Example 6 The rules of a "level-up game" stipulate: On the $n$-th level, a die must be rolled $n$ times. If the sum of the points obtained from these $n$ rolls is greater than $2^{n}$, the level is considered passed. Questions:
(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 number of points on the face that lands up after rolling the die is the result of the roll.)
Solution: Since the dice is a uniform cube, the possibility of each number appearing after a roll is equal.
(1) Because the maximum number that can appear on a single dice is 6, and $6 \times 1 > 2^4, 6 \times 5 < 2$.
Therefore, when $n \geqslant 5$, the sum of the numbers that appear after rolling the dice $n$ times cannot be greater than 2, which means this is an impossible event, and the probability of passing is 0.
So, 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 "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 or 2),
So 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 number of positive integer solutions to the equation $x+y=a$ when $a$ takes the values $2,3,4$. That is, $C_{1}^{3}+C_{3}^{1}+C_{3}^{1}=1+2+3=6$ (cases).
So 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 number of basic events contained in event $A_{3}$ is the sum of the number of positive integer solutions to the equation $x+y+z=a$ when $a$ takes the values $3,4,5,6,7,8$. That is, $C_{2}^{3}+C_{3}^{3}+C_{4}^{2}+C_{5}^{3}+C_{6}^{3}+C_{5}^{2}=1+3+6+10+15+21=56$ (cases).
So the probability of passing this level is: $P\left(\bar{A}_{3}\right)=1-P\left(A_{3}\right)=1-\frac{56}{6^{3}}=\frac{20}{27}$
Therefore, the probability of passing the first 3 levels consecutively is: $P\left(\bar{A}_{1}\right) \cdot P\left(\bar{A}_{2}\right) \cdot P\left(\bar{A}_{3}\right)=\frac{2}{3} \times \frac{5}{6} \times \frac{20}{27}=\frac{100}{243}$