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
$$P_1 = \frac{12}{25}p+\frac{13}{25}\cdot \frac 12\\
P_2 = \frac {12}{51} p + \frac{39}{51} \cdot \frac 12\\
\implies P_1 - P_2 = \left(\frac{12}{25}- \frac{12}{51} \right) p + \frac 12 \left( \frac{13}{25}-\frac{39}{51}\right)\\
> \left(\frac{12}{25}- \frac{12}{51} \right) \frac 12 + \frac 12 \left( \frac{13}{25}-\frac{39}{51}\right) = 0. \blacksquare
$$
(a sketch, too detailed for the comments)
Suppose you had $N$ cards.
For each $k\in \{1,2,\cdots, N\}$ we consider the strategy $\mathscr S_k$ to be "take the maximum of $k$ selected cards". We note that we can achieve this strategy by paying $k-1$ dollars, and sticking with the current max. $\mathscr S_1$ is, of course, just blind guessing. Letting $E_k$ denote the expected max of the $k$ selected cards, the value of $\mathscr S_k$ is $S_k=E_k-(k-1)$.
Now consider the special case $N=10$. Of course $S_1=5.5$ but we can do better. One round of information gets us to $E_2= 7.\overline 3$ so $S_2=6.\overline 3$ which is better than $S_1$.
Can we do even better? Well, let's consider $\mathscr S_3$. To get the distribution of the max $M$ it is easiest to compute the probability $P(M≤i)$ for $i\in \{3,4,5,6,7,8,9,10\}$. Then, of course, $P(M=i)=P(M≤i)-P(M≤i-1)$. Routine calculations yield $$\{P(M=i)\}_{i=3}^{10}=\{.008333,.025,.05,.08333,.125,.175,.2333,.3\}$$
from which we deduce that $E_3=8.25$ which implies that $S_3=6.25$ which is less than $S_2$.
This leads me to believe that $\mathscr S_2$ is optimal.
To stress: this is not complete. I believe, but have not shown, that the strategies increase to a max and then decrease thereafter. In this particular case, of course, all you need to do is to consider the other $\mathscr S_k$, which would be straight forward, if a bit tedious.
Note: complicating matters considerably, you are able to improve on $\mathscr S_3$ by only paying the second round selectively. For instance, if the first choice gets you a min of $9$, then you should obviously just take the $10$ and stop. Similarly, if the first choice gets you a min of $8$ (or maybe even less) it might well be better to stop. That sort of optional exercise can best be handled by backwards induction. though that's a bit messy even in the case $N=10$.
Best Answer
Here is a basic strategy that should do better than just flipping one card and staying with it:
Always flip a second card after the first. If the second card has a higher value than the first, stick with the second card, otherwise flip the third card.
Why does this better than staying with just the one first card?
Well, as you say, just flipping one card has an expected value of $n+1$
But for the other strategy:
There are 6 cases to consider, each of which is equally likely:
First card is $n$, second is $n+1$. The strategy says to stick with the second card, so outcome is $n+1$
First card is $n$, second is $n+2$. Stick with card. Outcome is $n+2$
$n+1$ followed by $n$. Strategy says to flip third card. Outcome $n+2$
$n+1$ followed by $n+2$. Stick with card: $n+2$
$n+2$ then $n$. Flip third card: outcome is $n+1$
$n+2$ then $n+1$. Flip third card: $n$
Since each of these events is equally likely, the expected outcome of this stratey is just the average of these outcomes, which is $n+\frac{4}{3}$
OK, so that's indeed a better strategy. Is it optimal? I don't know.