12 Use red, yellow, and blue to color the vertices of a regular pentagon, then the probability that no two adjacent vertices have the same color is $\qquad$ .
(12) $\frac{10}{81}$ Hint: From the condition, there must exist a point whose color is different from any other point, meaning there must be a color that is used for only one point. First, select a point, then select a color, such that this point has a different color from the rest, which gives $5 \times 3=15$ ways to choose. For the remaining four points, use the other two colors, where the two points adjacent to the first selected point are colored with two different colors, giving 2 ways. Once the colors of these three points are determined, there is only 1 way to color the remaining two points. Therefore, the number of ways that meet the condition is $5 \times 3 \times 2=30$, thus $P=\frac{30}{3^{5}}=\frac{10}{81}$.