Expected value. Problem with series.












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We have pawn on the infinite board. We roll a dice. When we roll six we can roll dice again. We move pawn as many times as we threw dots. Pawn move's What's the expected value of pawn moves in one turn ?



I figure out that expected value will be equal to :



$$ E(X)=frac16 cdot (1+2+3+4+5)+frac{1}{36} cdot (7+8+9+10+11) + frac{1}{216} cdot (13+14+15+16+17)+... $$



And i have some trouble to calculate above series.






probability dice







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    0












    $begingroup$


    We have pawn on the infinite board. We roll a dice. When we roll six we can roll dice again. We move pawn as many times as we threw dots. Pawn move's What's the expected value of pawn moves in one turn ?



    I figure out that expected value will be equal to :



    $$ E(X)=frac16 cdot (1+2+3+4+5)+frac{1}{36} cdot (7+8+9+10+11) + frac{1}{216} cdot (13+14+15+16+17)+... $$



    And i have some trouble to calculate above series.






    probability dice







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      We have pawn on the infinite board. We roll a dice. When we roll six we can roll dice again. We move pawn as many times as we threw dots. Pawn move's What's the expected value of pawn moves in one turn ?



      I figure out that expected value will be equal to :



      $$ E(X)=frac16 cdot (1+2+3+4+5)+frac{1}{36} cdot (7+8+9+10+11) + frac{1}{216} cdot (13+14+15+16+17)+... $$



      And i have some trouble to calculate above series.






      probability dice







      share|cite|improve this question











      $endgroup$




      We have pawn on the infinite board. We roll a dice. When we roll six we can roll dice again. We move pawn as many times as we threw dots. Pawn move's What's the expected value of pawn moves in one turn ?



      I figure out that expected value will be equal to :



      $$ E(X)=frac16 cdot (1+2+3+4+5)+frac{1}{36} cdot (7+8+9+10+11) + frac{1}{216} cdot (13+14+15+16+17)+... $$



      And i have some trouble to calculate above series.






      probability dice




      probability dice






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 26 at 14:39







      Lucian

















      asked Jan 26 at 13:55









      LucianLucian

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          2 Answers
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          0












          $begingroup$

          First note that $1+2+3+4+5 = 15$. So your series becomes:



          begin{align*}
          frac{1}{6}cdot 15 + frac{1}{6^2} cdot (6cdot 5+15)+ frac{1}{6^3}(12cdot 5+15)+cdots &= sum_{k=1}^infty frac{1}{6^k}(6cdot(k-1)cdot 5+15)\
          &=30sum_{k=1}^infty frac{(k-1)}{6^k} + 15sum_{k=1}^inftyfrac{1}{6^k}\
          &=5sum_{k=0}^infty frac{k}{6^k}+frac{15}{6}sum_{k=0}^infty frac{1}{6^k}.
          end{align*}



          The last term is just a geometric series. The sum of $frac{k}{6^k}$ can be found in the following way, assuming you know it converges. Let $S=sum_{k=0}^infty frac{k}{6^k}=0+frac{1}{6}frac{2}{6^2}+frac{3}{6^3}+cdots$. Then $frac{1}{6}S=0+frac{1}{6^2}+frac{2}{6^3}+frac{3}{6^4}+cdots$, so we have $S-frac{1}{6}S=frac{1}{6}+frac{1}{6^2}+frac{1}{6^3}+cdots = frac{1}{6}sum_{k=0}^infty frac{1}{6^k}.$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think you mean that 5+4+3+2+1=15.
            $endgroup$
            – Lucian
            Jan 26 at 14:09










          • $begingroup$
            Corrected, thank you.
            $endgroup$
            – kccu
            Jan 26 at 14:10



















          2












          $begingroup$

          I assume that the pawn moves as many spaces as the value shown by the die? Then, let $E$ denote the answer. Considering the first toss we see that $$E=frac 16 times (1+2+3+4+5)+frac 16times (E+6)implies E=frac {21}5$$



          Where we use the observation that, if you roll a $6$ you are back to the start, the pawn having moved $6$ spaces.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes, thank you very much !
            $endgroup$
            – Lucian
            Jan 26 at 14:07











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          2 Answers
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          2 Answers
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          active

