Probability that a five-card poker hand contains two pairs












2














What is the probability that a five-card poker hand contains two pairs (that is, two of each of two different ranks and a fifth card of a third rank)?



My attempt:



Let us first pick the 3 different ranks. There are ${13choose 3}$ ways of doing this.
Out of each rank consisting of 4 suits, we must pick 2 cards, 2 cards and 1 card respectively.
So, no. of ways $={13choose 3}cdot {4choose 2}cdot {4choose 2}cdot {4choose 1}$



Total no. of ways of selecting a five-card poker hand $={52choose 5}$



$p=dfrac{{13choose 3}cdot {4choose 2}cdot {4choose 2}cdot {4choose 1}}{{52choose 5}}$



This doesn't match the answer given in the textbook. Where have I gone wrong?






probability







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  • 4




    You forgot to choose which of the three ranks would be the one with only a single card.
    – Henning Makholm
    Nov 14 '15 at 18:59












  • Undercounting, which is unusual. If done in your style It should be $binom{13}{2}binom{11}{1}$.
    – André Nicolas
    Nov 14 '15 at 19:00
















2














What is the probability that a five-card poker hand contains two pairs (that is, two of each of two different ranks and a fifth card of a third rank)?



My attempt:



Let us first pick the 3 different ranks. There are ${13choose 3}$ ways of doing this.
Out of each rank consisting of 4 suits, we must pick 2 cards, 2 cards and 1 card respectively.
So, no. of ways $={13choose 3}cdot {4choose 2}cdot {4choose 2}cdot {4choose 1}$



Total no. of ways of selecting a five-card poker hand $={52choose 5}$



$p=dfrac{{13choose 3}cdot {4choose 2}cdot {4choose 2}cdot {4choose 1}}{{52choose 5}}$



This doesn't match the answer given in the textbook. Where have I gone wrong?






probability







share|cite|improve this question


















  • 4




    You forgot to choose which of the three ranks would be the one with only a single card.
    – Henning Makholm
    Nov 14 '15 at 18:59












  • Undercounting, which is unusual. If done in your style It should be $binom{13}{2}binom{11}{1}$.
    – André Nicolas
    Nov 14 '15 at 19:00














2












2








2







What is the probability that a five-card poker hand contains two pairs (that is, two of each of two different ranks and a fifth card of a third rank)?



My attempt:



Let us first pick the 3 different ranks. There are ${13choose 3}$ ways of doing this.
Out of each rank consisting of 4 suits, we must pick 2 cards, 2 cards and 1 card respectively.
So, no. of ways $={13choose 3}cdot {4choose 2}cdot {4choose 2}cdot {4choose 1}$



Total no. of ways of selecting a five-card poker hand $={52choose 5}$



$p=dfrac{{13choose 3}cdot {4choose 2}cdot {4choose 2}cdot {4choose 1}}{{52choose 5}}$



This doesn't match the answer given in the textbook. Where have I gone wrong?






probability







share|cite|improve this question













What is the probability that a five-card poker hand contains two pairs (that is, two of each of two different ranks and a fifth card of a third rank)?



My attempt:



Let us first pick the 3 different ranks. There are ${13choose 3}$ ways of doing this.
Out of each rank consisting of 4 suits, we must pick 2 cards, 2 cards and 1 card respectively.
So, no. of ways $={13choose 3}cdot {4choose 2}cdot {4choose 2}cdot {4choose 1}$



Total no. of ways of selecting a five-card poker hand $={52choose 5}$



$p=dfrac{{13choose 3}cdot {4choose 2}cdot {4choose 2}cdot {4choose 1}}{{52choose 5}}$



This doesn't match the answer given in the textbook. Where have I gone wrong?






probability




probability






share|cite|improve this question













share|cite|improve this question











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asked Nov 14 '15 at 18:52









ThomasThomas

730416




730416








  • 4




    You forgot to choose which of the three ranks would be the one with only a single card.
    – Henning Makholm
    Nov 14 '15 at 18:59












  • Undercounting, which is unusual. If done in your style It should be $binom{13}{2}binom{11}{1}$.
    – André Nicolas
    Nov 14 '15 at 19:00














  • 4




    You forgot to choose which of the three ranks would be the one with only a single card.
    – Henning Makholm
    Nov 14 '15 at 18:59












  • Undercounting, which is unusual. If done in your style It should be $binom{13}{2}binom{11}{1}$.
    – André Nicolas
    Nov 14 '15 at 19:00








4




4




You forgot to choose which of the three ranks would be the one with only a single card.
– Henning Makholm
Nov 14 '15 at 18:59






You forgot to choose which of the three ranks would be the one with only a single card.
– Henning Makholm
Nov 14 '15 at 18:59














Undercounting, which is unusual. If done in your style It should be $binom{13}{2}binom{11}{1}$.
– André Nicolas
Nov 14 '15 at 19:00




Undercounting, which is unusual. If done in your style It should be $binom{13}{2}binom{11}{1}$.
– André Nicolas
Nov 14 '15 at 19:00










2 Answers
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3














You're pretty close, but there is a problem: you do have to choose 3 ranks, but they're not all going to be treated the same. One will be a single, and two others will be pairs. If you multiply by a factor of $binom{3}{2}$ I think you'll have it.






share|cite|improve this answer





























    1














    In total, there are $52choose5$ ways to draw a hand (this is our |S|).



