In the book of Shiryaev, Probability, it is given in page $88-89$, while explaning the probability of ruin of random walk which described by the sum of independent Bernoulli trials, the following is given:
\begin{array}{l}{\text { Next suppose that the rules of the game are changed: the original bankrolls }} \\ {\text { of the players are still }(-A) \text { and } B \text { , but the payoff for each player is now } \frac{1}{2}}\end{array} \begin{array}{l}{\text { rather than } 1 \text { as before. In other words, now let } P\left(\xi_{i}=\frac{1}{2}\right)=p, P\left(\xi_{i}=-\frac{1}{2}\right)=} \\ {q \text { . In this case let us denote the limiting probability of ruin for the first player }} \\ {\text { by } \alpha_{1 / 2} \text { . Then }} \\ {\qquad \alpha_{1 / 2}=\frac{(q / p)^{28}-1}{(q / p)^{2}-(q / p)^{2} A}} \\ {\text { and therefore }} \\ {\qquad \alpha_{1 / 2}=\alpha \cdot \frac{(q / p)^{s}+1}{(q / p)^{8}+(q / p)^{4}}<\alpha} \\ {\text { if } q>p} \\ {\text { Hence we can draw the following conclusion. if the game is urfacable }} \\ { \text { to the first player (i.e., }q>p) \text { then doubling the stake decreases the probability }} \\ {\text { of ruin. }}\end{array}
However, if $\xi_i = 1/2$, then at each step, a player either gains or loses 1/2; meaning that the stake is halved, not doubled, and we obtain that $\alpha_{1/2} < \alpha_1 = \alpha$, so this contradicts with what the author is stating in the end. I am confused at this point.