I have attempted this question from Feller's Introduction to Probability Theory and Its Applications (chapter 5, question 14, page 141), but I keep coming up with a slightly different answer. The question is below but the context is that you have players a, b, and c. Two play, one sits out and then winner stays on. The tournament ends when a player gets two consecutive wins. The probability of winning a game is $\frac{1}{2}$ for all:
In example (2.a) let $x_n$ be the conditional probability that the winner of the $n^{th}$ trial wins the entire game given that the game does not terminate at the $n^{th}$ trial; let $y_n$ and $z_n$ be the corresponding probabilities of victory for the losing and pausing player, respectively, of the $n^{th}$ trial. (a) Show that: $x_n = \frac{1}{2} + \frac{1}{2}y_{n+1}$; $y_n = \frac{1}{2}z_{n+1}$; $z_n = \frac{1}{2}x_{n+1}$
I can get the first two but I am struggling with the third. I had that it should be $z_n = \frac{1}{2}x_{n+1} + \frac{1}{2}y_{n+1}$ because the person sitting out can either have win-win (game ends), win-lose (win later) or lose (win later). It seems like $\frac{1}{2}x_{n+1}$ term only covers the WW and WL scenarios so we would need the y term.
What have I misunderstood here?
Thanks