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











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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?










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    up vote
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    favorite
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    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
























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      down vote

      favorite
      1









      up vote
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      down vote

      favorite
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      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?







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          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.






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            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.






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              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














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