In a tournament, a team plays 10 matches with probabilities of winning and losing each match as $$\frac{1}{3}$$ and $$\frac{2}{3}$$ respectively. Let $$x$$ be the number of matches that the team wins, and $$y$$ be the number of matches that team loses. If the probability $$P(|x - y| \leq 2)$$ is $$p$$, then $$3^9 p$$ equals _____

A team plays 10 matches with probability of winning each match $$p = \frac{1}{3}$$ and losing $$q = \frac{2}{3}$$. Let $$x$$ denote the number of wins and $$y$$ the number of losses. Since $$x + y = 10$$, it follows that $$y = 10 - x$$.

Substituting into the condition $$|x - y| \le 2$$ gives $$|x - (10 - x)| \le 2 \implies |2x - 10| \le 2 \implies -2 \le 2x - 10 \le 2 \implies 4 \le x \le 6$$, so the possible values of $$x$$ are $$4, 5, 6$$.

Since $$X \sim \mathrm{Binomial}(10,1/3)$$, the probability mass function is $$P(X = r) = \binom{10}{r} \left(\frac{1}{3}\right)^r \left(\frac{2}{3}\right)^{10-r} = \frac{\binom{10}{r}\cdot2^{10-r}}{3^{10}}\,. $$

For $$r = 4$$ this yields $$P(X=4) = \frac{\binom{10}{4}\cdot2^6}{3^{10}} = \frac{210 \times 64}{3^{10}} = \frac{13440}{3^{10}}\,. $$

For $$r = 5$$ one finds $$P(X=5) = \frac{\binom{10}{5}\cdot2^5}{3^{10}} = \frac{252 \times 32}{3^{10}} = \frac{8064}{3^{10}}\,. $$

For $$r = 6$$ we have $$P(X=6) = \frac{\binom{10}{6}\cdot2^4}{3^{10}} = \frac{210 \times 16}{3^{10}} = \frac{3360}{3^{10}}\,. $$

Adding these probabilities gives $$p = P(|x - y| \le 2) = \frac{13440 + 8064 + 3360}{3^{10}} = \frac{24864}{3^{10}}\,. $$

Multiplying by $$3^9$$ yields $$3^9 p = 3^9 \cdot \frac{24864}{3^{10}} = \frac{24864}{3} = 8288\,. $$

Therefore, the final answer is $$\boxed{8288}$$.