Expected number of die rolls

Question

The question reads: You roll a fair die until you obtain a 6. Your opponent then rolls the same die until he or she rolls an even number. Find the probability that you roll the die more times than your opponent.

My attempt: Let X be the number of rolls until a six is shown, and let Y be the number of rolls until an even number is shown. Then, X and Y both follow geometric distributions with p = 1/6 and p = 1/2, respectively. I found the expected value (6 and 2, respectively), but now I'm stuck. Any help on where to go from here or whether I should just start over?

Answer 1

Let $E_j$ be the event that your opponent rolls exactly $j$ times, and you roll more than $j$ times. Then $\mathbb P(E_j) = (1/2)^j (5/6)^j = (5/12)^j$. Hence the answer is $$\mathbb P \left(\bigcup_{j\geq 1} E_j\right) = \sum_{j\geq 1}\mathbb P(E_j) = \sum_{j\geq 1}(5/12)^j = \frac{12}{7}-1 = \frac{5}{7}$$

Answer 2

Call roll $k$ "relevant" if either you rolled a six, or your opponent rolled an even number. This has chance $1/6+1/2-1/{12}=7/12$. Call roll $k$ "bad" if your opponent rolled an even number, but you did not roll a six. This has chance $1/2-1/{12}=5/12$. Consider the first relevant roll. The chance that it is bad is $(5/{12})/(7/{12})=5/7$.