8. Six people form a circle to play a coin-tossing game (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, while those who get heads up do not have to perform. The probability that no two performers are adjacent is $\qquad$ .
If no one performs, the corresponding probability is $p_{0}=\frac{1}{2^{6}}$; if 1 person performs, the corresponding probability is $p_{1}=6 \cdot \frac{1}{2^{6}}$; if 2 people perform, the corresponding probability is $p_{2}=9 \cdot \frac{1}{2^{6}}$; if 3 people perform, the corresponding probability is $p_{3}=2 \cdot \frac{1}{2^{6}}$, hence the required probability is $P=p_{0}+p_{1}+p_{2}+p_{3}=\frac{1+6+9+2}{64}=\frac{9}{32}$.