Indicator Random Variables for Card Game (2 Red in a row)












2












$begingroup$


Working through some practice problems in Introduction to Probability, Blitzstein. (I found similar problems to this one but want to make sure I understand this concept.)




You have a well-shuffed 52-card deck. On average, how many pairs of adjacent cards are there such that both cards are red?




Create indicator random variable I, where I=1 if both red, else 0



Cards could be RR, RB, BB, BR, so probability of indicator variable success is:




  • P(I=0)=$frac{3}{4}$

  • P(I=1)=$frac{1}{4}$


You don't have to check the 1st card alone, so there are 52-1 cards to check. Expectation is therefore:



$E(I) = sum_{i=1}^{51} P(I_i=1) = 51(0*frac{3}{4}+1frac{1}{4}) = frac{51}{4}$



Questions:
1. Is the logic correct?
2. Can someone verify and further explain why sum from 1 to 51 is correct (or incorrect if I'm wrong)
3. Is the nomenclature for what to sum correct?



(Edit: edited P(I=1) and P(I=0) per comment below)






random-variables card-games







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Working through some practice problems in Introduction to Probability, Blitzstein. (I found similar problems to this one but want to make sure I understand this concept.)




    You have a well-shuffed 52-card deck. On average, how many pairs of adjacent cards are there such that both cards are red?




    Create indicator random variable I, where I=1 if both red, else 0



    Cards could be RR, RB, BB, BR, so probability of indicator variable success is:




    • P(I=0)=$frac{3}{4}$

    • P(I=1)=$frac{1}{4}$


    You don't have to check the 1st card alone, so there are 52-1 cards to check. Expectation is therefore:



    $E(I) = sum_{i=1}^{51} P(I_i=1) = 51(0*frac{3}{4}+1frac{1}{4}) = frac{51}{4}$



    Questions:
    1. Is the logic correct?
    2. Can someone verify and further explain why sum from 1 to 51 is correct (or incorrect if I'm wrong)
    3. Is the nomenclature for what to sum correct?



    (Edit: edited P(I=1) and P(I=0) per comment below)






    random-variables card-games







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Working through some practice problems in Introduction to Probability, Blitzstein. (I found similar problems to this one but want to make sure I understand this concept.)




      You have a well-shuffed 52-card deck. On average, how many pairs of adjacent cards are there such that both cards are red?




      Create indicator random variable I, where I=1 if both red, else 0



      Cards could be RR, RB, BB, BR, so probability of indicator variable success is:




      • P(I=0)=$frac{3}{4}$

      • P(I=1)=$frac{1}{4}$


      You don't have to check the 1st card alone, so there are 52-1 cards to check. Expectation is therefore:



      $E(I) = sum_{i=1}^{51} P(I_i=1) = 51(0*frac{3}{4}+1frac{1}{4}) = frac{51}{4}$



      Questions:
      1. Is the logic correct?
      2. Can someone verify and further explain why sum from 1 to 51 is correct (or incorrect if I'm wrong)
      3. Is the nomenclature for what to sum correct?



      (Edit: edited P(I=1) and P(I=0) per comment below)






      random-variables card-games







      share|cite|improve this question











      $endgroup$




      Working through some practice problems in Introduction to Probability, Blitzstein. (I found similar problems to this one but want to make sure I understand this concept.)




      You have a well-shuffed 52-card deck. On average, how many pairs of adjacent cards are there such that both cards are red?




      Create indicator random variable I, where I=1 if both red, else 0



      Cards could be RR, RB, BB, BR, so probability of indicator variable success is:




      • P(I=0)=$frac{3}{4}$

      • P(I=1)=$frac{1}{4}$


      You don't have to check the 1st card alone, so there are 52-1 cards to check. Expectation is therefore:



      $E(I) = sum_{i=1}^{51} P(I_i=1) = 51(0*frac{3}{4}+1frac{1}{4}) = frac{51}{4}$



      Questions:
      1. Is the logic correct?
      2. Can someone verify and further explain why sum from 1 to 51 is correct (or incorrect if I'm wrong)
      3. Is the nomenclature for what to sum correct?



      (Edit: edited P(I=1) and P(I=0) per comment below)






      random-variables card-games




      random-variables card-games






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 20:27







      user603569

















      asked Dec 1 '18 at 19:58









      user603569user603569

      728




      728






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Apart from what looks likes a typo flipping $P(I_i = 1)$ and $P(I_i=0)$, this looks okay. You should also be careful with being explicit about your random variables; as you define $I$, you should have $E(I) = frac{1}{4}$ (what you really want is to define a separate $I_i$ for each $1 leq i leq 51$).



