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 the die is rolled and comes to rest 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 $2^{n}$, 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: $P\left(\overline{A_{\mathrm{i}}}\right)=1-P\left(A_{1}\right)=1-\frac{2}{6}=\frac{2}{3}$
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, $C_{1}^{\mathfrak{t}}+C_{3}^{1}+C_{3}^{1}=1+2+3=6$ (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 $A$ 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, $C_{2}^{e}+C_{3}^{2}+C_{1}^{2}+C_{5}^{2}+C_{6}^{2}+C_{7}^{2}=1+3+6+10+15+21=56$ (cases).
So the probability of passing this level is: $P\left(\bar{A}_{3}\right)=1-P\left(A_{i}\right)=1-\frac{56}{6^{2}}=\frac{20}{27}$
Therefore, the probability of passing the first two levels is: $P\left(\bar{A}_{1}\right) \cdot P\left(\bar{A}_{2}\right) \cdot P\left(\bar{A}_{3}\right)=\frac{2}{3} \times \frac{5}{6} \times \frac{20}{27}=\frac{100}{243}$