Answer: 754

Solution:

Each roll is an Alice roll, a Bob roll, or a Carol roll, each with probability $\frac{1}{3}.$ The event succeeds exactly when the rolls before the first Carol roll include at least two Alice rolls and at least two Bob rolls. The first Carol roll is roll $k + 1$ with probability $\left(\frac{2}{3}\right)^k \frac{1}{3},$ and given this, the first $k$ rolls form one of $2^k$ equally likely Alice/Bob strings. For $k \ge 3$ the bad strings — at most one Alice, or at most one Bob — number $(k + 1) + (k + 1) = 2k + 2,$ and no string is bad in both ways. Hence $$\begin{aligned} P &= \sum_{k \ge 4} \frac{1}{3}\left(\frac{2}{3}\right)^k \\ &\quad {}\cdot \left(1 - \frac{2k + 2}{2^k}\right). \end{aligned}$$

The first piece is $\sum_{k \ge 4} \left(\frac{2}{3}\right)^k = \frac{16}{27}.$ For the second, $\sum_{k \ge 0} \frac{k + 1}{3^k} = \frac{1}{(1 - 1/3)^2} = \frac{9}{4},$ so $$\begin{aligned} &\sum_{k \ge 4} \frac{2k + 2}{3^k} \\ &= 2\left(\frac{9}{4} - 1 - \frac{2}{3} - \frac{1}{3} - \frac{4}{27}\right) \\ &= 2 \cdot \frac{11}{108} = \frac{11}{54}. \end{aligned}$$

Therefore $P = \frac{1}{3}\left(\frac{16}{27} - \frac{11}{54}\right)$ $= \frac{1}{3} \cdot \frac{21}{54} = \frac{7}{54},$ and $100m + n = 700 + 54 = 754.$