Best strategy to optimize probability of winning in card drawing game with opponent
Update: Notice that if we had only two suits to begin with, the two strategies would yield the same winning probability. So in effect, having two more suits diluted the weight of $p$ in $P_2$.
There's little calculation needed.
Denote $P_i$: probability you win if you choose suit after seeing $i$ card(s).
Denote $p$: probability you win if you have already seen two cards of your suit, and no cards from opponent's suit. Then $p>\frac 12$.
And
$$P_1 = \frac{12}{25}p+\frac{13}{25}\cdot \frac 12\\ P_2 = \frac {12}{51} p + \frac{39}{51} \cdot \frac 12\\ \implies P_1 - P_2 = \left(\frac{12}{25}- \frac{12}{51} \right) p + \frac 12 \left( \frac{13}{25}-\frac{39}{51}\right)\\ > \left(\frac{12}{25}- \frac{12}{51} \right) \frac 12 + \frac 12 \left( \frac{13}{25}-\frac{39}{51}\right) = 0. \blacksquare $$
Let $f(i, j)$ be probability of you winning if there already are $i$ of your cards and $j$ of your opponents ($0 \leq i, j \leq 5$, at most one of $i$ and $j$ is equal to $5$). $f(5, j) = 1$ and $f(i, 5) = 0$ by definition, also $f(i, i) = \frac{1}{2}$ by symmetry.
Probability of winning if you choose before any card draw is $f(0, 0)$, if you choose after the first - $f(1, 0)$ and after the second - $p \cdot f(1, 1) + (1 - p) \cdot f(2, 0)$, where $p = \frac{39}{51}$ is probability of first two cards having different suits.
Your question is if $f(1, 0) > \frac{39}{102} + \frac{24}{102} \cdot f(2, 0)$.
It's not too hard to calculate using $f(i, j) = \frac{13 - i}{26 - i - j} \cdot f(i + 1, j) + \frac{13 - j}{26 - i - j} \cdot f(i, j + 1)$. In theory, it can be calculated by hands if needed, but I'll cheat a bit and just print the table:
$$\begin{array}{|c|c|c|c|c|c|} \hline j\backslash i & 4 & 3 & 2 & 1 & 0 \\ \hline 4 & \frac{1}{2} & \frac{5}{19} & \frac{11}{76} & \frac{11}{133} & \frac{13}{266} \\ \hline 3 & \frac{14}{19} & \frac{1}{2} & \frac{44}{133} & \frac{29}{133} & \frac{442}{3059} \\ \hline 2 & \frac{65}{76} & \frac{89}{133} & \frac{1}{2} & \frac{1117}{3059} & \frac{923}{3496} \\ \hline 1 & \frac{122}{133} & \frac{104}{133} & \frac{1942}{3059} & \frac{1}{2} & \frac{169}{437} \\ \hline 0 & \frac{253}{266} & \frac{2617}{3059} & \frac{2573}{3496} & \frac{268}{437} & \frac{1}{2} \\ \hline \end{array}$$
(we actually need just left-bottom half of the table)
So $f(1, 0) = \frac{268}{437} \approx 0.61$, $f(2, 0) = \frac{2573}{3496}$ and $\frac{39}{102} + \frac{24}{102} \cdot f(2, 0) \approx 0.56$.