Example 15 (2004 National High School Competition Question) The rules of a "level-passing game" stipulate: on the $n$-th level, a die must be rolled $n$ times. If the sum of the points from these $n$ rolls is greater than $2^{n}$, then the level is passed. The questions are:
(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 points numbered $1,2,3,4,5,6$ on its faces. The point number on the face that lands up after the die is rolled and comes to rest is the point number that appears.)
Solution: Since the dice is a uniform cube, the possibility of each number appearing after tossing is the same.
(1) Because the maximum number on the dice is 6, and $6 \cdot 4 > 2^{4}, 6 \cdot 5 < 2^{5}$, therefore when $n \geqslant 5$, the sum of the numbers appearing $n$ times being greater than $2^{n}$ is impossible. Hence, this is an impossible event, and the probability of ultimately passing the level is 0. Therefore, the maximum number of levels that can be passed consecutively is 4.
(2) Let event $A_{n}$ be "failing the $n$-th level", then the complementary event $\bar{A}$ 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 1 or 2 appears). Therefore, the probability of passing this level is
$$
P\left(\bar{A}_{1}\right)=1-P\left(A_{1}\right)=1-\frac{2}{6}=\frac{2}{3} .
$$

Level 2: The number of 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 $2,3,4$, which is $C_{1}^{1}+C_{2}^{1}+C_{3}^{1}=6$ (cases). Therefore, the probability of passing this level is
$$
P\left(\bar{A}_{2}\right)=1-P\left(A_{2}\right)=1-\frac{6}{6^{2}}=\frac{5}{6} .
$$

Level 3: The basic events contained in event $A_{3}$ are the positive integer solutions of the equation $x+y+z=a$ when $a$ takes the values $3,4,5,6,7,8$, which is $C_{2}^{2}+C_{3}^{2}+C_{4}^{2}+C_{5}^{2}+C_{6}^{2}+C_{7}^{2}=56$ (cases). Therefore, the probability of passing this level is $P\left(\bar{A}_{3}\right)=1-P\left(A_{3}\right)=1-\frac{56}{6^{3}}=\frac{20}{27}$.
Thus, the probability of passing the first three levels consecutively is
$$
P\left(\bar{A}_{1}\right) \cdot P\left(\bar{A}_{2}\right) \cdot P\left(\bar{A}_{3}\right)=\frac{2}{3} \cdot \frac{5}{6} \cdot \frac{20}{27}=\frac{100}{243} .
$$

Note: The number of basic events in the 2nd and 3rd levels can also be listed.