It is to prove that if A begins, then
$$P(A_{wins})= \frac{6}{11}$$ Similarly, if B begins,
$$P(A_{wins})= \frac{5}{11}$$ Now, for the given problem, P= P(A wins)= P(A begins and wins or B begins and A wins) But (A begins and wins) and (B begins and A wins) are mutually exclusive events.
So P= P(A begins and wins) + P(B begins and A wins)$= \frac{6}{11}+ \frac{5}{11}= 1.$
This means that A will surely win and B will surely lose, which is clearly incorrect. So what's the mistake in all of this?
On
A begins and probability of A wins, it will be an infinity geometric series such that:
$$\frac{1}{6}+(\frac{5}{6})^2\frac{1}{6}+(\frac{5}{6})^4\frac{1}{6}+...=\frac{1}{6}(1+(\frac{5}{6})^2+(\frac{5}{6})^4+...)=\frac{1}{6}(\frac{1}{1-(\frac{5}{6})^2})=\frac{6}{11}$$
B begins and probability of A wins, it will be an infinity geometric series such that:
$$(\frac{5}{6})(\frac{1}{6})+(\frac{5}{6})^3\frac{1}{6}+(\frac{5}{6})^5\frac{1}{6}+...=\frac{5}{36}(1+(\frac{5}{6})^2+(\frac{5}{6})^4+...)=\frac{5}{36}(\frac{1}{1-(\frac{5}{6})^2})=\frac{5}{11}$$
$\frac{6}{11}+\frac{5}{11}=1$ does not mean A always wins.
We have an event and two possibilities
$$P(A_{\text{first and wins}})=1- P(B_{\text{second and loses}})=\frac{6}{11}$$
$$P(A_{\text{second and wins}})=1- P(B_{\text{first and loses}})=\frac{5}{11}$$
One of them always wins.
This is not true. Either A throws it first or B throws it first. You have to assume one of them happens. If not, meaning you don't know who throws first, then you need to assign prior probabilities. Let's say $\pi (\text{A begins first}) = q, \pi(\text{B begins first})=1-q$, then
$$\Pr(\text{A wins}) = \frac{6}{11} q+\frac{5}{11}(1-q)$$
If it's completely random, $q=\frac 12$ then the above is also equal to $\frac 12.$