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          active

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          0












          $begingroup$

          First note that $1+2+3+4+5 = 15$. So your series becomes:



          begin{align*}
          frac{1}{6}cdot 15 + frac{1}{6^2} cdot (6cdot 5+15)+ frac{1}{6^3}(12cdot 5+15)+cdots &= sum_{k=1}^infty frac{1}{6^k}(6cdot(k-1)cdot 5+15)\
          &=30sum_{k=1}^infty frac{(k-1)}{6^k} + 15sum_{k=1}^inftyfrac{1}{6^k}\
          &=5sum_{k=0}^infty frac{k}{6^k}+frac{15}{6}sum_{k=0}^infty frac{1}{6^k}.
          end{align*}



          The last term is just a geometric series. The sum of $frac{k}{6^k}$ can be found in the following way, assuming you know it converges. Let $S=sum_{k=0}^infty frac{k}{6^k}=0+frac{1}{6}frac{2}{6^2}+frac{3}{6^3}+cdots$. Then $frac{1}{6}S=0+frac{1}{6^2}+frac{2}{6^3}+frac{3}{6^4}+cdots$, so we have $S-frac{1}{6}S=frac{1}{6}+frac{1}{6^2}+frac{1}{6^3}+cdots = frac{1}{6}sum_{k=0}^infty frac{1}{6^k}.$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think you mean that 5+4+3+2+1=15.
            $endgroup$
            – Lucian
            Jan 26 at 14:09










          • $begingroup$
            Corrected, thank you.
            $endgroup$
            – kccu
            Jan 26 at 14:10
















          0












          $begingroup$

          First note that $1+2+3+4+5 = 15$. So your series becomes:



          begin{align*}
          frac{1}{6}cdot 15 + frac{1}{6^2} cdot (6cdot 5+15)+ frac{1}{6^3}(12cdot 5+15)+cdots &= sum_{k=1}^infty frac{1}{6^k}(6cdot(k-1)cdot 5+15)\
          &=30sum_{k=1}^infty frac{(k-1)}{6^k} + 15sum_{k=1}^inftyfrac{1}{6^k}\
          &=5sum_{k=0}^infty frac{k}{6^k}+frac{15}{6}sum_{k=0}^infty frac{1}{6^k}.
          end{align*}



          The last term is just a geometric series. The sum of $frac{k}{6^k}$ can be found in the following way, assuming you know it converges. Let $S=sum_{k=0}^infty frac{k}{6^k}=0+frac{1}{6}frac{2}{6^2}+frac{3}{6^3}+cdots$. Then $frac{1}{6}S=0+frac{1}{6^2}+frac{2}{6^3}+frac{3}{6^4}+cdots$, so we have $S-frac{1}{6}S=frac{1}{6}+frac{1}{6^2}+frac{1}{6^3}+cdots = frac{1}{6}sum_{k=0}^infty frac{1}{6^k}.$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think you mean that 5+4+3+2+1=15.
            $endgroup$
            – Lucian
            Jan 26 at 14:09










          • $begingroup$
            Corrected, thank you.
            $endgroup$
            – kccu
            Jan 26 at 14:10














          0












          0








          0





          $begingroup$

          First note that $1+2+3+4+5 = 15$. So your series becomes:



          begin{align*}
          frac{1}{6}cdot 15 + frac{1}{6^2} cdot (6cdot 5+15)+ frac{1}{6^3}(12cdot 5+15)+cdots &= sum_{k=1}^infty frac{1}{6^k}(6cdot(k-1)cdot 5+15)\
          &=30sum_{k=1}^infty frac{(k-1)}{6^k} + 15sum_{k=1}^inftyfrac{1}{6^k}\
          &=5sum_{k=0}^infty frac{k}{6^k}+frac{15}{6}sum_{k=0}^infty frac{1}{6^k}.
          end{align*}



          The last term is just a geometric series. The sum of $frac{k}{6^k}$ can be found in the following way, assuming you know it converges. Let $S=sum_{k=0}^infty frac{k}{6^k}=0+frac{1}{6}frac{2}{6^2}+frac{3}{6^3}+cdots$. Then $frac{1}{6}S=0+frac{1}{6^2}+frac{2}{6^3}+frac{3}{6^4}+cdots$, so we have $S-frac{1}{6}S=frac{1}{6}+frac{1}{6^2}+frac{1}{6^3}+cdots = frac{1}{6}sum_{k=0}^infty frac{1}{6^k}.$






          share|cite|improve this answer











          $endgroup$



          First note that $1+2+3+4+5 = 15$. So your series becomes:



          begin{align*}
          frac{1}{6}cdot 15 + frac{1}{6^2} cdot (6cdot 5+15)+ frac{1}{6^3}(12cdot 5+15)+cdots &= sum_{k=1}^infty frac{1}{6^k}(6cdot(k-1)cdot 5+15)\
          &=30sum_{k=1}^infty frac{(k-1)}{6^k} + 15sum_{k=1}^inftyfrac{1}{6^k}\
          &=5sum_{k=0}^infty frac{k}{6^k}+frac{15}{6}sum_{k=0}^infty frac{1}{6^k}.
          end{align*}



