China Mathematical Competition (Hainan)

Since the die is fair, the probability of any of the six numbers appearing is the same.

(1) Since the highest point of a die is $6$, and $6\times 4>2^4$, $6\times 5<2^5$, it is impossible that the sum of points appearing in $n$ tosses is bigger than $2^n$ if $n\ge 5$. This means it is an impossible event, and the probability of crossing the obstacle is $0$. Therefore at most $4$ obstacles that a person can cross.

(2) We denote $A_n$ the event "at the $n$th obstacle the person fails to cross", the complementary event $\overline{A_n}$ is "at the $n$th obstacle the person crosses successfully". At the $n$th obstacle of this game the number of all possible outcomes is $6^n$.

The first obstacle: event $A_1$ contains $2$ possible outcomes (i.e., the outcomes in which the number appearing is $1$ or $2$). So the probability of crossing the obstacle is $$P(\overline{A_1})=1-P(A_1)=1-\frac{2}{6}=\frac{2}{3}.$$

The second obstacle: the number of outcomes contained in event $A_2$ is the total number of positive integer solution sets of the equation $x+y=a$ where $a$ is taken to be $2$, $3$ and $4$ respectively. Thus the number of outcomes equals $C_1^1+C_2^1+C_3^1=1+2+3=6$, and the probability of crossing the obstacle is $$P(\overline{A_2})=1-P(A_2)=1-\frac{6}{6^2}=\frac{5}{6}.$$

The third obstacle: the number of outcomes contained in event $A_3$ is the total number of positive integer solution sets of the equation $x+y+z=a$ where $a$ is taken to be $3$, $4$, $5$, $6$, $7$ and $8$ respectively. Thus the number of outcomes equals $$C_2+C_3+C_4+C_5+C_6+C_7=1+3+6+10+15+21=56,$$ and the probability of crossing the obstacle is $$P(\overline{A_3})=1-P(A_3)=1-\frac{56}{6^3}=\frac{20}{27}.$$

Consequently, the probability that a person crosses the first three obstacles is $$ P(\overline{A_1}) \times P(\overline{A_2}) \times P(\overline{A_3}) = \frac{2}{3} \times \frac{5}{6} \times \frac{20}{27} = \frac{100}{243}.