Expected Number of Rolls needed to get a 3 and then a 4 consecutively of an unbiased dice











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Not to be confused with the number of rolls needed to get two of the same number, suppose you keep rolling a dice until you get a 3 and then 4 consecutively. Calculate the expected number of rolls required.



My logic was that the probability of rolling a three is 1/6 and the probability of rolling a four is 1/6, so we can multiply the two to get 1/36. Then since the distribution is geometric, the expected value of the geometric distribution is $1/p = 1/(1/36) = 36$.



However, multiple people have gotten an answer of either 9 or 12. I get how one would arrive at 12, but how do you get 9 rolls out of this? Just what does this problem have to do with Harmonic Numbers? Any help would be appreciated.










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    You can take a look here. I don't think $9$ or $12$ is correct.
    – VHarisop
    Nov 14 at 21:40















up vote
0
down vote

favorite












Not to be confused with the number of rolls needed to get two of the same number, suppose you keep rolling a dice until you get a 3 and then 4 consecutively. Calculate the expected number of rolls required.



My logic was that the probability of rolling a three is 1/6 and the probability of rolling a four is 1/6, so we can multiply the two to get 1/36. Then since the distribution is geometric, the expected value of the geometric distribution is $1/p = 1/(1/36) = 36$.



However, multiple people have gotten an answer of either 9 or 12. I get how one would arrive at 12, but how do you get 9 rolls out of this? Just what does this problem have to do with Harmonic Numbers? Any help would be appreciated.










share|cite|improve this question


















  • 1




    You can take a look here. I don't think $9$ or $12$ is correct.
    – VHarisop
    Nov 14 at 21:40













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Not to be confused with the number of rolls needed to get two of the same number, suppose you keep rolling a dice until you get a 3 and then 4 consecutively. Calculate the expected number of rolls required.



My logic was that the probability of rolling a three is 1/6 and the probability of rolling a four is 1/6, so we can multiply the two to get 1/36. Then since the distribution is geometric, the expected value of the geometric distribution is $1/p = 1/(1/36) = 36$.



However, multiple people have gotten an answer of either 9 or 12. I get how one would arrive at 12, but how do you get 9 rolls out of this? Just what does this problem have to do with Harmonic Numbers? Any help would be appreciated.










share|cite|improve this question













Not to be confused with the number of rolls needed to get two of the same number, suppose you keep rolling a dice until you get a 3 and then 4 consecutively. Calculate the expected number of rolls required.



My logic was that the probability of rolling a three is 1/6 and the probability of rolling a four is 1/6, so we can multiply the two to get 1/36. Then since the distribution is geometric, the expected value of the geometric distribution is $1/p = 1/(1/36) = 36$.



However, multiple people have gotten an answer of either 9 or 12. I get how one would arrive at 12, but how do you get 9 rolls out of this? Just what does this problem have to do with Harmonic Numbers? Any help would be appreciated.







probability probability-theory






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asked Nov 14 at 21:23









WhatsDUI

214111




214111








  • 1




    You can take a look here. I don't think $9$ or $12$ is correct.
    – VHarisop
    Nov 14 at 21:40














  • 1




    You can take a look here. I don't think $9$ or $12$ is correct.
    – VHarisop
    Nov 14 at 21:40








1




1




You can take a look here. I don't think $9$ or $12$ is correct.
– VHarisop
Nov 14 at 21:40




You can take a look here. I don't think $9$ or $12$ is correct.
– VHarisop
Nov 14 at 21:40










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Just to have a different method, let's do it with states. There are only three states of interest. Either you are just starting or the last throw was not a three (call it $mathscr S_{emptyset}$) or the last throw was a $3$, call it $mathscr S_{3}$, or you are done.



We also denote by $E_{emptyset}$ ansd $E_3$ the expected number of moves it should take to finish from there. Of course $E_{emptyset}$ is the answer we seek.



