TITLE: Two people A and B throwing dice
QUESTION [6 upvotes]: A begins by throwing a dice until he gets $6$, then B does the same thing.
What is the probability that A throws more times than B?
I try to solve it, but I got 2 different answers:
1. We mark: $X=$How many times A throws, $Y$=How many times B throws.
$$P(X>n|Y=n)=\frac{P(X>n)\cdot P(Y=n)}{P(Y=n)}=P(X>n)\\ \sum_{k=n}^\infty\left(\frac56\right)^{k-1}\cdot \left(\frac16\right)=\left(\frac65\right)^{1-n}=\left(\frac56\right)^{n-1}$$
2. The second way is: We can say that "B throw less times than A", so let's assume that A throws the dice $n$ times. So:
$$P(Y<n)=\sum_{k=1}^{n-1}\left(\frac56\right)^{k-1}\cdot \left(\frac16\right)=1-\left(\frac56\right)^{n}$$
They are opposite, and I don't know which one of them is correct or both of them are incorrect..
Please help with this.
Thank you!

REPLY [2 votes]: Note that if A doesn't get a 6 first roll (probability 5/6), then the probability of B getting a 6 first is the same as the probability of A getting a 6 first was before the first roll.
If $p_A $ is the probability of A getting a 6 first, $p_B$, of B first. Then we have
$$p_A + p_B = 1\text{, and }\;p_B  = \dfrac{5}{6} p_A$$
So
$$\dfrac{6}{5} p_B + p_B = 1$$
hence
$$p_B = \dfrac{5}{11}$$