8. Six boys and $a$ girls stand in a row at random, with each boy being adjacent to at least one other boy, and the probability that at least four boys stand together is $p$. If $p \leqslant \frac{1}{100}$, then the minimum value of $a$ is . $\qquad$
8.594.

Notice that,
$$
\begin{aligned}
p & =\frac{2 \mathrm{C}_{a+1}^{2}+(a+1)}{\mathrm{C}_{a+1}^{3}+3 \mathrm{C}_{a+1}^{2}+(a+1)} \\
& =\frac{6 a+6}{a^{2}+8 a+6} \leqslant \frac{1}{100} .
\end{aligned}
$$

Therefore, $f(a)=a^{2}-592 a-594 \geqslant 0$.
Also, $f(0)0$, thus, the minimum value of $a$ is 594.