting this as two different possibilities, but they yield the same hands.



        In your second alternative, you're undercounting. You're just choosing three card values, but you're not saying which of them are pairs and which is the singlet, so even though $(7,7,8,8,9)$, $(7,7,8,9,9)$ and $(7,8,8,9,9)$ all yield different hands, you're counting all three as one.






        share|cite|improve this answer









        $endgroup$
















          8












          8








          8





          $begingroup$

          In your first alternative, you're overcounting. You could first choose two eights and then choose two sevens, or first choose two sevens and then choose two eights, and you'd be counting this as two different possibilities, but they yield the same hands.



          In your second alternative, you're undercounting. You're just choosing three card values, but you're not saying which of them are pairs and which is the singlet, so even though $(7,7,8,8,9)$, $(7,7,8,9,9)$ and $(7,8,8,9,9)$ all yield different hands, you're counting all three as one.






          share|cite|improve this answer









          $endgroup$



          In your first alternative, you're overcounting. You could first choose two eights and then choose two sevens, or first choose two sevens and then choose two eights, and you'd be counting this as two different possibilities, but they yield the same hands.



          In your second alternative, you're undercounting. You're just choosing three card values, but you're not saying which of them are pairs and which is the singlet, so even though $(7,7,8,8,9)$, $(7,7,8,9,9)$ and $(7,8,8,9,9)$ all yield different hands, you're counting all three as one.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 29 '15 at 23:17









          jorikijoriki

          171k10189351




          171k10189351























              3












              $begingroup$

              Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:



              $$
              binom{13}{2}
              cdot binom{4}{2}
              cdot binom{4}{2}
              cdot binom{13 - 2}{1}
              cdot binom{4}{1}
              = 123,552
              $$



              Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:



                $$
                binom{13}{2}
                cdot binom{4}{2}
                cdot binom{4}{2}
                cdot binom{13 - 2}{1}
                cdot binom{4}{1}
                = 123,552
                $$



                Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:



                  $$
                  binom{13}{2}
                  cdot binom{4}{2}
                  cdot binom{4}{2}
                  cdot binom{13 - 2}{1}
                  cdot binom{4}{1}
                  = 123,552
                  $$



                  Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in






                  share|cite|improve this answer









                  $endgroup$



                  Reduce your problem to counting a simple unique description of the hand. A hand with two pairs of different value can be described as selecting 2 values for the pairs out of 13, 2 suits out of 4 for the lesser pair, 2 suits out of 4 for the higher pair, the value of the fifth card out of the remaining 11 values, and its suit out of 4. Each of these is independent of the others. In all:



                  $$
                  binom{13}{2}
                  cdot binom{4}{2}
                  cdot binom{4}{2}
                  cdot binom{13 - 2}{1}
                  cdot binom{4}{1}
                  = 123,552
                  $$



                  Breaking down complex problems into such easy-to-count descriptions makes it easier to convince yourself (or others!) that no mistake snuck in







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 1 '15 at 12:59









                  vonbrandvonbrand

                  20k63260




                  20k63260























                      0












                      $begingroup$

                      First, let's look the steps in the process of forming a 5-card hand pairs, say 2 eight and 2 aces:




                      1. Choose two face values out of 13 possible face values for the pairs (eights and aces):
                        13 choose 2 = 78

                      2. Choose two cards from the smaller face value (8(clubs), 8(spades)):
                        4 choose 2 = 6

                      3. Choose two cards from the larger face value (A(spades),A(clubs));
                        4 choose 2 = 6

                      4. Choose one card from those remaining (9(diamond)):
                        44 choose 1 = 44 (eliminating the 8 face cards already considered: 52-8=44)


                      By the product rule, H=78*6*6*44=123552






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Welcome to MSE. Your answer adds nothing new to the already existing answers.
                        $endgroup$
                        – José Carlos Santos
                        Dec 13 '18 at 14:26










                      • $begingroup$
                        Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
                        $endgroup$
                        – hardmath
                        Dec 13 '18 at 14:28


















