In a normal deck of cards, you can either reveal the top card or guess whether that card is black. If you reveal the top card, you get to see what the card is and the game continues with one less card in the deck. If you were to make a guess, the game ends, and you get paid out $\$100$ if the card is black and $\$0$ if it's red.
What is the optimal strategy of this game and its expected value?
The lower bound has to be $0.5\times \$100 = \$50$ since you can just guess on the first card.
There really is no optimal strategy: If you start with an equal number of black and red cards, you have a 50:50 chance of winning (hence an expected value of $50) no matter what you do (as long as you pick some card if the remaining deck gets down to all black). You can prove it by induction.
Given a deck consisting of $b$ black cards and $r$ red cards, the probability of winning with an optimal strategy, $P(b,r)$, satisfies the recursion
$$P(b,r)=\max\left\{{b\over b+r},{b\over b+r}P(b-1,r)+{r\over b+r}P(b,r-1) \right\}$$
That is, you can either guess the next card is black, in which case you win with probability $b/(b+r)$, or you can pass, in which case you'll be left with either one fewer black card hence probability of winning $P(b-1,r)$ or one few fewer red card hence probability of winning $P(b,r-1)$, with probabilities $b/(b+r)$ and $r/(b+r)$ respectively; optimality amounts to making the choice that gives the higher probability, i.e., taking the max. In particular, if the two choices are equal, then it doesn't matter which choice you make. And that's what we want to prove:
$${b\over b+r}P(b-1,r)+{r\over b+r}P(b,r-1)={b\over b+r}$$
Now it's clear that $P(1,0)=1$ and $P(0,1)=0$. This is the base case. So if we assume, inductively, that $P(b-1,r)=(b-1)/(b-1+r)$ and $P(b,r-1)=b/(b+r-1)$, then we have
$$\begin{align} {b\over b+r}P(b-1,r)+{r\over b+r}P(b,r-1) &={b(b-1)\over(b+r)(b-1+r)}+{rb\over(b+r)(b+r-1)}\\ &={b(b-1+r)\over(b+r)(b-1+r)}\\ &={b\over b+r} \end{align}$$