10. (20 points) Two players, A and B, take turns rolling a die, with A rolling first, then B, and so on. After a roll, if the sum of the last three rolls is a multiple of 2, and the sum of the last two rolls is not a multiple of 3, the game ends and A wins. If the sum of the last two rolls is a multiple of 3, and the sum of the last three rolls is not a multiple of 2, the game also ends and B wins. In all other cases, the game continues. Find the probability that B wins.
Note: If the number of rolls is less than three, the sum of the "last three" rolls is not a multiple of 2.
10. Let the probability of A winning after the $n$-th coin toss be $p_{n}$, and the probability of B winning be $q_{n}$, and denote
$$
P=\sum_{n=1}^{\infty} p_{n}, \quad Q=\sum_{n=1}^{\infty} q_{n} .
$$

According to the problem, we have $p_{1}=p_{2}=0, q_{1}=0$. On the second roll of the dice, B wins if the sum of the points from the first and second rolls is a multiple of 3. Among the numbers $1,2,3,4,5,6$, there are two each that leave remainders of $0,1,2$ when divided by 3, so $q_{2}=\frac{2}{6}=\frac{1}{3}$.

For the $n(n \geqslant 3)$-th roll of the dice, the points that make the sum of the last three rolls even are 3 in number (1,3,5 or 2,4,6). These three points leave different remainders when divided by 3, so two of them make A win. Similarly, the points that make the sum of the last three rolls not even are also 3 in number, and leave different remainders when divided by 3, so one of them makes B win, i.e., $q_{n}=\frac{1}{2} p_{n}(n \geqslant 3)$, and $Q-q_{2}=\frac{1}{2} P$.
The probability that the game continues indefinitely is $\lim _{n \rightarrow \infty}\left(1-\frac{1}{3}\right) \times\left(\frac{3}{6}\right)^{n-2}=0$, so $P+Q=1$, and $Q-\frac{1}{3}=\frac{1}{2} P$, thus the probability of B winning $Q=\frac{5}{9}$.