The two best players in the final

Idea. The strongest player wins every match, and the second strongest loses only to the strongest. So they meet in the final unless they are paired earlier. With random pairings, this is the same as placing the 64 players in a random bracket and asking whether the two strongest land in different halves.

Step 1: a random bracket. Random pairings in each round give the same distribution of matches as a bracket filled in at random at the start, because the stronger player always wins and the pairings are uniform each round.

Step 2: the halves. Put the strongest player somewhere. The second strongest is equally likely to be in any of the other 63 slots, and 32 of them are in the other half of the bracket.

Step 3. $$\mathbb{P}(\text{they meet in the final})=\frac{32}{63}\approx \mathrm{0.508.}$$

In general. With $2^k$ players the answer is $\frac{2^{k-1}}{2^k-1}$, always a little above $\frac{1}{2}$.

Round by round. The same answer comes from the rounds: the two avoid each other in the first round with probability $\frac{62}{63}$, in the second with $\frac{30}{31}$, and so on: $\frac{62}{63}\cdot \frac{30}{31}\cdot \frac{14}{15}\cdot \frac{6}{7}\cdot \frac{2}{3}=\frac{32}{63}$.