Middle School Probability Problem 
Sven and Ole take turns rolling a standard six-sided die. The first person to roll
  a six wins. If Sven goes first, what is the probability that he will win the game?

This is a problem from a worksheet for middle schoolers.  I see that one could solve it by summing the infinite series
$$P(\text{Sven wins}) = P(\text{Sven wins in 1st round}) + P(\text{Sven wins in 2nd round}) + \cdots$$ but that's surely not how it is supposed to be solved.  How else could one solve this?
 A: I don't see a middle school approach.  As you say, you can sum series, or you can do it recursively:
Let $p$ be the answer, then $1-p$ is the probability that the player who tosses second will win.  Sven tosses.  He wins immediately (probability $\frac 16$) or he doesn't (probability $\frac 56$).  If he doesn't, the game restarts only now Sven is the second player.  Thus:  $$p = \frac 16\,1\;+\;\frac 56\,(1-p)\;\Rightarrow\;p=\frac 6{11}$$
Easier than series, perhaps, but middle school?
A: I'd do it by noticing that the game is time-independent. Each round, Sven has chance 1/6 of winning, and Ole has chance 5/6 * 1/6, or 5/36. Then the game restarts.
Therefore Sven's chance of winning before Ole is $\frac{1/6}{1/6 + 5/36}$, or 6/11.
A: Let $A$ be the event in which Sven roll $6$ and $B$ be the event in which Ole rool $6$
So $\displaystyle P(A)=\frac{1}{6}$ and $\displaystyle P(\bar{A}) = \frac{5}{6}$ and $\displaystyle P(B)=\frac{1}{6}$ and $\displaystyle P(\bar{B}) = \frac{5}{6}$ 
So Required Probability 
$$\displaystyle = P(A)+P(\bar{A})\cdot P(\bar{B})\cdot P(A)+P(\bar{A})\cdot P(\bar{B})\cdot P(\bar{A)}\cdot p(\bar{B})\cdot P(A)+.......\infty$$
So required probability $$\displaystyle = \frac{1}{6}+\frac{5}{6}\cdot \frac{5}{6}\cdot \frac{1}{6}+\frac{5}{6}\cdot \frac{5}{6}\cdot \frac{5}{6}\cdot \frac{5}{6}\cdot \frac{1}{6}+.....\infty$$
So Required Probability $$\displaystyle = \frac{1}{6}\left[\frac{1}{1-\left(\frac{5}{6}\right)^2}\right] = \frac{6}{11}.$$
