card-gamesexpectation
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.
As Ross rightly points out, the cuts are just to seemingly make it random though it doesn't affect anything. Irrespective of the cuts, note that there are $15$ cards between the first card the contestant places and the second card the contestant places. Similarly, irrespective of the cuts, note that there are $15$ cards between the second card the contestant places and the third card the contestant places.
Let us label the cards as follows. Let $a_k^{j}$ be the $k^{th}$ card from top in the hand of the performer after the $j^{th}$ up-down phase. Initially, i.e. after the contestant places the cards and before the first up-down starts, $j=0$.
Now the cards in the last pile i.e. the pile containing $9$ cards (the only pile on which the contestant doesn't place any card) be $a_1^{0}, a_2^{0}, \ldots, a_9^{0}$ starting from the top most card. Let the third card the contestant places on the third pile be $a_{10}^{0}$. Then there are $15$ cards followed by the second card the contestant places on the second pile. Accounting for the $15$ cards in between, the second card is $a_{26}^{0}$. Now there are $15$ cards followed by the first card the contestant places on the first pile. Accounting for the $15$ cards in between, the first card is $a_{42}^{0}$. Hence, now the contestant cards are $\color{red}{a_{10}^{0}, a_{26}^{0}, a_{42}^{0}}$.
Now the performer moves $4$ cards to the rear. Hence, now the contestant cards are $a_6^{0}, a_{22}^{0}$ and $a_{38}^{0}$.
$$\color{red}{\{a_{10}^{0}, a_{26}^{0}, a_{42}^{0}\} \to \{a_{6}^{0}, a_{22}^{0}, a_{38}^{0}\}}$$
Now in the first up-down phase all the odd number cards are eliminated i.e. $a_{2k-1}^{0}$ gets eliminated. However, on the pile with the cards closed, the order has reversed i.e. $a_2^{0}$ is the bottom most card, followed by $a_4^{0}$ and so on and the top-most card is $a_{52}^{0}$. Now reordering the card so that the topmost card is now $a_1^{1}$, we find that the card $a_{2k}^{0}$ gets mapped to $a_{27-k}^{1}$. Hence, the contestant cards are now at $a_{24}^{1}, a_{16}^{1}$ and $a_8^{1}$. $$\color{red}{\{a_{6}^{0}, a_{22}^{0}, a_{38}^{0}\} \to \{a_{24}^{1}, a_{16}^{1}, a_8^{1}\}}$$ There are now $26$ cards left.
Now in the second up-down phase all the odd number cards are eliminated i.e. $a_{2k-1}^{1}$ gets eliminated. As before, on the pile with the cards closed, the order has reversed i.e. $a_2^{1}$ is the bottom most card, followed by $a_4^{1}$ and so on and the top-most card is $a_{26}^{1}$. Now reordering the card so that the topmost card is now $a_1^{2}$, we find that the card $a_{2k}^{1}$ gets mapped to $a_{14-k}^{2}$. Hence, the contestant cards are now at $a_{2}^{2}, a_{6}^{2}$ and $a_{10}^{2}$. $$\color{red}{\{a_{24}^{1}, a_{16}^{1}, a_8^{1}\} \to \{a_{2}^{2}, a_{6}^{2}, a_{10}^{2}\}}$$
There are now $13$ cards left.
Now in the third up-down phase all the odd number cards are eliminated i.e. $a_{2k-1}^{2}$ gets eliminated. As before, on the pile with the cards closed, the order has reversed i.e. $a_2^{2}$ is the bottom most card, followed by $a_4^{2}$ and so on and the top-most card is $a_{6}^{2}$. Now reordering the card so that the topmost card is now $a_1^{3}$, we find that the card $a_{2k}^{2}$ gets mapped to $a_{7-k}^{3}$. Hence, the contestant cards are now at $a_{6}^{3}, a_{4}^{3}$ and $a_{2}^{3}$. $$\color{red}{\{a_{2}^{2}, a_{6}^{2}, a_{10}^{2}\} \to \{a_6^3,a_4^3, a_2^3\}}$$ There are now $6$ cards left.
Hence, the last up-down has the open cards as $a_1^{3}$, $a_3^{3}$ and $a_5^{3}$; the closed cards being $a_2^{3}$, $a_4^{3}$ and $a_6^{3}$, which are precisely the contestant cards.
EDIT
Below is an attempt to explain this pictorially. The document was created using $\LaTeX$ and below is a screenshot.
Let $v(r,b)$ be the expected value of the game for the player, assuming optimal play, if the remaining deck has $r$ red cards and $b$ black cards.
Then $v(r,b)$ satisfies the recursion $$ v(r,b) = \begin{cases} 0&\;\text{if}\;r=0\\[4pt] r&\;\text{if}\;b=0\;\text{and}\;r>0\\[4pt] \max(0,f(r,b))&\;\text{if}\;r,b>0\\[4pt] \end{cases} $$ where $$ f(r,b) = \left({\small{\frac{r}{r+b}}}\right)(1+v(r-1,b)) + \left({\small{\frac{b}{r+b}}}\right)(-1+v(r,b-1)) $$ The stopping rule is simple: Stop when $v(r,b)=0$.
To explain the recursion . . .
If $r,b>0$, and the player elects to play a card, then:
Thus, if $r,b>0$, electing to play a card yields the value $f(r,b)$.
But the player always has the option to quit, hence, if $r,b>0$, we get $v(r,b) = \max(0,f(r,b))$.
Implementing the recursion in Maple, the value of the game is $$v(26,26) = \frac{41984711742427}{15997372030584}\approx 2.624475549$$ and the optimal stopping strategy is as follows . . .
Best Answer
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.