Two persons are playing a game where they take turns rolling a die (so A rolls first, then B, then A again and so on). The first person to roll a $6$ wins the game. What is the probability that the person who started the game (rolled the die first) wins?
This was the question that I was given, but I feel like the probability depends on the number of turns played? My approach was that the probability of rolling a 6 at any particular turn will be $\frac{1}{6}$. So, the probability of winning in $n$ turns will be the probability of not rolling a $6$ on the first $n-1$ turns (since if a $6$ had been rolled, that would have been the last, i.e. the $nth$ turn) and then rolling a $6$ on the $nth$ turn. The required probability would then be $$\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot...(n-1\text{ times})\cdot\frac{1}{6} = \frac{5^{n-1}}{6^n}$$
Is this train of thought correct? Or have I misunderstood something?