# Expected number of die rolls

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?

• I would start by finding the probability (a) you roll a $6$ and your opponent an even, (b) you roll a $6$ and your opponent an odd, (c) you roll a non-$6$ and your opponent an even, (d) you roll a non-$6$ and your opponent an odd. In case (d) you have to go on rolling to find the result so the conditional probability thereafter is the same as the original marginal probability Commented Oct 27, 2022 at 0:28

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 $$\mathbb P \left(\bigcup_{j\geq 1} E_j\right) = \sum_{j\geq 1}\mathbb P(E_j) = \sum_{j\geq 1}(5/12)^j = \frac{12}{7}-1 = \frac{5}{7}$$
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 $$(5/{12})/(7/{12})=5/7$$.