We have three players A, B and C. They roll an honest die, first A, then B, and lastly C until someone is the first to get an even number... 
We have three players A, B and C. They roll an honest die, first A, then B, and lastly C until someone is the first to get an even number. This player will be the winner of the game. If after all 3 players have played no one gets the desired result, they start rolling the dice again in the same order until they get it. What is the probability that A wins the game?

Probability of getting even is 0.5 = p of getting odd is 0.5 = 1-p
A can win in the first round so p=0.5
A can win in the second round so we have $1-p \cdot 1-p\cdot 1-p \cdot p = (1-p)^3 \cdot p$
A can win in the third round so $1-p \cdot 1-p\cdot 1-p \cdot 1-p \cdot 1-p \cdot 1-p \cdot p = (1-p)^6 \cdot p$
A can win in the nth round $(1-p)^{3(n-1)} \cdot p$
$0.5+ (1-p)^3 \cdot p + (1-p)^6 \cdot p + .... +(1-p)^{3(n-1)} \cdot p = 0.57$

Your answer is correct to the number of decimal places to which you rounded. The probability that player $A$ wins is
\begin{align*}
\sum_{k = 1}^{\infty} \frac{1}{2}\left(1 - \frac{1}{2}\right)^{3(k - 1)} & = \frac{1}{2}\sum_{k = 1}^{\infty} \left(\frac{1}{2}\right)^{3(k - 1)}\\
& = \frac{1}{2}\sum_{k = 1}^{\infty} \left(\frac{1}{8}\right)^{k - 1}\\
& = \frac{1}{2} \cdot \frac{1}{1 - \frac{1}{8}}\\
& = \frac{1}{2} \cdot \frac{1}{\frac{7}{8}}\\
& = \frac{1}{2} \cdot \frac{8}{7}\\
& = \frac{4}{7}
\end{align*}
Here is another approach.  Let $p$ be the probability that player $A$ wins.  Either player $A$ wins on the first roll with probability $1/2$ or all three players must fail in the first round, in which case we are back in the original situation.  Hence,
\begin{align*}
p & = \frac{1}{2} + \left(1 - \frac{1}{2}\right)^3p\\
p & = \frac{1}{2} + \left(\frac{1}{2}\right)^3p\\
p & = \frac{1}{2} + \frac{1}{8}p\\
\frac{7}{8}p & = \frac{1}{2}\\
p & = \frac{4}{7}
\end{align*}
