Whatever Alice guesses, the deck evolves the same way; only the immediate success probability depends on her guess, so it is optimal to guess a color with the most cards remaining. Let $E(r, b)$ be the expected number of correct guesses from a state with $r$ red and $b$ black cards left. Then $E(r, 0) = r,$ $E(0, b) = b,$ and $$\begin{aligned} E(r, b) &= \frac{\max(r, b)}{r + b} \\ &\quad {}+ \frac{r\,E(r-1,b)}{r+b} \\ &\quad {}+ \frac{b\,E(r,b-1)}{r+b}. \end{aligned}$$

By symmetry $E(r, b) = E(b, r).$ Computing upward: $E(1,1) = \frac{3}{2},$ $E(2,1) = \frac{2}{3} + \frac{2 \cdot \frac{3}{2} + 2}{3} = \frac{7}{3},$ $E(2,2) = \frac{1}{2} + \frac{7}{3} = \frac{17}{6},$ $E(3,1) = \frac{3}{4} + \frac{3 \cdot \frac{7}{3} + 3}{4} = \frac{13}{4},$ $E(3,2) = \frac{3}{5} + \frac{3 \cdot \frac{17}{6} + 2 \cdot \frac{13}{4}}{5} = \frac{18}{5},$ and finally $E(3,3) = \frac{1}{2} + \frac{18}{5} = \frac{41}{10}.$

So the expected number of correct guesses is $\frac{41}{10},$ and $m + n = 41 + 10 = 51.$