A and B flip a fair dice until 6 occurs. What's the expectation of flipping conditioned on A wins.
My solution is to calculate it by
$$\begin{align*}E(6\text{ occurs}) &= E(E(6 \text{ occurs}|A \text{ wins})) \\[0.3cm] &= E(6 \text{ occurs}|A \text{ wins}) \cdot P(A \text{ wins}) + E(6 \text{ occurs}|B \text{ wins}) \cdot P(B \text{ wins}) \end{align*}$$
For $P(A \text{ wins})$, it's simply
$$P(A \text{ wins}) = 1/6 + 5/6 \cdot (1-P(A \text{ wins}))=6/11$$
However, how to tackle $E(6 \text{ occurs}|B \text{ wins})$?
Thanks in advance.
I’ll write $X$ for the number of rolls (since “$6$ occurs” sounds more like an event than a random variable).
Conditional on $B$ winning, $A$’s first roll is irrelevant, since it’s known not to be a $6$. After that roll, $B$ goes first, so the expected number of the remaining rolls conditional on $B$ winning is simply $\mathsf E[X\mid\text{A wins}]$. Thus, $\mathsf E[X\mid\text{B wins}]=1+\mathsf E[X\mid\text{A wins}]$, and
\begin{eqnarray*} \mathsf E[X] &=& \mathsf P(\text{A wins})\mathsf E[X|\text{A wins}]+\mathsf P(\text{B wins})\mathsf E[X|\text{B wins}] \\ &=& \frac6{11}\mathsf E[X|\text{A wins}]+\frac5{11}(1+\mathsf E[X|\text{A wins}]) \\ &=& E[X|\text{A wins}]+\frac5{11}\;. \end{eqnarray*}
So
\begin{eqnarray*} E[X|\text{A wins}] &=& \mathsf E[X]-\frac5{11} \\ &=& 6-\frac5{11} \\ &=& 5\frac6{11}\;. \end{eqnarray*}