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.
On
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 $$E=\frac 16 \times (1+2+3+4+5)+\frac 16\times (E+6)\implies E=\frac {21}5$$
Where we use the observation that, if you roll a $6$ you are back to the start, the pawn having moved $6$ spaces.
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 (6\cdot 5+15)+ \frac{1}{6^3}(12\cdot 5+15)+\cdots &= \sum_{k=1}^\infty \frac{1}{6^k}(6\cdot(k-1)\cdot 5+15)\\ &=30\sum_{k=1}^\infty \frac{(k-1)}{6^k} + 15\sum_{k=1}^\infty\frac{1}{6^k}\\ &=5\sum_{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 $S-\frac{1}{6}S=\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+\cdots = \frac{1}{6}\sum_{k=0}^\infty \frac{1}{6^k}.$