8. Six people are playing a coin-tossing game in a circle (the coin is fair), with each person tossing the coin once. The rule is: those who get the coin landing tails up have to perform a show, while those who get heads up do not have to perform. What is the probability that no two performers are adjacent? $\qquad$
8. $\frac{9}{32}$.

There are a total of $2^{6}=64$ possible scenarios, with the number of performers being at most three.

It is easy to see that the number of scenarios where the number of performers is exactly $0$, $1$, or $3$ and meets the conditions are $1$, $6$, and $2$, respectively. The number of scenarios where exactly two performers meet the conditions is $\mathrm{C}_{6}^{2}-6=9$.
Therefore, the required probability is
$$
P=\frac{1+6+2+9}{64}=\frac{9}{32} \text {. }
$$