Probability brainteaser Normal 52 card deck. Cards are dealt one-by-one. You get to say when to stop. After you say stop you win a dollar if the next card is red, lose a dollar if the next is black. Assuming you use the optimal stopping strategy, how much would you be willing to pay to play? Proof?
If there are $r$ red cards on a total of $n$ cards let $p(r,n)$ be the probability to win with the optimal strategy. 
If you stop immediately you win with probability $r/n$.
If you don't stop you can get a red card with probability $r/n$ and then you win with probability $p(r-1,n-1)$. If instead you get a black card you win with probability $p(r,n-1)$. Hence with the best strategy you win with probability
$$
p(r,n) = \max\left\{\frac r n, \frac r n p(r-1,n-1) + \frac{n-r} n p(r,n-1)\right\}.
$$
Let's prove by induction that $p(r,n)=r/n$. Suppose it is true for $n-1$ that $p(r,n-1)=r/(n-1)$ then 
$$
p(r,n) = \max\left\{\frac r n, \frac r n \frac{r-1}{n-1}+\frac{n-r}n\frac{r}{n-1} \right\}
= \max\left\{\frac r n, \frac{nr-r}{n(n-1)}\right\} 
= \frac r n.
$$
In the case $n=1$ you immediately find that $p(0,1)=0$ and $p(1,1)=1$.

So at anytime you stop you always get the same probability to win. If you start with the same number of red and black cards, you win with probability $1/2$ hence you should pay nothing to play.
You should pay 0.5 because you can decide just before the last card is dealt. The probability that the last card is red is 1/2 and you will win 1; otherwise you don't say stop. Thus, the expectation of this game should be 0.5