smc

We will subtract from one the probability that the first condition is violated and the probability that only the second condition is violated, being careful not to double-count the probability that both conditions are violated. For the first condition to be violated, all four dice must read $4$ or less, which happens with probability ${\left(\frac{2}{3}\right)}^4=\frac{16}{81}$. For the first condition to be met but the second condition to be violated, at least one of the dice must read greater than $4$, but less than two of the dice can read greater than $2$. Therefore, one of the four die must read $5$ or $6$, while the remaining three dice must read $2$ or less, which happens with probability $\left(\genfrac{}{}{0ex}{}{4}{1}\right)\left(\frac{1}{3}\right){\left(\frac{1}{3}\right)}^3=4\cdot \frac{1}{3}\cdot \frac{1}{27}=\frac{4}{81}$. Therefore, the overall probability of meeting both conditions is $1-\frac{16}{81}-\frac{4}{81}=\boxed{\frac{61}{81}}$.