Throwing of non-fair dice

Question

Given that Mark has a non-fair die with a probability of $p$ of getting a $6$, and that the probability of Alex getting a $6$ is $2p$, $(0 < p \leq 0.5)$, what is the probability of Mark winning the game, when the game rules are that each one throws his die once according to their turn, that whoever gets a $6$ first wins, and that Mark starts the game?

So, how do we get to this question? What is $\Omega$ here? It is quite confusing to me.

I started thinking about it and came up with this:

The game can go on infinitely. The game ends when either Mark or Alex gets a $6$. Since we have two dice, we will look at our $\Omega$ as all possibilities $(i,j), 1 \leq i,j \leq 6$ such as $i$ is for Mark and $j$ is for Alex.

Is that a correct first approach?

Answer 1

An alternate answer I figured out:

Let $A$ be the event that mark wins. $P(A) = p + (1-p)(1-2p)p + (1-p)^2(1-2p)^2p +... = p \cdot [ 1+ (1-p)(1-2p) + ...] = \frac{1}{3-2p}$

Answer 2

Hint: Let $m$ be the probability that Mark wins. If Mark and Alex both roll and neither gets a $6$, we are back at the start of the game, so Mark again has chance $m$ to win. Write the equation $m=$ (chance Mark wins on the first throw) + (chance neither wins on their first throws)(chance Mark wins ever)