Probability of Increasing Sequence of Die Rolls Remaining Increasing












0














Context



I was talking with a friend about this simple problem and we couldn't figure it out easily. I spent a little while with paper and pen and googled a bit and checked previous questions but have not been successful finding a solution. I think I am missing some language/terms.



Problem



Let $D_i$ be the $i$th outcome of a 20-sided die event. That is, $D_iin[1,20]$. What is the probability that $D_1<D_2<D_3$ for three sequential die events?



Scratch Work



My pen & paper work thus far (I feel like I am abusing notation):



$P(D_1)=1$. $P(D_2>D_1)=frac{20-D_1}{20}$. $P(D_3>D_2)=frac{20-D_2}{20}$.



$P(D_1<D_2<D_3)=P(D_1)P(D_2>D_1)P(D_3>D_2)=(1)(frac{20-D_1}{20})(frac{20-D_2}{20})=1-frac{D_1+D_2}{20}+frac{D_1D_2}{400}$.



$D_1+D_2in[2,40]$; $D_1D_2stackrel{?}{in}[1,400]$.



Closing



I know that I am doing something gravely wrong and I want to know what it is. All help appreciated, especially generalizations.






probability conditional-probability







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  • You need to sum over all valid pairs of $D_1,D_2$, but otherwise I think this approach works.
    – platty
    Nov 28 '18 at 3:09
















0














Context



I was talking with a friend about this simple problem and we couldn't figure it out easily. I spent a little while with paper and pen and googled a bit and checked previous questions but have not been successful finding a solution. I think I am missing some language/terms.



Problem



Let $D_i$ be the $i$th outcome of a 20-sided die event. That is, $D_iin[1,20]$. What is the probability that $D_1<D_2<D_3$ for three sequential die events?



Scratch Work



My pen & paper work thus far (I feel like I am abusing notation):



$P(D_1)=1$. $P(D_2>D_1)=frac{20-D_1}{20}$. $P(D_3>D_2)=frac{20-D_2}{20}$.



$P(D_1<D_2<D_3)=P(D_1)P(D_2>D_1)P(D_3>D_2)=(1)(frac{20-D_1}{20})(frac{20-D_2}{20})=1-frac{D_1+D_2}{20}+frac{D_1D_2}{400}$.



$D_1+D_2in[2,40]$; $D_1D_2stackrel{?}{in}[1,400]$.



Closing



I know that I am doing something gravely wrong and I want to know what it is. All help appreciated, especially generalizations.






probability conditional-probability







share|cite|improve this question






















  • You need to sum over all valid pairs of $D_1,D_2$, but otherwise I think this approach works.
    – platty
    Nov 28 '18 at 3:09














0












0








0







Context



I was talking with a friend about this simple problem and we couldn't figure it out easily. I spent a little while with paper and pen and googled a bit and checked previous questions but have not been successful finding a solution. I think I am missing some language/terms.



Problem



Let $D_i$ be the $i$th outcome of a 20-sided die event. That is, $D_iin[1,20]$. What is the probability that $D_1<D_2<D_3$ for three sequential die events?



Scratch Work



My pen & paper work thus far (I feel like I am abusing notation):



$P(D_1)=1$. $P(D_2>D_1)=frac{20-D_1}{20}$. $P(D_3>D_2)=frac{20-D_2}{20}$.



$P(D_1<D_2<D_3)=P(D_1)P(D_2>D_1)P(D_3>D_2)=(1)(frac{20-D_1}{20})(frac{20-D_2}{20})=1-frac{D_1+D_2}{20}+frac{D_1D_2}{400}$.



$D_1+D_2in[2,40]$; $D_1D_2stackrel{?}{in}[1,400]$.



Closing



I know that I am doing something gravely wrong and I want to know what it is. All help appreciated, especially generalizations.






probability conditional-probability







share|cite|improve this question













Context



I was talking with a friend about this simple problem and we couldn't figure it out easily. I spent a little while with paper and pen and googled a bit and checked previous questions but have not been successful finding a solution. I think I am missing some language/terms.



Problem



Let $D_i$ be the $i$th outcome of a 20-sided die event. That is, $D_iin[1,20]$. What is the probability that $D_1<D_2<D_3$ for three sequential die events?



Scratch Work



My pen & paper work thus far (I feel like I am abusing notation):



$P(D_1)=1$. $P(D_2>D_1)=frac{20-D_1}{20}$. $P(D_3>D_2)=frac{20-D_2}{20}$.



$P(D_1<D_2<D_3)=P(D_1)P(D_2>D_1)P(D_3>D_2)=(1)(frac{20-D_1}{20})(frac{20-D_2}{20})=1-frac{D_1+D_2}{20}+frac{D_1D_2}{400}$.



$D_1+D_2in[2,40]$; $D_1D_2stackrel{?}{in}[1,400]$.



