When do players #1 and #2 meet only in the final?

Single-elimination tournament with 2n players, random pairings each round, and the better player always wins.
Tournament size
16 players
Choose n so total players = 2n

P = 2n-1 / (2n - 1)
Half containing player #1
If player #2 is placed anywhere in this half, then #1 and #2 must meet before the final.
Opposite half
If player #2 is placed anywhere in this half, then #1 and #2 can only meet in the final.

Main idea

Fix player #1 somewhere in the bracket. There are 15 remaining positions for player #2.

Exactly 8 of those are in the opposite half of the bracket. Those are the only positions that make them meet only in the final.

Probability = good positions / all possible positions = 8 / 15

Probability that #1 and #2 meet only in the final
8 / 15
53.33%

Round-by-round view

Since player #1 beats everyone, #1 always reaches the final. Player #2 loses only to #1, so #2 also reaches the final iff #2 avoids #1 in every earlier round.

That happens exactly when #2 starts in the opposite half of the tournament tree.