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 obtained from these $n$ rolls is greater than $2^{n}$, it counts as passing 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?
(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.)
( I ) 4( II )
$$
\frac{100}{243}
$$
13.【Analysis and Solution】Since the dice are uniform cubes, 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. That is, it 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 $\bar{A}_{n}$ 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 numbers 1 and 2 appear in these two cases),
$\therefore$ The probability of passing this level is: $P\left(\bar{A}_{1}\right)\left(\bar{A}_{1}\right)=1-P\left(A_{1}\right)=1-\frac{2}{6}=\frac{2}{3}$.
Level 2: The number of basic events contained in event $A_{2}$ is the sum of the 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 $P:\left(\bar{A}_{2}\right)=1-P\left(\bar{A}_{2}\right)=1-\frac{6}{6^{2}}=\frac{5}{6}$. $\qquad$
Level 3: The basic events contained in event $A_{3}$ are 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, $\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\mathrm{C}_{4}^{2}+\mathrm{C}_{5}^{2}+\mathrm{C}_{6}^{2}+\mathrm{C}_{7}^{2}=1+3+6+10+15+21=56$ (cases).
$\therefore$ The probability of passing this level is: $P\left(\bar{A}_{3}\right)=1-P\left(\bar{A}_{3}\right)=1-\frac{56}{6^{3}}=\frac{20}{27}$. $\qquad$
Therefore, the probability of passing the first three levels is: $P\left(\bar{A}_{3}\right) \times P\left(\bar{A}_{2}\right) \times P\left(\bar{A}_{3}\right)=\frac{2}{3} \times \frac{5}{6} \times \frac{20}{27}=\frac{100}{243}$.
..... 20 points
(Note: The number of basic events in the 2nd and 3rd levels can also be listed out.)