Expected Payoff of the Red-Black Card Game

Worked Solution

How to Think About It: You have 50 red and 50 black cards. You draw one at a time and can stop whenever you want -- your payoff is (fraction of reds drawn) times $\$2$. The tempting argument is: "the fraction of reds is always $1/2$ in expectation, so you can't beat $\$1$." That argument is wrong. It confuses $E[R_k/k]$ for a fixed draw count (which is indeed $1/2$) with $E[R_\tau/\tau]$ for an adaptive stopping time $\tau$. The optional stopping theorem would only rule out a gain if $R_k/k$ were a martingale, and it is not: after a red first card ($R_1/1 = 1$) the expected fraction after the next draw is $(1 + 49/99)/2 \approx 0.75$, not $1$. So you CAN beat $\$1$ by stopping at favorable moments.

Quick Estimate: Try the simplest smart strategy. Draw one card. If it is red, stop -- your fraction is $1/1 = 1$, payoff $= \$2$. If it is black, you are stuck with fraction $0/1$, so just draw all remaining 99 cards and end at $50/100$, payoff $= \$1$. Each case happens with probability $1/2$, so this gives:

$$E[\text{payoff}] = \frac{1}{2}(\$2) + \frac{1}{2}(\$1) = \$1.50$$

Already beats $\$1$. Can we do better? Yes: the plan above gives up too early after a black first card. At that point you hold $0/1$, but the deck has $50$ reds and only $49$ blacks left, and the running fraction wanders as you draw. Any later moment where it sits above $1/2$ (say $2$ reds out of $3$, or $6$ out of $10$) is worth stopping at, and you can still fall back on drawing everything for $\$1$. Exploiting those moments lifts the value after a black first card from $\$1$ to about $\$1.17$, so the total is about $\tfrac12(\$2) + \tfrac12(\$1.17) \approx \$1.58$.

Approach: Backward induction over the state space $(r, b)$ = (reds remaining, blacks remaining). At each state, compare the payoff from stopping to the expected value of drawing one more card.

Formal Solution:

Let $V(r, b)$ be the game value when $r$ reds and $b$ blacks remain in the deck. Cards drawn so far: $k = 100 - r - b$. Reds drawn: $j = 50 - r$.

If you stop, you collect $2j/k$. If you draw, the next card is red with probability $r/(r+b)$:

$$V(r,b) = \max\!\left(\frac{2(50 - r)}{100 - r - b},\;\frac{r}{r+b}\,V(r-1,b) + \frac{b}{r+b}\,V(r,b-1)\right)$$

Boundary: $V(0,0) = 2 \cdot 50/100 = 1$ (drew everything). At the start ($k = 0$, no cards drawn), the payoff $0/0$ is undefined so we must draw:

$$V(50,50) = \tfrac{1}{2}\,V(49,50) + \tfrac{1}{2}\,V(50,49)$$

Running this DP over the $51 \times 51 = 2{,}601$ states gives:

$$V(50,50) \approx \$1.5849$$

The optimal policy is a state-dependent threshold: stop when the red fraction $j/k$ exceeds a cutoff $\theta(k)$ that depends on how many cards have been drawn. Early on ($k = 1, 2$), you need fraction $= 1$ (all reds) to stop. As $k$ grows, the threshold falls steadily toward $1/2$ (about $0.60$ by $k = 20$ and about $0.52$ by $k = 80$) because there is less optionality left with fewer remaining cards.

Answer: The expected payoff under optimal play is approximately $\$1.5849$. The naive claim of $\$1$ is wrong -- $R_k/k$ is not a martingale, so the optional stopping theorem does not apply. Even the dead-simple "stop if the first card is red" strategy already yields $\$1.50$.