<s>[INST] Geometric series $ar^n$ where $n \ne 1,2,3,4 \cdots$ The probability of rolling a seven on two dice is as follows
$$p=1/6+1/6(5/6)^2+1/6(5/6)^4 + \cdots$$
what is the probability of rolling a $7$?
Is there an advantage to rolling first?
My attempt at a solution; $n=1,2,3,4,5 \cdots$
$$\sum_{n=1}^\infty \frac 1 6 \left ( \frac 5 6 \right)^{2n2} = \frac 1 6 + \frac 1 6 \left( \frac 5 6 \right)^2 + \frac 1 6 \left( \frac5 6 \right )^4 + \cdots$$
this is an example of a geometric series where $a=(1/6)$, $r=(5/6)$
we can say
$$1/6\sum_{n=1}^{\infty}(5/6)^{2n-2}=1/6+1/6(5/6)^2+1/6(5/6)^4...$$
let $$n_1=2n-2$$
therefore 
$$S_n=1/6\sum_{n_1=0}^{\infty}(5/6)^{n_1}$$
$$6S_n=\sum_{n_1=0}^{\infty}(5/6)^{n_1}=\frac{5/6}{1-5/6}$$
$$S_n=(1/6)\frac{5/6}{1-(5/6)}=\frac{5}{1-(5/6)}=\frac{5}{1/6}=30$$
The probability of rolling a seven is 30 percent. 
I think rolling first does give an advantage.
Am I understanding the rules of series properly??
Thanks
 [/INST] You are complicating things, note that :
$$p=\frac{1}{6} [1+(\frac{5}{6})^2+(\frac{5}{6})^4+\cdots ]$$
This is a geometric progression, where $a=1, r=\frac{25}{36}$ The infinite sum of which is:
$$\frac{a}{6(1-r)}=\frac{1}{6}\cdot\frac{36}{36-25}=\color {blue}{\frac{6}{11}}$$
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