Question 193, The rules of the Chinese Men's Basketball First Division League stipulate: Each game, the winner gets 2 points, the loser gets 1 point (each game, even if it goes to overtime, must have a winner). A certain men's first division team is very strong, with a probability of winning each game being $\frac{3}{4}$, and a probability of losing being $\frac{1}{4}$. Find the probability that the team will score $\mathrm{n}$ points through several games in the middle of the season (assuming the total number of games the team plays in this season is $\mathrm{S}$, here $0<\mathrm{n} \leq \mathrm{S}$).
Question 193, Answer: Let the probability of a team scoring $\mathrm{n}$ points through several matches in the middle of the season be $\mathrm{p}_{\mathrm{n}}$. According to the conditions: $\mathrm{p}_{1}=\frac{1}{4}, \mathrm{p}_{2}=\left(\frac{1}{4}\right)^{2}+\frac{3}{4}=\frac{13}{16}$. Next, we consider the relationship between $\mathrm{p}_{\mathrm{n}+2}$ and $\mathrm{p}_{\mathrm{n}+1}, \mathrm{p}_{\mathrm{n}}$.
For the men's top-tier basketball team, obtaining $\mathrm{n}+2$ points through several matches can only happen in two scenarios: (1) before obtaining $\mathrm{n}+2$ points, they scored $\mathrm{n}+1$ points in the previous matches, and then scored 1 point in the last match, with a probability of $\mathrm{p}_{\mathrm{n}+1} \cdot \frac{1}{4}$; or before obtaining $\mathrm{n}+2$ points, they scored $\mathrm{n}$ points in the previous matches, and then scored 2 points in the last match, with a probability of $\mathrm{p}_{\mathrm{n}} \cdot \frac{3}{4}$. Therefore,
$\mathrm{p}_{\mathrm{n}+2}=\mathrm{p}_{\mathrm{n}+1} \cdot \frac{1}{4}+\mathrm{p}_{\mathrm{n}} \cdot \frac{3}{4}$
The characteristic equation is $\mathrm{x}^{2}=\frac{\mathrm{x}}{4}+\frac{3}{4}$, solving which gives the two characteristic roots as $\mathrm{x}_{1}=1, \mathrm{x}_{2}=-\frac{3}{4}$. Let $\mathrm{p}_{\mathrm{n}}=\mathrm{A}+ B\left(-\frac{3}{4}\right)^{n}$. Then $\left\{\begin{array}{c}\frac{1}{4}=A-\frac{3}{4} B \\ \frac{13}{16}=A+\frac{9}{16} B\end{array}\right.$, solving which gives $\left\{\begin{array}{l}A=\frac{4}{7} \\ B=\frac{3}{7}\end{array}\right.$. Therefore, $p_{n}=\frac{4}{7}+\frac{3}{7}\left(-\frac{3}{4}\right)^{n}$.
Note: If the team wins $s$ matches and loses $\mathrm{t}$ matches, scoring exactly $\mathrm{n}$ points, then $2 \mathrm{~s}+\mathrm{t}=\mathrm{n}(\mathrm{s}, \mathrm{t} \in \mathrm{N})$, and the probability of such a match situation.