13 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 $2^{n}$, it counts as passing the level. 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. After rolling the die and it comes to rest, the number of points on the top face is the result of the roll.)
Solution: Since the dice is a uniform cube, the possibility of each number appearing after tossing is equal.
(1) 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 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. $\cdots 5$ points
(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
$$
P\left(\overline{A_{1}}\right)=1-P\left(A_{1}\right)=1-\frac{2}{6}=\frac{2}{3} \text{. }
$$

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 2, $3$, and 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
$$
P\left(\overline{A_{2}}\right)=1-P\left(A_{2}\right)=1-\frac{6}{6^{2}}=\frac{5}{6} . \cdots \cdots \cdots \cdot 10 \text{ points }
$$

Level 3: 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 3, $4$, $5$, $6$, $7$, and 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
$$
P\left(\overline{A_{3}}\right)=1-P\left(A_{3}\right)=1-\frac{56}{6^{3}}=\frac{20}{27} . \quad \cdots \cdots 15 \text{ points }
$$

Therefore, the probability of passing the first three levels consecutively is:
$$
P\left(\overline{A_{1}}\right) \times P\left(\overline{A_{2}}\right) \times P\left(\overline{A_{3}}\right)=\frac{2}{3} \times \frac{5}{6} \times \frac{20}{27}=\frac{100}{243} . \cdots \cdots 20 \text{ points }
$$
(Note: The number of basic events in the 2nd and 3rd levels can also be listed.)