    We want to choose 2 out of four cards of one value, 2 out of four cards of another value, and one other card not of the first two values (This will be our |E|).



    First we choose two values, there are 13 values (2 to A), so $13choose2$.



    Then we want to choose two cards of the first value out of four cards, $4choose 2$



    Again, we want to choose two cards of the second value out of four cards, $4choose 2$



    And finally, choose one card not of the previously selected types (we can’t choose the 4 cards of the first value and the 4 cards of the second value), ${52-8choose1} = {44choose1}$



    So we get:
    $${{{13choose2}times{4choose2}times{4choose2}times{44choose1} }over{52choose2}} = {198over4165} ≈ 0.0475$$






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      active

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      3














      You're pretty close, but there is a problem: you do have to choose 3 ranks, but they're not all going to be treated the same. One will be a single, and two others will be pairs. If you multiply by a factor of $binom{3}{2}$ I think you'll have it.






      share|cite|improve this answer


























        3














        You're pretty close, but there is a problem: you do have to choose 3 ranks, but they're not all going to be treated the same. One will be a single, and two others will be pairs. If you multiply by a factor of $binom{3}{2}$ I think you'll have it.






        share|cite|improve this answer
























          3












          3








          3






          You're pretty close, but there is a problem: you do have to choose 3 ranks, but they're not all going to be treated the same. One will be a single, and two others will be pairs. If you multiply by a factor of $binom{3}{2}$ I think you'll have it.






          share|cite|improve this answer












          You're pretty close, but there is a problem: you do have to choose 3 ranks, but they're not all going to be treated the same. One will be a single, and two others will be pairs. If you multiply by a factor of $binom{3}{2}$ I think you'll have it.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 14 '15 at 19:00









          Kevin LongKevin Long

          3,38121330




          3,38121330























              1














              In total, there are $52choose5$ ways to draw a hand (this is our |S|).



              We want to choose 2 out of four cards of one value, 2 out of four cards of another value, and one other card not of the first two values (This will be our |E|).



              First we choose two values, there are 13 values (2 to A), so $13choose2$.



              Then we want to choose two cards of the first value out of four cards, $4choose 2$



              Again, we want to choose two cards of the second value out of four cards, $4choose 2$



              And finally, choose one card not of the previously selected types (we can’t choose the 4 cards of the first value and the 4 cards of the second value), ${52-8choose1} = {44choose1}$



              So we get:
              $${{{13choose2}times{4choose2}times{4choose2}times{44choose1} }over{52choose2}} = {198over4165} ≈ 0.0475$$






              share|cite|improve this answer


























                1














                In total, there are $52choose5$ ways to draw a hand (this is our |S|).



                We want to choose 2 out of four cards of one value, 2 out of four cards of another value, and one other card not of the first two values (This will be our |E|).



                First we choose two values, there are 13 values (2 to A), so $13choose2$.



                Then we want to choose two cards of the first value out of four cards, $4choose 2$



                Again, we want to choose two cards of the second value out of four cards, $4choose 2$



                And finally, choose one card not of the previously selected types (we can’t choose the 4 cards of the first value and the 4 cards of the second value), ${52-8choose1} = {44choose1}$



                So we get:
                $${{{13choose2}times{4choose2}times{4choose2}times{44choose1} }over{52choose2}} = {198over4165} ≈ 0.0475$$






                share|cite|improve this answer
























                  1












                  1








                  1






                  In total, there are $52choose5$ ways to draw a hand (this is our |S|).



                  We want to choose 2 out of four cards of one value, 2 out of four cards of another value, and one other card not of the first two values (This will be our |E|).



                  First we choose two values, there are 13 values (2 to A), so $13choose2$.



                  Then we want to choose two cards of the first value out of four cards, $4choose 2$



                  Again, we want to choose two cards of the second value out of four cards, $4choose 2$



                  And finally, choose one card not of the previously selected types (we can’t choose the 4 cards of the first value and the 4 cards of the second value), ${52-8choose1} = {44choose1}$



                  So we get:
                  $${{{13choose2}times{4choose2}times{4choose2}times{44choose1} }over{52choose2}} = {198over4165} ≈ 0.0475$$






                  share|cite|improve this answer












                  In total, there are $52choose5$ ways to draw a hand (this is our |S|).



                  We want to choose 2 out of four cards of one value, 2 out of four cards of another value, and one other card not of the first two values (This will be our |E|).



                  First we choose two values, there are 13 values (2 to A), so $13choose2$.



                  Then we want to choose two cards of the first value out of four cards, $4choose 2$



                  Again, we want to choose two cards of the second value out of four cards, $4choose 2$



                  And finally, choose one card not of the previously selected types (we can’t choose the 4 cards of the first value and the 4 cards of the second value), ${52-8choose1} = {44choose1}$



                  So we get:
                  $${{{13choose2}times{4choose2}times{4choose2}times{44choose1} }over{52choose2}} = {198over4165} ≈ 0.0475$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 3:52









                  The RoomThe Room

                  15017




                  15017






























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