Calling the colour of each card

Idea. Calling the colour with more cards left maximises the chance of each call, and since calls do not change which cards come next, it also maximises the total. Then compute the value by working backwards over the cards left.

Step 1: the recursion. With $r$ red and $b$ black cards left, you call the majority colour and are right with probability $\frac{\max (r,b)}{r+b}$. Then $$\begin{array}{rl}V(r,b)=\frac{\max (r,b)}{r+b}& +\frac{r}{r+b}V(r-1,b)\\ & +\frac{b}{r+b}V(r,b-1),\end{array}$$ with $V(r,0)=r$ and $V(0,b)=b$: once only one colour is left, every call is right.

Step 2: small decks.

• $V(1,1)=\frac{1}{2}+\frac{1}{2}\cdot 1+\frac{1}{2}\cdot 1=\frac{3}{2}$.
• $V(2,1)=\frac{2}{3}+\frac{2}{3}\cdot V(1,1)+\frac{1}{3}\cdot V(2,0)=\frac{2}{3}+1+\frac{2}{3}=\frac{7}{3}$, and $V(1,2)=\frac{7}{3}$ by symmetry.

Step 3: the full deck. $$\begin{array}{rl}V(2,2)& =\frac{1}{2}+\frac{1}{2}\cdot \frac{7}{3}+\frac{1}{2}\cdot \frac{7}{3}\\ & =\frac{1}{2}+\frac{7}{3}=\frac{17}{6}\approx \mathrm{2.83.}\end{array}$$

Compare. Calling at random wins 2 dollars on average. Memory adds $\frac{5}{6}$ of a dollar here. With a full deck of 26 red and 26 black cards, the same strategy is worth about 30.04, compared with 26 for random calls.