Closing



I know that I am doing something gravely wrong and I want to know what it is. All help appreciated, especially generalizations.






probability conditional-probability




probability conditional-probability






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asked Nov 28 '18 at 3:02









Dohleman

309110




309110












  • You need to sum over all valid pairs of $D_1,D_2$, but otherwise I think this approach works.
    – platty
    Nov 28 '18 at 3:09


















  • You need to sum over all valid pairs of $D_1,D_2$, but otherwise I think this approach works.
    – platty
    Nov 28 '18 at 3:09
















You need to sum over all valid pairs of $D_1,D_2$, but otherwise I think this approach works.
– platty
Nov 28 '18 at 3:09




You need to sum over all valid pairs of $D_1,D_2$, but otherwise I think this approach works.
– platty
Nov 28 '18 at 3:09










3 Answers
3






active

oldest

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2














Hint: Partition over the middle value:
$$begin{align}mathsf P(D_1<D_2<D_3)&=sum_{t=2}^{19}mathsf P(D_1<t)mathsf P(D_2=t)mathsf P(D_3>t)\[1ex]&=sum_{t=2}^{19}dfrac{(t-1)(20-t)}{20^3}end{align}$$






share|cite|improve this answer





















  • this is the style of solution i was looking for, thank you
    – Dohleman
    Nov 28 '18 at 3:16



















2














Note that, if all three rolls are distinct, there is a $frac{1}{3!}$ chance they come in increasing order (if they are not distinct, there is no chance). The probability of at least two of them matching is $3 times dfrac{1}{20} times dfrac{19}{20} + 20 times dfrac{1}{20^3}$ (splitting into cases where exactly 2 match and all 3 match). The desired probability is then $dfrac{1}{6} times left( 1 - left( 3 times dfrac{1}{20} times dfrac{19}{20} + 20 times dfrac{1}{20^3} right) right) = boxed{dfrac{57}{400}}$.






share|cite|improve this answer





















  • this is a clever solution that I also appreciate
    – Dohleman
    Nov 28 '18 at 3:16



















2














$$ P(mbox{Increasing}) = P(mbox{Increasing}|mbox{All distinct})P(mbox{All distinct})$$



$$P(mbox{All distinct})=frac{20*19*18}{20*20*20}= frac{19*9}{200}.$$



As argued above,
$$ P(mbox{Increasing}|mbox{All distinct}) = frac{1}{3!}.$$



Therefore, the answer is $$ frac{19*9}{200} times frac{1}{6} = frac{57}{400}.$$






share|cite|improve this answer





















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    3 Answers
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    3 Answers
    3






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    active

    oldest

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    2














    Hint: Partition over the middle value:
    $$begin{align}mathsf P(D_1<D_2<D_3)&=sum_{t=2}^{19}mathsf P(D_1<t)mathsf P(D_2=t)mathsf P(D_3>t)\[1ex]&=sum_{t=2}^{19}dfrac{(t-1)(20-t)}{20^3}end{align}$$






    share|cite|improve this answer





















    • this is the style of solution i was looking for, thank you
      – Dohleman
      Nov 28 '18 at 3:16
















    2














    Hint: Partition over the middle value:
    $$begin{align}mathsf P(D_1<D_2<D_3)&=sum_{t=2}^{19}mathsf P(D_1<t)mathsf P(D_2=t)mathsf P(D_3>t)\[1ex]&=sum_{t=2}^{19}dfrac{(t-1)(20-t)}{20^3}end{align}$$






    share|cite|improve this answer





















    • this is the style of solution i was looking for, thank you
      – Dohleman
      Nov 28 '18 at 3:16














    2












    2








    2






    Hint: Partition over the middle value:
    $$begin{align}mathsf P(D_1<D_2<D_3)&=sum_{t=2}^{19}mathsf P(D_1<t)mathsf P(D_2=t)mathsf P(D_3>t)\[1ex]&=sum_{t=2}^{19}dfrac{(t-1)(20-t)}{20^3}end{align}$$






    share|cite|improve this answer












    Hint: Partition over the middle value:
    $$begin{align}mathsf P(D_1<D_2<D_3)&=sum_{t=2}^{19}mathsf P(D_1<t)mathsf P(D_2=t)mathsf P(D_3>t)\[1ex]&=sum_{t=2}^{19}dfrac{(t-1)(20-t)}{20^3}end{align}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 28 '18 at 3:13









    Graham Kemp

    84.7k43378




    84.7k43378












    • this is the style of solution i was looking for, thank you
      – Dohleman
      Nov 28 '18 at 3:16


















    • this is the style of solution i was looking for, thank you
      – Dohleman
      Nov 28 '18 at 3:16
















    this is the style of solution i was looking for, thank you
    – Dohleman
    Nov 28 '18 at 3:16




    this is the style of solution i was looking for, thank you
    – Dohleman
    Nov 28 '18 at 3:16











