Series and probability The task at hand:
Suppose there are 3 players (A,B,C), who have 3 different dices with the following chances (per roll) to win:
Player A has a dice with a $\frac{1}{3}$ chance to win 
Player B has a dice with a $\frac{2}{5}$ chance to win 
Player C has a dice with a $\frac{4}{7}$ chance to win 
Player A starts rolling the dice, followed by Player B, then C and then starting with A all over again until one player wins.
Calculate the chance of player A to win. 
My question is wether the following approach is valid or has a logical error:
$\sum_{n=0}^\infty (\frac{2}{3}\frac{3}{5}\frac{3}{7})^n(\frac{1}{3}) = \frac{1}{3}\sum_{n=0}^\infty (\frac{2\cdot3\cdot3}{3\cdot5\cdot7})^n= \frac{1}{3}\sum_{n=0}^\infty (\frac{18}{105})^n= \frac{1}{3}\frac{1}{1-(\frac{18}{105})}= \frac{1}{3}\frac{1}{(\frac{87}{105})} =\frac{105}{261} \sim 0.4023$

Let $x$ be the probability that A wins. So $x$ equals P(A rolls a win) + P(B does not roll a win and C does not roll a win)*$x$. In other words:$$x = \frac{1}{3} + \frac{2}{3}\frac{3}{5}\frac{3}{7}x$$
and so 
$$x = \frac{35}{3*29} \approx 0.4023.$$
