For example, 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 $2^{n}$, the player passes the level. Questions:
( I ) What is the maximum number of levels a person can pass in this game?
( II ) What is the probability that he can pass the first three levels consecutively?
(2004 National High School Mathematics Competition)
(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 is the result of the roll.)
Solution: Since the dice is a uniform cube, the possibility of each number appearing after tossing is equal.
(I) Because the maximum number on the dice 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 being greater than $2^{n}$ is impossible, i.e., 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.
(II) 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 $6^{n}$. 
First level: The number of basic events contained in event $A_{1}$ is 2 (i.e., the cases where the number is 1 or 2). Therefore, the probability of passing this level is $P\left(\overline{A_{1}}\right)=1-P\left(A_{1}\right)=1-\frac{2}{6}=\frac{2}{3}$. 
Second level: 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, $C_{1}^{1}+C_{2}^{1}+C_{3}^{1}=1+2+3=6$ (cases). Therefore, the probability of passing this level is $P\left(\overline{A_{2}}\right)=1-P\left(A_{2}\right)=1-\frac{6}{6^{2}}=\frac{5}{6}$. 
Third level: The number of basic events contained in event $A_{3}$ is 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, $C_{2}^{2}+C_{3}^{2}+C_{4}^{2}+C_{5}^{2}+C_{6}^{2}+C_{7}^{2}=1+3+6+10+15+21=56$ (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 $P\left(\overline{A_{1}}\right) \times P\left(\overline{A_{2}}\right) \times P\left(\overline{A_{3}}\right)=\frac{2}{3} \times \frac{5}{6} \times \frac{20}{27}=\frac{100}{243}$.
/Comment The difficulty of this problem lies in calculating the number of basic events.