guess the color of the next card, what is the payoff of this game? There are 26 red cards and 26 black cards on the table which are randomly shuffled and are facing down onto the table. The host turns up the cards one at a time.  You can stop the game any time (even at the beginning of the game). Once you stops the game, the next card is turned up: if it is red, you get $1; otherwise you pay the host one dollar. What is the payoff of this game?

[Thanks to Brian Scott for pointing out my original misinterpretation of this problem.]
The value of the game is zero.
Let $f(r,b)$ denote the value of the game if there are $r$ red cards and $b$ black cards remaining.  If we stop at that point, the expected value is $(r-b)/(r+b)$; but as we also have the option of continuing, we have the recurrence
$$f(r,b)=\max\left((r-b)/(r+b),\frac{r}{r+b}f(r-1,b)+\frac{b}{r+b}f(r,b-1)\right).$$
The boundary conditions are $f(0,0)=0$, $f(1,0)=1$, $f(0,1)=-1$, and $f(r,b)=0$ if either $r$ or $b$ is negative.  Calculating individual values of $f(r,b)$ is easy using dynamic programming.  Here are values of $f(r,b)$ for small $r$ and $b$:
$$
\begin{array}{c|cccccc}
 r\,\backslash b & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline
 0 & 0 & -1 & -1 & -1 & -1 & -1 \\
 1 & 1 & 0 & -\frac{1}{3} & -\frac{1}{2} & -\frac{3}{5} & -\frac{2}{3} \\
 2 & 1 & \frac{1}{3} & 0 & -\frac{1}{5} & -\frac{1}{3} & -\frac{3}{7} \\
 3 & 1 & \frac{1}{2} & \frac{1}{5} & 0 & -\frac{1}{7} & -\frac{1}{4} \\
 4 & 1 & \frac{3}{5} & \frac{1}{3} & \frac{1}{7} & 0 & -\frac{1}{9} \\
 5 & 1 & \frac{2}{3} & \frac{3}{7} & \frac{1}{4} & \frac{1}{9} & 0 \\
\end{array}
$$
Apparently $f(r,b)=-f(b,r)$; this is easy to prove from the recurrence. Accordingly $f(26,26)=0$.
[Actually it follows from the recurrence (and is evident from the table] that $f(r,b)=(r-b)/(r+b)$ -- in other words, it is never optimal to continue playing.]
