The payment/stopping rule in the link is quite different from the rule presented in the question here. I'm going to answer the question here.
It turns out there is no strategy that will improve your odds of winning a game in which you are allowed to end the game by guessing that the next card is red; the expected value is $0$, so you shouldn't be willing to pay anything to play. (I am assuming that the cards have been shuffled, so that no one knows the status of any card until it is revealed.)
Here's a way to see that the expected value is $0$. Imagine the cards are spread out, face down, from left to right. Start by placing your finger on the rightmost card, to indicate that you tentatively intend to stop at the very end. Obviously, this card has a $50\%$ chance of being red. Now, before each card, starting from the left, is turned over, you are allowed to change your mind and stop with it instead. Whether you do or don't doesn't matter: the rightmost card and the current still-face-down leftmost card have equal probability of being red, so you may as well stick with your initial tentative decision. Although your current assessment of how much you stand to gain (or lose) will change each time a leftmost card is revealed, that has no bearing on whether to stop. In other words, even though it may feel as is you have some control over your fate, you really don't.
Remark: This answer is similar to an answer I gave to a "number battle" problem.
Theorem: Even if you cheat, you cannot do better than chance. In particular, your expected number of points is equal to your probability of winning a single round between a random card of yours and a random card of your opponent's, times your total number of cards.
Proof: The proof is by induction. Suppose you and your opponent each have a deck of $n=2$ cards in a fixed but arbitrary order. Consider the payout matrix $\mathbf{P} = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ which has a $1$ in entry $(i,j)$ if your $i$th card beats their $j$th card, and a 0 otherwise. If you choose row 1, your opponent will randomly choose a column $j$. Your payout will either be $a+d$ (if they choose column 1), or $b+c$ (if they choose column 2). Your expected payout on choosing row 1 is $(a+b+c+d)/2$. A similar argument shows the same expected payout on choosing row 2. This establishes the result that your choices do not influence the expected score when $n=2$.
In the inductive case, suppose you have an $(n+1)\times (n+1)$ payout matrix $\mathbf{P}$. Removing cards corresponds to removing a row/column from the matrix — let $\mathbf{P}^{i,\times}$ denote $\mathbf{P}$ with row $i$ deleted; $\mathbf{P}^{\times,j}$ with column $j$ deleted, and $\mathbf{P}^{i,j}$ with both deleted. Finally, let $|\mathbf{P}|$ denote the sum of all the entries of $\mathbf{P}$.
Now consider your strategy of choosing row $i$. Your opponent chooses
a column $j$. For a given column $j$, your payout will be $P_{i,j}$,
plus the expected payout resulting from continuing the game; by
hypothesis, it is $\frac{1}{n}|\mathbf{P}^{i,j}|$. Hence when choosing
row $i$, your expected payout over all opponent responses $j$ is:
$$\frac{1}{n+1} \sum_{j=1}^{n+1}\left[ P_{i,j} + \frac{1}{n}|\mathbf{P}^{i,j}|\right]$$
Note that the term $|\mathbf{P}^{i,j}|$ occurs once for each $j$ from
1 to $n+1$. Hence we can assemble the deleted columns into a single
negative term $-|\mathbf{P}^{i,\times}|$.
$$\frac{1}{n+1} \sum_{j=1}^{n+1}\left[ P_{i,j} + \frac{1}{n}|\mathbf{P}^{i,j}|\right] = \frac{1}{n+1} \left[ \sum_{j=1}^{n+1} P_{i,j} + \frac{1}{n}\sum_{j=1}^{n+1}|\mathbf{P}^{i,j}|\right] = \frac{1}{n+1} \left[ \sum_{j=1}^{n+1} P_{i,j} + \frac{1}{n}\left(-|\mathbf{P}^{i,\times}| +\sum_{j=1}^{n+1}|\mathbf{P}^{i,\times}|\right)\right] = \frac{1}{n+1}\left[|\mathbf{P}^{i,\times}|+\sum_{j=1}^{n+1} P_{i,j}\right] = \frac{1}{n+1}|\mathbf{P}|$$
This establishes the result for the inductive case.
Best Answer
Suppose that a fraction $p_i$ of the cards are labelled $i$, for $1\leq i\leq 5$.
The crucial insight is that your best strategy will have $$1p_1=2p_2=\dots=5p_5,$$ so that the opponent is indifferent to which card they guess. You can then see that the expected value is $\frac{60}{137}$.
Indeed, if we don't follow this distribution, then some $j$ would have $jp_j>\frac{60}{137}$, so the opponent could always guess card $j$ and have a higher EV.