19. (12 points) A bag contains 2 white balls and $n$ red balls $\left(n \geqslant 2\right.$, and $\left.n \in \mathbf{Z}_{+}\right)$, each time two balls are drawn from the bag (the two balls are returned to the bag after each draw). If the two balls drawn are of the same color, it is considered a win; otherwise, it is a loss.
(1) Express the probability of winning in one draw, $p_{1}$, as an algebraic expression in terms of $n$;
(2) If $n=3$, find the probability of winning exactly once in three draws;
(3) Let the probability of winning exactly once in three draws be $f(p)$. For what value of $n$ is $f(p)$ maximized?
19. (1) When selecting two balls from $n+2$ balls, there are $\mathrm{C}_{n+2}^{2}$ ways to choose, among which the ways to select two balls of the same color are $\mathrm{C}_{n}^{2}+\mathrm{C}_{2}^{2}$.
Thus, the probability of winning in one draw is
$$
p_{1}=\frac{\mathrm{C}_{n}^{2}+\mathrm{C}_{2}^{2}}{\mathrm{C}_{n+2}^{2}}=\frac{n^{2}-n+2}{n^{2}+3 n+2} \text {. }
$$
(2) If $n=3$, then the probability of winning in one draw is $\frac{2}{5}$.

Since three draws are independent repeated experiments, the probability of winning exactly once in three draws is
$$
P_{3}(1)=\mathrm{C}_{3}^{1} P(1-P)^{2}=\frac{54}{125} \text {. }
$$
(3) Let the probability of winning in one draw be $p$, then the probability of winning exactly once in three draws is
$$
f(p)=\mathrm{C}_{3}^{1} p(1-p)^{2}=3 p^{3}-6 p^{2}+3 p(0<p<1) \text {. }
$$

Notice that,
$$
f^{\prime}(p)=9 p^{2}-12 p+3=3(p-1)(3 p-1) \text {. }
$$

Thus, $f(p)$ is an increasing function on $\left(0, \frac{1}{3}\right)$ and a decreasing function on $\left(\frac{1}{3}, 1\right)$.
Therefore, when $p=\frac{1}{3}$, $f(p)$ reaches its maximum value, at which time,
$$
p=\frac{n^{2}-n+2}{n^{2}+3 n+2}=\frac{1}{3}(n \geqslant 2, n \in \mathbf{Z}) \text {. }
$$

Solving for $n$ yields $n=2$.
Hence, when $n=2$, the probability of winning exactly once in three draws is maximized.