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







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














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











share|cite|improve this question




share|cite|improve this question










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
2






active

oldest

votes


















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





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1528964%2fprobability-that-a-five-card-poker-hand-contains-two-pairs%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1528964%2fprobability-that-a-five-card-poker-hand-contains-two-pairs%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      1300-talet

                      Image
                      Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
                      Read more

                      1300-talet

                      Image
                      Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
                      Read more

                      Display a custom attribute below product name in the front-end Magento 1.9.3.8

                      Image
                      0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
                      Read more