          The last term is just a geometric series. The sum of $frac{k}{6^k}$ can be found in the following way, assuming you know it converges. Let $S=sum_{k=0}^infty frac{k}{6^k}=0+frac{1}{6}frac{2}{6^2}+frac{3}{6^3}+cdots$. Then $frac{1}{6}S=0+frac{1}{6^2}+frac{2}{6^3}+frac{3}{6^4}+cdots$, so we have $S-frac{1}{6}S=frac{1}{6}+frac{1}{6^2}+frac{1}{6^3}+cdots = frac{1}{6}sum_{k=0}^infty frac{1}{6^k}.$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 26 at 14:10

























          answered Jan 26 at 14:08









          kccukccu

          10.6k11229




          10.6k11229












          • $begingroup$
            I think you mean that 5+4+3+2+1=15.
            $endgroup$
            – Lucian
            Jan 26 at 14:09










          • $begingroup$
            Corrected, thank you.
            $endgroup$
            – kccu
            Jan 26 at 14:10


















          • $begingroup$
            I think you mean that 5+4+3+2+1=15.
            $endgroup$
            – Lucian
            Jan 26 at 14:09










          • $begingroup$
            Corrected, thank you.
            $endgroup$
            – kccu
            Jan 26 at 14:10
















          $begingroup$
          I think you mean that 5+4+3+2+1=15.
          $endgroup$
          – Lucian
          Jan 26 at 14:09




          $begingroup$
          I think you mean that 5+4+3+2+1=15.
          $endgroup$
          – Lucian
          Jan 26 at 14:09












          $begingroup$
          Corrected, thank you.
          $endgroup$
          – kccu
          Jan 26 at 14:10




          $begingroup$
          Corrected, thank you.
          $endgroup$
          – kccu
          Jan 26 at 14:10











          2












          $begingroup$

          I assume that the pawn moves as many spaces as the value shown by the die? Then, let $E$ denote the answer. Considering the first toss we see that $$E=frac 16 times (1+2+3+4+5)+frac 16times (E+6)implies E=frac {21}5$$



          Where we use the observation that, if you roll a $6$ you are back to the start, the pawn having moved $6$ spaces.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes, thank you very much !
            $endgroup$
            – Lucian
            Jan 26 at 14:07
















          2












          $begingroup$

          I assume that the pawn moves as many spaces as the value shown by the die? Then, let $E$ denote the answer. Considering the first toss we see that $$E=frac 16 times (1+2+3+4+5)+frac 16times (E+6)implies E=frac {21}5$$



          Where we use the observation that, if you roll a $6$ you are back to the start, the pawn having moved $6$ spaces.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes, thank you very much !
            $endgroup$
            – Lucian
            Jan 26 at 14:07














          2












          2








          2





          $begingroup$

          I assume that the pawn moves as many spaces as the value shown by the die? Then, let $E$ denote the answer. Considering the first toss we see that $$E=frac 16 times (1+2+3+4+5)+frac 16times (E+6)implies E=frac {21}5$$



          Where we use the observation that, if you roll a $6$ you are back to the start, the pawn having moved $6$ spaces.






          share|cite|improve this answer









          $endgroup$



          I assume that the pawn moves as many spaces as the value shown by the die? Then, let $E$ denote the answer. Considering the first toss we see that $$E=frac 16 times (1+2+3+4+5)+frac 16times (E+6)implies E=frac {21}5$$



          Where we use the observation that, if you roll a $6$ you are back to the start, the pawn having moved $6$ spaces.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 26 at 14:03









          lulululu

          43.1k25080




          43.1k25080












          • $begingroup$
            Yes, thank you very much !
            $endgroup$
            – Lucian
            Jan 26 at 14:07


















          • $begingroup$
            Yes, thank you very much !
            $endgroup$
            – Lucian
            Jan 26 at 14:07
















          $begingroup$
          Yes, thank you very much !
          $endgroup$
          – Lucian
          Jan 26 at 14:07




          $begingroup$
          Yes, thank you very much !
          $endgroup$
          – Lucian
          Jan 26 at 14:07


















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