Suppose we roll two die, find the mass of the largest roll











up vote
1
down vote

favorite
1












roll two dice and let $X$ be the largest value obtained and $Y$ the smallest. find pmf of $X$ and $Y$.



Im trying to find $P(X=i)$. I start with easy cases. For example, $P(X=1)$ this can only occur when we have $(1,1)$ so we have $frac{1}{36}$. For $P(X=2)$ we have now $(2,1), (1,2), (2,2)$ so now we have $frac{3}{36}$. Also $P(X=3)$ occurs when $(3,2),(3,1),(3,3), (3,1),(3,2)$ so this gives $P(X=3) = frac{5}{36}$ so the pattern is that we have odd numbers of event for $X$ bein largest and so we have



$$ P(X=i) = frac{2i-1}{36} $$



Isnt is $P(Y=j) = frac{2i-1}{26}$ the same?



Is my reasoning correct? How can I approach this problem without doing the easy cases first? Is there any general strategy to handle this type of problems?










share|cite|improve this question


























    up vote
    1
    down vote

    favorite
    1












    roll two dice and let $X$ be the largest value obtained and $Y$ the smallest. find pmf of $X$ and $Y$.



    Im trying to find $P(X=i)$. I start with easy cases. For example, $P(X=1)$ this can only occur when we have $(1,1)$ so we have $frac{1}{36}$. For $P(X=2)$ we have now $(2,1), (1,2), (2,2)$ so now we have $frac{3}{36}$. Also $P(X=3)$ occurs when $(3,2),(3,1),(3,3), (3,1),(3,2)$ so this gives $P(X=3) = frac{5}{36}$ so the pattern is that we have odd numbers of event for $X$ bein largest and so we have



    $$ P(X=i) = frac{2i-1}{36} $$



    Isnt is $P(Y=j) = frac{2i-1}{26}$ the same?



    Is my reasoning correct? How can I approach this problem without doing the easy cases first? Is there any general strategy to handle this type of problems?










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      roll two dice and let $X$ be the largest value obtained and $Y$ the smallest. find pmf of $X$ and $Y$.



      Im trying to find $P(X=i)$. I start with easy cases. For example, $P(X=1)$ this can only occur when we have $(1,1)$ so we have $frac{1}{36}$. For $P(X=2)$ we have now $(2,1), (1,2), (2,2)$ so now we have $frac{3}{36}$. Also $P(X=3)$ occurs when $(3,2),(3,1),(3,3), (3,1),(3,2)$ so this gives $P(X=3) = frac{5}{36}$ so the pattern is that we have odd numbers of event for $X$ bein largest and so we have



      $$ P(X=i) = frac{2i-1}{36} $$



      Isnt is $P(Y=j) = frac{2i-1}{26}$ the same?



      Is my reasoning correct? How can I approach this problem without doing the easy cases first? Is there any general strategy to handle this type of problems?










      share|cite|improve this question













      roll two dice and let $X$ be the largest value obtained and $Y$ the smallest. find pmf of $X$ and $Y$.



      Im trying to find $P(X=i)$. I start with easy cases. For example, $P(X=1)$ this can only occur when we have $(1,1)$ so we have $frac{1}{36}$. For $P(X=2)$ we have now $(2,1), (1,2), (2,2)$ so now we have $frac{3}{36}$. Also $P(X=3)$ occurs when $(3,2),(3,1),(3,3), (3,1),(3,2)$ so this gives $P(X=3) = frac{5}{36}$ so the pattern is that we have odd numbers of event for $X$ bein largest and so we have



      $$ P(X=i) = frac{2i-1}{36} $$



      Isnt is $P(Y=j) = frac{2i-1}{26}$ the same?



      Is my reasoning correct? How can I approach this problem without doing the easy cases first? Is there any general strategy to handle this type of problems?







      probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked yesterday









      Neymar

      36713




      36713






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          Your work for $X$ is correct, but the distribution for $Y$ is not the same, though it is the mirror image. You can calculate it in a similar way: $P(Y=6) = 1/36,$ $P(Y=5) = 3/36,ldots$ $P(Y=1) = 11/36.$



          It is actually a pretty good approach in general to do the easy cases first. You do them and then think about how to generalize, like how you caught the pattern was $(2i-1)/36.$ There are plenty of methods and tricks you'll learn for dealing with combinatorics and probability problems, but I would still approach this problem exactly as you did here.






          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',
            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%2f2997805%2fsuppose-we-roll-two-die-find-the-mass-of-the-largest-roll%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








            up vote
            2
            down vote



            accepted










            Your work for $X$ is correct, but the distribution for $Y$ is not the same, though it is the mirror image. You can calculate it in a similar way: $P(Y=6) = 1/36,$ $P(Y=5) = 3/36,ldots$ $P(Y=1) = 11/36.$



            It is actually a pretty good approach in general to do the easy cases first. You do them and then think about how to generalize, like how you caught the pattern was $(2i-1)/36.$ There are plenty of methods and tricks you'll learn for dealing with combinatorics and probability problems, but I would still approach this problem exactly as you did here.






            share|cite|improve this answer



























              up vote
              2
              down vote



              accepted










              Your work for $X$ is correct, but the distribution for $Y$ is not the same, though it is the mirror image. You can calculate it in a similar way: $P(Y=6) = 1/36,$ $P(Y=5) = 3/36,ldots$ $P(Y=1) = 11/36.$



              It is actually a pretty good approach in general to do the easy cases first. You do them and then think about how to generalize, like how you caught the pattern was $(2i-1)/36.$ There are plenty of methods and tricks you'll learn for dealing with combinatorics and probability problems, but I would still approach this problem exactly as you did here.






              share|cite|improve this answer

























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                Your work for $X$ is correct, but the distribution for $Y$ is not the same, though it is the mirror image. You can calculate it in a similar way: $P(Y=6) = 1/36,$ $P(Y=5) = 3/36,ldots$ $P(Y=1) = 11/36.$



                It is actually a pretty good approach in general to do the easy cases first. You do them and then think about how to generalize, like how you caught the pattern was $(2i-1)/36.$ There are plenty of methods and tricks you'll learn for dealing with combinatorics and probability problems, but I would still approach this problem exactly as you did here.






                share|cite|improve this answer














                Your work for $X$ is correct, but the distribution for $Y$ is not the same, though it is the mirror image. You can calculate it in a similar way: $P(Y=6) = 1/36,$ $P(Y=5) = 3/36,ldots$ $P(Y=1) = 11/36.$



                It is actually a pretty good approach in general to do the easy cases first. You do them and then think about how to generalize, like how you caught the pattern was $(2i-1)/36.$ There are plenty of methods and tricks you'll learn for dealing with combinatorics and probability problems, but I would still approach this problem exactly as you did here.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited yesterday

























                answered yesterday









                spaceisdarkgreen

                31.2k21552




                31.2k21552






























                     

                    draft saved


                    draft discarded



















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997805%2fsuppose-we-roll-two-die-find-the-mass-of-the-largest-roll%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

                    Plaza Victoria

                    In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                    How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...