                      0












                      $begingroup$

                      First, let's look the steps in the process of forming a 5-card hand pairs, say 2 eight and 2 aces:




                      1. Choose two face values out of 13 possible face values for the pairs (eights and aces):
                        13 choose 2 = 78

                      2. Choose two cards from the smaller face value (8(clubs), 8(spades)):
                        4 choose 2 = 6

                      3. Choose two cards from the larger face value (A(spades),A(clubs));
                        4 choose 2 = 6

                      4. Choose one card from those remaining (9(diamond)):
                        44 choose 1 = 44 (eliminating the 8 face cards already considered: 52-8=44)


                      By the product rule, H=78*6*6*44=123552






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Welcome to MSE. Your answer adds nothing new to the already existing answers.
                        $endgroup$
                        – José Carlos Santos
                        Dec 13 '18 at 14:26










                      • $begingroup$
                        Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
                        $endgroup$
                        – hardmath
                        Dec 13 '18 at 14:28
















                      0












                      0








                      0





                      $begingroup$

                      First, let's look the steps in the process of forming a 5-card hand pairs, say 2 eight and 2 aces:




                      1. Choose two face values out of 13 possible face values for the pairs (eights and aces):
                        13 choose 2 = 78

                      2. Choose two cards from the smaller face value (8(clubs), 8(spades)):
                        4 choose 2 = 6

                      3. Choose two cards from the larger face value (A(spades),A(clubs));
                        4 choose 2 = 6

                      4. Choose one card from those remaining (9(diamond)):
                        44 choose 1 = 44 (eliminating the 8 face cards already considered: 52-8=44)


                      By the product rule, H=78*6*6*44=123552






                      share|cite|improve this answer









                      $endgroup$



                      First, let's look the steps in the process of forming a 5-card hand pairs, say 2 eight and 2 aces:




                      1. Choose two face values out of 13 possible face values for the pairs (eights and aces):
                        13 choose 2 = 78

                      2. Choose two cards from the smaller face value (8(clubs), 8(spades)):
                        4 choose 2 = 6

                      3. Choose two cards from the larger face value (A(spades),A(clubs));
                        4 choose 2 = 6

                      4. Choose one card from those remaining (9(diamond)):
                        44 choose 1 = 44 (eliminating the 8 face cards already considered: 52-8=44)


                      By the product rule, H=78*6*6*44=123552







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 13 '18 at 14:08









                      Shamsul HudaShamsul Huda

                      12




                      12












                      • $begingroup$
                        Welcome to MSE. Your answer adds nothing new to the already existing answers.
                        $endgroup$
                        – José Carlos Santos
                        Dec 13 '18 at 14:26










                      • $begingroup$
                        Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
                        $endgroup$
                        – hardmath
                        Dec 13 '18 at 14:28




















                      • $begingroup$
                        Welcome to MSE. Your answer adds nothing new to the already existing answers.
                        $endgroup$
                        – José Carlos Santos
                        Dec 13 '18 at 14:26










                      • $begingroup$
                        Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
                        $endgroup$
                        – hardmath
                        Dec 13 '18 at 14:28


















                      $begingroup$
                      Welcome to MSE. Your answer adds nothing new to the already existing answers.
                      $endgroup$
                      – José Carlos Santos
                      Dec 13 '18 at 14:26




                      $begingroup$
                      Welcome to MSE. Your answer adds nothing new to the already existing answers.
                      $endgroup$
                      – José Carlos Santos
                      Dec 13 '18 at 14:26












                      $begingroup$
                      Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
                      $endgroup$
                      – hardmath
                      Dec 13 '18 at 14:28






                      $begingroup$
                      Welcome to Math.SE. While your post is a clear enough explanation of applying the principle of dependent choices, the Question here was not so much about how to count the hands as it was about why the two alternative methods give different results. In any case when responding to a three year old Question with previous upvoted and/or Accepted Answers, it is a good practice to read those carefully in order to highlight what new information you are trying to add.
                      $endgroup$
                      – hardmath
                      Dec 13 '18 at 14:28




















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