Considering the first toss we see that $$E_{emptyset}=frac 16times (E_3+1)+frac 56times (E_{emptyset}+1)$$



And if you are in $mathscr S_{3}$ we see that $$E_3=frac 16times 1+frac 16times (E_3+1)+frac 46times (E_{emptyset}+1)$$



This system is easy to solve and we get $$E_{emptyset}=36quad E_3=30$$



Which, of course, confirms your result.






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    Just to have a different method, let's do it with states. There are only three states of interest. Either you are just starting or the last throw was not a three (call it $mathscr S_{emptyset}$) or the last throw was a $3$, call it $mathscr S_{3}$, or you are done.



    We also denote by $E_{emptyset}$ ansd $E_3$ the expected number of moves it should take to finish from there. Of course $E_{emptyset}$ is the answer we seek.



    Considering the first toss we see that $$E_{emptyset}=frac 16times (E_3+1)+frac 56times (E_{emptyset}+1)$$



    And if you are in $mathscr S_{3}$ we see that $$E_3=frac 16times 1+frac 16times (E_3+1)+frac 46times (E_{emptyset}+1)$$



    This system is easy to solve and we get $$E_{emptyset}=36quad E_3=30$$



    Which, of course, confirms your result.






    share|cite|improve this answer



























      up vote
      2
      down vote













      Just to have a different method, let's do it with states. There are only three states of interest. Either you are just starting or the last throw was not a three (call it $mathscr S_{emptyset}$) or the last throw was a $3$, call it $mathscr S_{3}$, or you are done.



      We also denote by $E_{emptyset}$ ansd $E_3$ the expected number of moves it should take to finish from there. Of course $E_{emptyset}$ is the answer we seek.



      Considering the first toss we see that $$E_{emptyset}=frac 16times (E_3+1)+frac 56times (E_{emptyset}+1)$$



      And if you are in $mathscr S_{3}$ we see that $$E_3=frac 16times 1+frac 16times (E_3+1)+frac 46times (E_{emptyset}+1)$$



      This system is easy to solve and we get $$E_{emptyset}=36quad E_3=30$$



      Which, of course, confirms your result.






      share|cite|improve this answer

























        up vote
        2
        down vote










        up vote
        2
        down vote









        Just to have a different method, let's do it with states. There are only three states of interest. Either you are just starting or the last throw was not a three (call it $mathscr S_{emptyset}$) or the last throw was a $3$, call it $mathscr S_{3}$, or you are done.



        We also denote by $E_{emptyset}$ ansd $E_3$ the expected number of moves it should take to finish from there. Of course $E_{emptyset}$ is the answer we seek.



        Considering the first toss we see that $$E_{emptyset}=frac 16times (E_3+1)+frac 56times (E_{emptyset}+1)$$



        And if you are in $mathscr S_{3}$ we see that $$E_3=frac 16times 1+frac 16times (E_3+1)+frac 46times (E_{emptyset}+1)$$



        This system is easy to solve and we get $$E_{emptyset}=36quad E_3=30$$



        Which, of course, confirms your result.






        share|cite|improve this answer














        Just to have a different method, let's do it with states. There are only three states of interest. Either you are just starting or the last throw was not a three (call it $mathscr S_{emptyset}$) or the last throw was a $3$, call it $mathscr S_{3}$, or you are done.



        We also denote by $E_{emptyset}$ ansd $E_3$ the expected number of moves it should take to finish from there. Of course $E_{emptyset}$ is the answer we seek.



        Considering the first toss we see that $$E_{emptyset}=frac 16times (E_3+1)+frac 56times (E_{emptyset}+1)$$



        And if you are in $mathscr S_{3}$ we see that $$E_3=frac 16times 1+frac 16times (E_3+1)+frac 46times (E_{emptyset}+1)$$



        This system is easy to solve and we get $$E_{emptyset}=36quad E_3=30$$



        Which, of course, confirms your result.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 14 at 22:55

























        answered Nov 14 at 21:41









        lulu

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