【Question 6】Person A and Person B stand facing each other 30 meters apart, playing "Rock, Paper, Scissors". The winner moves forward 3 meters, the loser moves back 2 meters, and in the event of a tie, both move forward 1 meter. After 15 rounds, Person A is 17 meters from the starting point, and Person B is 2 meters from the starting point. How many times did Person A win $\qquad$?
Analysis:
(Method 1) In games with a winner and a loser, the distance between the two people shortens by 1 meter; in a tie, the distance shortens by 2 meters. After 15 games, the distance between the two people will shorten by 15 to 30 meters.
(1) If both people retreat in the end, the distance between them will not shorten but will increase, which contradicts the above analysis.
(2) If one person advances and the other retreats in the end, the distance between them will shorten by $17-2=15$ meters. If the distance between them shortens by 15 meters, it can only be that all 15 games were "win-lose games."
Assuming Player A wins all 15 games, he will advance 45 meters. Each time a "winner" is replaced by a "loser," he will advance 5 meters less. 45 minus any number of 5s cannot equal 17. This situation does not hold.
(4) If both people advance in the end, the distance between them will shorten by $17+2=19$ meters. Assuming all 15 games were "win-lose games," the distance between them will shorten by 15 meters. Each time a "win-lose game" is replaced by a tie, the distance between them will shorten by 1 meter more. Using the "chicken and rabbit in the same cage" method, we can find that there were 11 "win-lose games" and 4 ties.
In the 4 ties, Player A advanced 4 meters. Assuming Player A won the remaining 11 games, he would have advanced $33+4=37$ meters in total. Each time a win is changed to a loss, he will retreat 5 meters. Changing 4 games will meet the requirement, and he will advance $37-5 \times 4=17$ meters in total.
Therefore, Player A has 7 wins, 4 ties, and 4 losses.
(Method 2) Let Player A win $x$ games, lose $y$ games, and tie $z$ games (where $x, y, z$ are natural numbers).
From the problem, we have
\[
\left\{
\begin{array}{l}
x+y+z=15 \\
3x-2y+z=17 \\
3y-2x+z=2
\end{array}
\right.
\]
or
\[
\left\{
\begin{array}{l}
x+y+z=15 \\
3x-2y+z=17 \\
2x-(3y+z)=2
\end{array}
\right.
\]
or
\[
\left\{
\begin{array}{l}
x+y+z=15 \\
2y-(3x+z)=17 \\
3y-2x+z=2
\end{array}
\right.
\]
or
\[
\left\{
\begin{array}{l}
x+y+z=15 \\
2y-(3x+z)=17 \\
2x-(3y+z)=2
\end{array}
\right.
\]
Solving these, we get
\[
\left\{
\begin{array}{l}
x=7 \\
y=4 \\
z=4
\end{array}
\right.
\]
or
\[
\left\{
\begin{array}{l}
x=\frac{47}{5} \\
y=\frac{28}{5} \\
z=0
\end{array}
\quad
\right.
\]
or
\[
\left\{
\begin{array}{l}
x=\frac{103}{5} \\
y=\frac{122}{5} \\
z=-30
\end{array}
\quad
\right.
\]
or
\[
\left\{
\begin{array}{l}
x=23 \\
y=26 \\
z=-34
\end{array}
\right.
\]
Therefore, Player A has 7 wins, 4 ties, and 4 losses.