I'd like to know which is the tree diagram of a chess tournament among 4 players. In particular, given 4 players A B C D, we know that there are the drawings of the players in order to create the couple of players. The 3 possibilities of drawing are (uniformly distributed so each of them has a probability of $1/3$):
$$ T_1: (A \, versus \, B) (C \, versus \, D) \\ T_2: (A \, versus \, C) (B \, versus \, D) \\ T_3: (A \, versus \, D) (B \, versus \, C) $$
We know also that A is stronger than B, B is stronger than C, and C is stronger than D; in a challenge between 2 players, the stronger player has the win probability of $2/3$, the other player of $1/3$.
The goal is to get the win probability of C. If I consider the first tournament $T_1$ (its drawing probability is $\color{red}{1/3}$), the win probability of C in the first challenge (C vs. D) is $\color{lightgreen}{2/3}$, instead the win probability of C in the second challenge (C vs. A or C vs. B) is $\color{orange}{1/3}$, so:
$$ P(C | T_1) = \color{red}{\frac{1}{3}} \color{lightgreen}{\frac{2}{3}} \color{orange}{\frac{1}{3}} $$
Tree diagram of the above partial situation ($T_1$ tournament): what is the tree diagram?
Assuming it is a knockout tournament.
From $T_1$ you need two branches, "$C$ wins" with probability $\frac{2}{3}$ and "$C$ loses" with probability $\frac{1}{3}$. From the end of the first of these you need another two branches, "$C$ wins" with probability $\frac{1}{3}$ and "$C$ loses" with probability $\frac{2}{3}$.
The probability of $C$ winning from $T_1$ is therefore $$\frac{2}{3}\times \frac{1}{3}=\frac{2}{9}.$$