We roll a die 30 times. Determine the probability, to 8 decimal places, that among the numbers rolled there will be a six, and that before the first six, neither a five nor a four appears.
First, let's examine the number of favorable cases! Suppose the first six appears on the $k$-th roll (1 $\leq k \leq 30$). In this case, the outcomes of the preceding $(k-1)$ rolls can only be 1, 2, or 3, while the $30-k$ rolls following the six can be any of the six possible values. Thus, there are exactly $3^{k-1} \cdot 6^{30-k}$ sequences in which the $k$-th roll is a six, and no 4, 5, or 6 values appear before it (1 $\leq k \leq 30$).

The number of favorable cases is therefore the sum of these values, which can be calculated using the formula for the sum of a geometric series:

$$
\begin{aligned}
3^{0} \cdot 6^{29} + 3^{1} \cdot 6^{28} + \ldots + 3^{28} \cdot 6 + 3^{29} = & 3^{29} \left(2^{29} + 2^{28} + \ldots + 2 + 1\right) = \\
& = 3^{29} \cdot \frac{2^{30} - 1}{2 - 1} = 3^{29} \left(2^{30} - 1\right)
\end{aligned}
$$

The total number of cases is the number of 30-element sequences containing the digits 1 through 6: $6^{30}$. Thus, the desired probability is:

$$
P = \frac{3^{29} \left(2^{30} - 1\right)}{6^{30}} = \frac{3^{29} \left(2^{30} - 1\right)}{3^{30} \cdot 2^{30}} = \frac{1}{3} - \frac{1}{3 \cdot 2^{30}}
$$

Since $3 \cdot 10^{-10} < \frac{1}{3 \cdot 2^{30}} < 4 \cdot 10^{-10}$, subtracting $\frac{1}{3 \cdot 2^{30}}$ from $\frac{1}{3}$ does not change the first eight decimal places. Therefore, the desired probability, to eight decimal places, is $\frac{1}{3}$.