A and B flip a fair dice until 6 occurs. I have read this post and tried to work out the expectation of flipping on A wins. But it seems something goes wrong. Following is my work:
A wins iff $6$ occurs in odd round, so the EXP flippings on A wins is therefore: $$E = \frac{1}{6}*1 + (\frac{5}{6})^2*\frac{1}{6}*3+....(\frac{5}{6})^{2k}*\frac{1}{6}*(2k+1)+...$$ It is straightforward to get $E=\frac{366}{121}$, where did I go wrong?
$\displaystyle\sum_{k=0}^\infty (2k+1)(\tfrac 56)^{2k}\tfrac 16+\sum_{k=1}^\infty (2k)(\tfrac 56)^{2k-1}\tfrac 16$ is the expected count for rolls until the game ends: $\mathsf E(X)$.
The first term counts rolls where the game ends on an odd roll, and the second counts rolls where the game ends on and even roll.
$$\mathsf E(X)=\mathsf E(X\,\mathbf 1_{\{X\in2\Bbb N+1\}})+\mathsf E(X\,\mathbf 1_{\{X\in 2\Bbb N\}})\tag{Linearity}$$
So, you have evaluated: $\displaystyle\mathsf E(X\,\mathbf 1_{\{X\in2\Bbb N+1\}})=\sum_{k=0}^\infty (2k+1)(\tfrac 56)^{2k}\tfrac 16$
However, you want the expected count given that the game ends on an odd count.
$$\begin{align}\mathsf E(X\mid X\in2\Bbb N+1) &= \dfrac{\mathsf E(X~\mathbf 1_{\{X\in2\Bbb N+1\}})}{\mathsf P(X\in2\Bbb N+1)} \\[1ex] &= \dfrac{(1/6)\sum\limits_{k=0}^\infty (2k+1)(5/6)^{2k}}{(1/6)\sum\limits_{k=0}^\infty (5/6)^{2k}}\\[1ex] &= \dfrac{366}{121}\div\dfrac6{11}\\[1ex]&= \dfrac{61}{11}\end{align}$$