$A, B, C, D$ are four teams in a tournament where $A$ faces $B$ and $C$ faces $D$ in the first round and the winners of those rounds face each other. $A$ has a $0.7$ probability of beating any other team, while $B, C, D$ have a $0.5$ probability of beating each other. If the games are independent, what is the probability that each team wins the tournament?
My idea is $P(\text{$A$ wins})=P[A \text{ beats } B $ and ($A \text{ beats } C$ or $A \text{ beats D}$)$]$ $=P[(A \text{ beats B and A beats C) or (A beats B and A beats D)}]$ $=0.7*0.7 + 0.7*0.7 =98$%, and repeating this process for each $B,C,D$ but this seems incorrect.
Is there an easier way to go about this?
Since $A$ has probability $0.7$ of beating all teams, it doesn't matter who wins in the $C$ vs. $D$ match. We have $$P(A\text{ wins})= P(A\text{ wins twice})=0.7\cdot0.7=0.49$$ Similarly, since $B$ has probability $0.5$ of beating both $C$ and $D$, we can write: $$P(B\text{ wins}) = P(B\text{ beats }A)P(B\text{ wins next game}) = 0.3 \cdot 0.5 = 0.15$$ By symmetry, the probability that $C$ wins is the same as the probability that $D$ wins. We have: \begin{align} P(C\text{ wins}) &= P(C\text{ beats }D)\left[P(A\text{ beats }B)P(C\text{ beats }A) + P(B\text{ beats }A)P(C\text{ beats }B)\right]\\ &=0.5(0.7\cdot0.3+0.3\cdot0.5) \\ &=0.5\cdot 0.36 \\ &= 0.18\\ &=P(D\text{ wins}) \end{align} An easier way to get $P(C\text{ wins})$ is to divide the remaining probability $1 - (0.49 + 0.15) = 0.36$ by $2$, since $C$ and $D$ have the same chance of winning the tournament.