9. A chess piece starts from point $A$ in the figure, and each step is determined by the number that appears when a die is rolled. If the result of the roll is 1 or 2, the chess piece remains stationary; if the result is 3 or 4, the chess piece moves up one square; if the result is 5 or 6, the chess piece moves right one square.
(1) What is the probability that the chess piece reaches point $B$ within 8 steps (including 8 steps)?
(2) If it is stipulated that when there are two consecutive steps of remaining stationary, the chess piece moves right one square. The chess piece continues to move indefinitely, what is the probability that the chess piece passes through point $C$?
9. Solution: (1) The probability of reaching point $B$ for the first time at the 6th step is $P_{1}=C_{6}^{2} \cdot \frac{1}{3^{6}}=\frac{15}{729}$,

The probability of reaching point $B$ for the first time at the 7th step is $P_{2}=C_{6}^{1} \cdot C_{6}^{2} \cdot \frac{1}{3^{7}}=\frac{30}{729}$,
The probability of reaching point $B$ for the first time at the 8th step is $P_{3}=C_{7}^{2} \cdot C_{6}^{2} \cdot \frac{1}{3^{8}}=\frac{35}{729}$,
Therefore, the probability of reaching point $B$ within 8 steps is $P=P_{1}+P_{2}+P_{3}=\frac{80}{729}$.
(2) Classify according to the number of dice rolls:

Rolling 2 times, there are two paths: right-up, up-right, so $P_{1}=2 \cdot \frac{1}{3^{2}}=\frac{2}{9}$,
Rolling 3 times, there are six paths: First category, staying once, then there are four cases: stay-right-up, stay-up-right, right-stay-up, up-stay-right;
Second category, staying twice, then there are two cases: stay-stay-up, up-stay-stay. So $P_{2}=6 \cdot \frac{1}{3^{3}}=\frac{2}{9}$,

Rolling 4 times, there are four paths: First category, staying twice, then there are two cases: stay-up-stay-right, stay-right-stay-up;
Second category, staying three times, then there are two cases: stay-stay-stay-up, stay-up-stay-stay, so $P_{3}=4 \cdot \frac{1}{3^{4}}=\frac{4}{81}$,

In summary, $P=P_{1}+P_{2}+P_{3}=\frac{40}{81}$.