Solution:

Answer: $\frac{7}{33}$.

Select any blue marble and consider the remaining eleven marbles, arranged in a line. The proportion of arrangements for which no two red marbles are adjacent will be the same as for the original twelve marbles, arranged in a ring. The total number of ways of arranging $4$ red marbles out of $11$ is $\binom{11}{4} = 330$.

To count the number of arrangements such that no two red marbles are adjacent, there must be one blue marble between each two would-be adjacent red marbles. Having fixed the positions of three blue marbles, we have four blue marbles to play with. So the number of ways we can arrange the remaining four marbles is $\binom{8}{4} = 70$ ways.

This yields a probability of $70 / 330 = 7 / 33$ as our final answer.