    2














    Note that, if all three rolls are distinct, there is a $frac{1}{3!}$ chance they come in increasing order (if they are not distinct, there is no chance). The probability of at least two of them matching is $3 times dfrac{1}{20} times dfrac{19}{20} + 20 times dfrac{1}{20^3}$ (splitting into cases where exactly 2 match and all 3 match). The desired probability is then $dfrac{1}{6} times left( 1 - left( 3 times dfrac{1}{20} times dfrac{19}{20} + 20 times dfrac{1}{20^3} right) right) = boxed{dfrac{57}{400}}$.






    share|cite|improve this answer





















    • this is a clever solution that I also appreciate
      – Dohleman
      Nov 28 '18 at 3:16
















    2














    Note that, if all three rolls are distinct, there is a $frac{1}{3!}$ chance they come in increasing order (if they are not distinct, there is no chance). The probability of at least two of them matching is $3 times dfrac{1}{20} times dfrac{19}{20} + 20 times dfrac{1}{20^3}$ (splitting into cases where exactly 2 match and all 3 match). The desired probability is then $dfrac{1}{6} times left( 1 - left( 3 times dfrac{1}{20} times dfrac{19}{20} + 20 times dfrac{1}{20^3} right) right) = boxed{dfrac{57}{400}}$.






    share|cite|improve this answer





















    • this is a clever solution that I also appreciate
      – Dohleman
      Nov 28 '18 at 3:16














    2












    2








    2






    Note that, if all three rolls are distinct, there is a $frac{1}{3!}$ chance they come in increasing order (if they are not distinct, there is no chance). The probability of at least two of them matching is $3 times dfrac{1}{20} times dfrac{19}{20} + 20 times dfrac{1}{20^3}$ (splitting into cases where exactly 2 match and all 3 match). The desired probability is then $dfrac{1}{6} times left( 1 - left( 3 times dfrac{1}{20} times dfrac{19}{20} + 20 times dfrac{1}{20^3} right) right) = boxed{dfrac{57}{400}}$.






    share|cite|improve this answer












    Note that, if all three rolls are distinct, there is a $frac{1}{3!}$ chance they come in increasing order (if they are not distinct, there is no chance). The probability of at least two of them matching is $3 times dfrac{1}{20} times dfrac{19}{20} + 20 times dfrac{1}{20^3}$ (splitting into cases where exactly 2 match and all 3 match). The desired probability is then $dfrac{1}{6} times left( 1 - left( 3 times dfrac{1}{20} times dfrac{19}{20} + 20 times dfrac{1}{20^3} right) right) = boxed{dfrac{57}{400}}$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 28 '18 at 3:13









    platty

    3,370320




    3,370320












    • this is a clever solution that I also appreciate
      – Dohleman
      Nov 28 '18 at 3:16


















    • this is a clever solution that I also appreciate
      – Dohleman
      Nov 28 '18 at 3:16
















    this is a clever solution that I also appreciate
    – Dohleman
    Nov 28 '18 at 3:16




    this is a clever solution that I also appreciate
    – Dohleman
    Nov 28 '18 at 3:16











    2














    $$ P(mbox{Increasing}) = P(mbox{Increasing}|mbox{All distinct})P(mbox{All distinct})$$



    $$P(mbox{All distinct})=frac{20*19*18}{20*20*20}= frac{19*9}{200}.$$



    As argued above,
    $$ P(mbox{Increasing}|mbox{All distinct}) = frac{1}{3!}.$$



    Therefore, the answer is $$ frac{19*9}{200} times frac{1}{6} = frac{57}{400}.$$






    share|cite|improve this answer


























      2














      $$ P(mbox{Increasing}) = P(mbox{Increasing}|mbox{All distinct})P(mbox{All distinct})$$



      $$P(mbox{All distinct})=frac{20*19*18}{20*20*20}= frac{19*9}{200}.$$



      As argued above,
      $$ P(mbox{Increasing}|mbox{All distinct}) = frac{1}{3!}.$$



      Therefore, the answer is $$ frac{19*9}{200} times frac{1}{6} = frac{57}{400}.$$






      share|cite|improve this answer
























        2












        2








        2






        $$ P(mbox{Increasing}) = P(mbox{Increasing}|mbox{All distinct})P(mbox{All distinct})$$



        $$P(mbox{All distinct})=frac{20*19*18}{20*20*20}= frac{19*9}{200}.$$



        As argued above,
        $$ P(mbox{Increasing}|mbox{All distinct}) = frac{1}{3!}.$$



        Therefore, the answer is $$ frac{19*9}{200} times frac{1}{6} = frac{57}{400}.$$






        share|cite|improve this answer












        $$ P(mbox{Increasing}) = P(mbox{Increasing}|mbox{All distinct})P(mbox{All distinct})$$



        $$P(mbox{All distinct})=frac{20*19*18}{20*20*20}= frac{19*9}{200}.$$



        As argued above,
        $$ P(mbox{Increasing}|mbox{All distinct}) = frac{1}{3!}.$$



        Therefore, the answer is $$ frac{19*9}{200} times frac{1}{6} = frac{57}{400}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 3:54









        Fnacool

        4,976511




        4,976511






























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