How much would you be willing to pay to play this card game? 
Consider a normal $52$-card deck. Cards are dealt one-by-one. You get to say when to stop. After you say "stop" you win $\$1$ if the next card is red, lose $\$1$ if the next card is black.  Assuming you use the optimal stopping strategy, how much would you be willing to pay to play?

Is there an optimal strategy? I found someone analysing it using this table. I can understand it, but I am still confused about how much I would be willing to pay to play.
 A: The payment/stopping rule in the link is quite different from the rule presented in the question here.  I'm going to answer the question here.
It turns out there is no strategy that will improve your odds of winning a game in which you are allowed to end the game by guessing that the next card is red; the expected value is $0$, so you shouldn't be willing to pay anything to play.  (I am assuming that the cards have been shuffled, so that no one knows the status of any card until it is revealed.)
Here's a way to see that the expected value is $0$.  Imagine the cards are spread out, face down, from left to right.  Start by placing your finger on the rightmost card, to indicate that you tentatively intend to stop at the very end.  Obviously, this card has a $50\%$ chance of being red.  Now, before each card, starting from the left, is turned over, you are allowed to change your mind and stop with it instead.  Whether you do or don't doesn't matter:  the rightmost card and the current still-face-down leftmost card have equal probability of being red, so you may as well stick with your initial tentative decision.  Although your current assessment of how much you stand to gain (or lose) will change each time a leftmost card is revealed, that has no bearing on whether to stop.  In other words, even though it may feel as is you have some control over your fate, you really don't.
Remark:  This answer is similar to an answer I gave to a "number battle" problem.
A: Problem:  A friend turns over the cards of a well shuffled deck one at a time. You can stop anytime you choose and bet that the next card is red. What is the best strategy?
Solution: 
For $n=0,1,\dots, 51$, let $R_n$ be the number of red cards left after $n$ cards have
been turned over. Then
$$R_{n+1}=\cases{R_n&with probability $1-p$\\  R_n-1&with probability $p$},$$
where $p=R_n/(52-n)$, the proportion of reds left. Taking expectations we get
$$\mathbb{E}(R_{n+1}\,|\, {\cal F}_n)=\mathbb{E}(R_{n+1}\,|\, R_n)=R_n-{R_n\over 52-n}=R_n\left({52-(n+1)\over 52-n}\right)$$
so that
$$\mathbb{E}\left({R_{n+1}\over 52-(n+1)}\,\Bigl|\, {\cal F}_n\right)={R_n\over 52-n}.$$ Here ${\cal F}_n:=\sigma(R_0,R_1,\dots, R_n).$
This means that $M_n:=R_n/(52-n)$ is a martingale.
Now let $T$ represent your stopping strategy.
Then $\mathbb{P}(T \mbox{ is successful}\,|\, {\cal F}_T)=M_T$, and by the optional sampling theorem,
$$\mathbb{P}(T\mbox{ is successful})=\mathbb{E}(M_T)=\mathbb{E}(M_0)=1/2.$$
Every strategy has a 50% chance of success!