          The justification for the sum comes from linearity of expectation. If you define the $I_i$'s as described, and let $I$ be the random variable denoting the total number of positions where this occurs, we have:
          $$Eleft(Iright) = Eleft(sum_{i=1}^{51} I_i right) = sum_{i=1}^{51} Eleft(I_i right) = sum_{i=1}^{51} Pr(I_i = 1) = frac{51}{4}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
            $endgroup$
            – user603569
            Dec 1 '18 at 20:26








          • 1




            $begingroup$
            Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
            $endgroup$
            – platty
            Dec 1 '18 at 20:27











          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%2f3021752%2findicator-random-variables-for-card-game-2-red-in-a-row%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Apart from what looks likes a typo flipping $P(I_i = 1)$ and $P(I_i=0)$, this looks okay. You should also be careful with being explicit about your random variables; as you define $I$, you should have $E(I) = frac{1}{4}$ (what you really want is to define a separate $I_i$ for each $1 leq i leq 51$).



          The justification for the sum comes from linearity of expectation. If you define the $I_i$'s as described, and let $I$ be the random variable denoting the total number of positions where this occurs, we have:
          $$Eleft(Iright) = Eleft(sum_{i=1}^{51} I_i right) = sum_{i=1}^{51} Eleft(I_i right) = sum_{i=1}^{51} Pr(I_i = 1) = frac{51}{4}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
            $endgroup$
            – user603569
            Dec 1 '18 at 20:26








          • 1




            $begingroup$
            Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
            $endgroup$
            – platty
            Dec 1 '18 at 20:27
















          0












          $begingroup$

          Apart from what looks likes a typo flipping $P(I_i = 1)$ and $P(I_i=0)$, this looks okay. You should also be careful with being explicit about your random variables; as you define $I$, you should have $E(I) = frac{1}{4}$ (what you really want is to define a separate $I_i$ for each $1 leq i leq 51$).



          The justification for the sum comes from linearity of expectation. If you define the $I_i$'s as described, and let $I$ be the random variable denoting the total number of positions where this occurs, we have:
          $$Eleft(Iright) = Eleft(sum_{i=1}^{51} I_i right) = sum_{i=1}^{51} Eleft(I_i right) = sum_{i=1}^{51} Pr(I_i = 1) = frac{51}{4}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
            $endgroup$
            – user603569
            Dec 1 '18 at 20:26








          • 1




            $begingroup$
            Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
            $endgroup$
            – platty
            Dec 1 '18 at 20:27














          0












          0








          0





          $begingroup$

          Apart from what looks likes a typo flipping $P(I_i = 1)$ and $P(I_i=0)$, this looks okay. You should also be careful with being explicit about your random variables; as you define $I$, you should have $E(I) = frac{1}{4}$ (what you really want is to define a separate $I_i$ for each $1 leq i leq 51$).



          The justification for the sum comes from linearity of expectation. If you define the $I_i$'s as described, and let $I$ be the random variable denoting the total number of positions where this occurs, we have:
          $$Eleft(Iright) = Eleft(sum_{i=1}^{51} I_i right) = sum_{i=1}^{51} Eleft(I_i right) = sum_{i=1}^{51} Pr(I_i = 1) = frac{51}{4}$$






          share|cite|improve this answer









          $endgroup$



          Apart from what looks likes a typo flipping $P(I_i = 1)$ and $P(I_i=0)$, this looks okay. You should also be careful with being explicit about your random variables; as you define $I$, you should have $E(I) = frac{1}{4}$ (what you really want is to define a separate $I_i$ for each $1 leq i leq 51$).



          The justification for the sum comes from linearity of expectation. If you define the $I_i$'s as described, and let $I$ be the random variable denoting the total number of positions where this occurs, we have:
          $$Eleft(Iright) = Eleft(sum_{i=1}^{51} I_i right) = sum_{i=1}^{51} Eleft(I_i right) = sum_{i=1}^{51} Pr(I_i = 1) = frac{51}{4}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 20:20









          plattyplatty

          3,370320




          3,370320












          • $begingroup$
            Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
            $endgroup$
            – user603569
            Dec 1 '18 at 20:26








          • 1




            $begingroup$
            Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
            $endgroup$
            – platty
            Dec 1 '18 at 20:27


















          • $begingroup$
            Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
            $endgroup$
            – user603569
            Dec 1 '18 at 20:26








          • 1




            $begingroup$
            Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
            $endgroup$
            – platty
            Dec 1 '18 at 20:27
















          $begingroup$
          Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
          $endgroup$
          – user603569
          Dec 1 '18 at 20:26






          $begingroup$
          Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
          $endgroup$
          – user603569
          Dec 1 '18 at 20:26






          1




          1




          $begingroup$
          Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
          $endgroup$
          – platty
          Dec 1 '18 at 20:27




          $begingroup$
          Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
          $endgroup$
          – platty
          Dec 1 '18 at 20:27


















          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021752%2findicator-random-variables-for-card-game-2-red-in-a-row%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

          mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

          Image
          0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
          Read more

          How to change which sound is reproduced for terminal bell?

          Image
          3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
          Read more

          Can I use Tabulator js library in my java Spring + Thymeleaf project?

          Image
          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
          Read more