Optimal stopping in red vs black card game deck of 52 cards

Solution 1:

Let $v(r,b)$ be the expected value of the game for the player, assuming optimal play, if the remaining deck has $r$ red cards and $b$ black cards.

Then $v(r,b)$ satisfies the recursion $$ v(r,b) = \begin{cases} 0&\;\text{if}\;r=0\\[4pt] r&\;\text{if}\;b=0\;\text{and}\;r>0\\[4pt] \max(0,f(r,b))&\;\text{if}\;r,b>0\\[4pt] \end{cases} $$ where $$ f(r,b) = \left({\small{\frac{r}{r+b}}}\right)(1+v(r-1,b)) + \left({\small{\frac{b}{r+b}}}\right)(-1+v(r,b-1)) $$ The stopping rule is simple: Stop when $v(r,b)=0$.

To explain the recursion . . .

If $r,b>0$, and the player elects to play a card, then:

• The revealed card is red with probability ${\large{\frac{r}{r+b}}}$, and in that case, the player gets a score of $+1$, and the new value is $v(r-1,b)$.$\\[4pt]$
• The revealed card is black with probability ${\large{\frac{b}{r+b}}}$, and in that case, the player gets a score of $-1$, and the new value is $v(r,b-1)$.

Thus, if $r,b>0$, electing to play a card yields the value $f(r,b)$.

But the player always has the option to quit, hence, if $r,b>0$, we get $v(r,b) = \max(0,f(r,b))$.

Implementing the recursion in Maple, the value of the game is $$v(26,26) = \frac{41984711742427}{15997372030584}\approx 2.624475549$$ and the optimal stopping strategy is as follows . . .

• If $24 \le b \le 26$, play while $r \ge b - 5$.$\\[4pt]$
• If $17 \le b \le 23$, play while $r \ge b - 4$.$\\[4pt]$
• If $11 \le b \le 16$, play while $r \ge b - 3$.$\\[4pt]$
• If $6 \le b \le 10$, play while $r \ge b - 2$.$\\[4pt]$
• If $3 \le b \le 5$, play while $r \ge b - 1$.$\\[4pt]$
• If $1 \le b \le 2$, play while $r \ge b$.$\\[4pt]$
• If $b = 0$, play while $r > 0$.