You are right that each throw of a die has a $(1/6)$ chance of rolling a $(3)$. Therefore, your addition is correct that if you roll the die $8$ times, the expected number of times that you will roll a $(3)$ is
$$8 \times (1/6) = (8/6) = (4/3) > 1.$$
Your intuition is also right, that if you roll the die $8$ times, the chance of at least one of the rolls coming up $(3)$ must be less than $(1)$.
This begs the question: if the distribution of $8$ rolls is such that you expect $(4/3)$ of the rolls to be a $3$, how can the chance of not rolling any $(3)$ in 8 rolls, still be positive.
It is because there is a (small) chance, that in $8$ rolls, you may have a $3$ appear two or more times. It is the possibility of a $3$ appearing two or more times that balances the fact that there is still a positive chance that there will be no $(3)$'s rolled.
However, this is all intuitive hand waving, which doesn't mean much, without math to back it up.
Suppose that you roll a die $8$ times. There are $(6)$ possibilities for each roll. Therefore, the total number of possible sequences of $8$ rolls is $6^8$.
For $k \in \{0,1,2,\cdots,8\}$, a natural question is: how many of the $6^8$ sequences will result in exactly $k$ of the rolls coming up $(3)$.
There are $\frac{8!}{k!(8-k)!} = \binom{8}{k}$ ways of selecting $k$ rolls out of $8$, so that those rolls (and only those rolls) come up $3$.
Once the $k$ rolls are selected, you then have $(8-k)$ rolls whose only constraint is that the roll is any number other than a $(3)$.
Therefore, there are exactly $\left[\binom{8}{k} \times 5^{(8-k)}\right]$ sequences of $8$ rolls, in which $3$ comes up exactly $k$ times.
Therefore, the probability of this happening is
$$P(k) = \frac{\binom{8}{k} \times 5^{(8-k)}}{6^8}.$$
You will find that :
- $\sum_{k = 0}^8 [k \times P(k)] = (8/6) = (4/3)$, as expected.
- $P(0) = \frac{\binom{8}{0} \times 5^{(8-0)}}{6^8} = \left(\frac{5}{6}\right)^8 > 0.$