If the sum of the points that appear in these $n$ dice rolls is greater than $2^{n}$, then it counts as passing. Ask:
(1) What is the maximum number of levels a person can pass in this game?
(2) What is the probability that he can pass the first three levels consecutively?
(Note: A die is a uniform cube with 1, 2, 3, 4, 5, 6 points on each of its faces. The number of points on the face that lands up after the die is rolled and comes to rest is the number of points that appear.)
Since the die is a uniform cube, the possibility of each number appearing after tossing is equal.
(1) Because the maximum number on the die is 6, and $6 \times 4 > 2^{4}, 6 \times 5 < 2^{5}$, therefore, when $n \geqslant 5$, “the sum of the numbers appearing $n$ times is greater than $2^{n}$” is impossible. That is, this is an impossible event, and the probability of passing is 0. Therefore, the maximum number of consecutive levels that can be passed is 4.
(2) Let event $A_{n}$ be “failing the $n$-th level”, then the complementary event $\overline{A_{n}}$ is “successfully passing the $n$-th level”.
In the $n$-th level game, the total number of basic events is $6^{n}$.
Level 1: The number of basic events contained in event $A_{1}$ is 2 (i.e., the cases where the number is 1 and 2). Therefore, the probability of passing this level is
$$
P\left(\overline{A_{1}}\right)=1-P\left(A_{1}\right)=1-\frac{2}{6}=\frac{2}{3} .
$$

Level 2: The number of basic events contained in event $A_{2}$ is the sum of the positive integer solutions of the equation $x+y=a$ when $a$ takes the values $2, 3, 4$. That is, $\mathrm{C}_{1}^{1}+\mathrm{C}_{2}^{1}+\mathrm{C}_{3}^{1}=1+2+3=6$ (cases). Therefore, the probability of passing this level is
$$
P\left(\overline{A_{2}}\right)=1-P\left(A_{2}\right)=1-\frac{6}{6^{2}}=\frac{5}{6} .
$$

Level 3: The basic events contained in event $A_{3}$ are the sum of the positive integer solutions of the equation $x+y+z=a$ when $a$ takes the values $3, 4, 5, 6, 7, 8$. That is, $\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\mathrm{C}_{4}^{2}+\mathrm{C}_{5}^{2}+\mathrm{C}_{6}^{2}+\mathrm{C}_{7}^{2}=1+3+6+10+15+21=56$ (cases). Therefore, the probability of passing this level is
$$
P\left(\overline{A_{3}}\right)=1-P\left(A_{3}\right)=1-\frac{56}{6^{3}}=\frac{20}{27} .
$$

Thus, the probability of passing the first three levels consecutively is:
$$
P\left(\overline{A_{1}}\right) \times P\left(\overline{A_{2}}\right) \times P\left(\overline{A_{3}}\right)=\frac{2}{3} \times \frac{5}{6} \times \frac{20}{27}=\frac{100}{243} .
$$