A dice is rolled until a $6$ occurs. What is the probability that the sum including the $6$ is even?

Let $p$ be the desired probability, and consider the first roll. It is either a $6$, in which case we're done and the sum is even, a $2$ or $4$, in which case we want the sum of the rest of the terms to be even, or a $1,3$, or $5$, in which case we want the sum of the rest to be odd.

Thus $$p = \frac{1}{6}+ \frac{1}{3}p+\frac{1}{2}(1-p)$$ which simplifies to $p=\frac{4}{7}$.

We need only consider the rolls before a $6$ is obtained, because rolling an even $6$ doesn't change the parity of our total. Let $p_n$ represent the probability that the sum of $n$ rolls, not including any $6$, is even. Then we have: $p_{n+1}=\frac25p_n + \frac35(1-p_n)$, because a $2$ or a $4$ keeps a previous even total even, while a $1$, $3$ or $5$ makes a previous odd total into an even total. This simplifies to: $p_{n+1}=\frac35-\frac15p_n$. We also have $p_0=1$. We can solve this recurrence, and find that

$$p_n=\frac12\left(1+\left(-\frac15\right)^n\right)$$

Now, let $x_n$ represent the probability of rolling $n$ non-6's before the first $6$, so $x_n=\frac16\left(\frac56\right)^n$. The number we need is:

$$\begin{align} \sum\limits_{n=0}^\infty x_np_n &= \sum\limits_{n=0}^\infty \left[\frac16\left(\frac56\right)^n\cdot\frac12\left(1+\left(-\frac15\right)^n\right)\right]\\ &=\frac1{12}\sum\limits_{n=0}^\infty \left[\left(\frac56\right)^n + \left(-\frac16\right)^n\right]\\ &=\frac1{12}\left(6 + \frac67\right) = \frac47 \end{align}$$

That said, @carmichael561's answer is much, much nicer.

Let $p$ be the probability that the sum of the die until(and including) the first six is even.

Let $R_1$ be the roll of the first die. So partitioning on this roll, and noticing the recurance:

$$p= \underline\qquad\,\mathsf P(R_1\in\{\underline\qquad\})+\underline\qquad\,\mathsf P(R_1\in\{\underline\qquad\})+\underline {~1~}\,\mathsf P(R_1=6)$$

Fill in the blanks, evalute the probabilities, and then solve for $p$.