In this paper, at page 302, the author calculates the expected score of a gamer who plays the game that a player guesses the colour of the next card, red of black, in a normal 52 deck, if it gets right, it scores 1, and not it gets 0.
The author defines a "game path" in n-n lattice such that one each vertex specifies the number of red/black card in the remaining deck. Then he argues that between $(n,n)$ and $(n-k, n-k)$, the expected score of the gamer is $k + 0,5$ provided that the contribution of the first card to the expected score is $0,5$.
But I couldn't understand how he did reasoned that the expected score is $k + 0,5$ between $(n,n)$ and $(n-k, n-k)$ provided that in this "interval" he always guesses the colour which is not the colour of the first card.
Any help is appreciated.
A paradoxical aspect to counting cards in "Red or Black" is that the expected score grows proportionally with the number of times that card counting gives no information. Those situations occur on the diagonal: red = black. Suppose that we started at $(n, n)$ and that the first card in the deck is black. We can assume that this first card contributes $1/2$ to our expected score. Now suppose that our game path meets the diagonal next at $(n - k,n - k)$. In the course of traversing this path we would continue to guess that the next card is red since red $>$ black in that portion of the path; therefore, we will be correct exactly $k$ times and incorrect $k - 1 $ times during this part of the game. In the path from $(n, n)$ to $(n - k, n - k)$, our expected score is then $k + 0.5$ of a possible $2k$.
Between two positions on the diagonal, one colour will continuoiusly be more likely than the other (namely the colour that is not the colour of the first card). So obviously, the best strategy to use suring that time is to always guess the more likely colour. On the way from $(n-1,n)$ or $(n-n-1)$ to $(n-k,n-k)$, this guess will be correct exactly $k$ times. Add the $\frac12$ from the first guess, and we have our $k+\frac12$.