Say "red" with playing cards
It doesn't matter what you do. There may be a simpler way to see this, but the straightforward way is by induction on the size of the deck.
I claim that your win probability with $n$ red and $m$ black cards is always $p_{n,m}=\frac{n}{n+m}$. This is certainly true when $n+m=1$, where you are obligated to say "red" immediately.
For larger deck sizes, you have two possible strategies. If you say "red" immediately, you will win with probability $\frac{n}{n+m}$. Alternately, you could wait one round, and then play from the resulting position. In this case you will win with probability $$ \frac{n}{n+m}p_{n-1,m}+\frac{m}{n+m}p_{n,m-1} $$
But by the induction hypothesis, this is equal to $$ \frac{n}{n+m}\frac{n-1}{n+m-1}+\frac{m}{n+m}\frac{n}{n+m-1}=\frac{n(n-1+m)}{(n+m)(n+m-1)}=\frac{n}{n+m} $$
So in fact, in either case you will win with probability $\frac{n}{n+m}$.
What we have to do is define hitting times.
A hitting time is a pair $(i,j)$ in which we should yell red when there are $i$ red and $j$ black remaining.
In order to determine hitting times efficiently we can do it recursively.
In order to do so we also calculate $f(i,j)$ recursively where $f(i,j)$ is the probability that we win when there are $i$ red and $j$ black balls if we follow the strategy perfectly.
Then we can calculate $f(i,j)$ as $\max( \frac{i}{i+j}, \frac{i}{i+j}f(i-1,j) + \frac{j}{i+j}f(i,j-1)$
The value $(i,j)$ is a hitting time precisely when the number on the left is larger.
One clearly has the values $f(0,j)=0$ and $f(i,0)=1$ for positive $i$ and $j$.
So we just have to run that recursion on the $27^2$ values and see what comes up (im doing this now)
Its pretty easy to prove that $f(i,j)=\frac{i}{i+j}$ by induction on $i+j$
This is simply because $\frac{i}{i+j}(\frac{i-1}{i+j-1}) + \frac{j}{i+j}(\frac{i}{i+j-1})= \frac{i^2 -i}{(i+j)(i+j-1)}+\frac{ij}{(i+j)(i+j-1)}=\frac{i(i+j-1)}{(i+j)(i+j-1)}=\frac{i}{i+j}$
So surprisingly it doesnt matter what you do, the probablity that the last ball comes out black is $\frac{i}{i+j}$ also !