The Dynamic Card Game

A standard deck of $52$ cards, $26$ red and $26$ black, is shuffled and dealt face up one at a time. Each red card drawn pays you one dollar and each black card costs you one dollar, and you may tell the dealer to stop at any moment. What is the optimal stopping rule, and how much is the game worth to you?

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#The value recursion

Suppose $r$ red and $b$ black cards remain. Whatever you have banked so far is settled, so let $f(r,b)$ be the expected additional profit under optimal play from this point. You may stop, for nothing more, or draw the next card. It is red with probability $\tfrac{r}{r+b}$ , paying $1$ and leaving $f(r-1,b)$ , or black with probability $\tfrac{b}{r+b}$ , costing $1$ and leaving $f(r,b-1)$ . So

f(r,b) = \max\!\left(0,\ \frac{r}{r+b}\big(1 + f(r-1,b)\big) + \frac{b}{r+b}\big(-1 + f(r,b-1)\big)\right), \tag{1}

with $f(0,b) = 0$ and $f(r,0) = r$ .

#Stop when the edge is gone

Keep drawing while the bracket is positive and stop the instant it reaches $0$ . Draining all remaining cards nets exactly $r - b$ , so once black outnumbers the red still to come the deck becomes a drag, but quitting at the right surplus still banks a profit.

Draining the whole deck nets zero, yet the freedom to stop at a favourable surplus has value. With k cards of each colour left it grows with the deck, reaching about 2.62 at a full 26 and 26.

#The value

f(26,26) \approx 2.62. \tag{2}

The full deck holds equal red and black, so playing stubbornly to the last card nets exactly zero. The entire worth of the game, about $2.62, is the value of the option to walk away once you are ahead.