Players ranked first and second meeting in the final

A knockout chess tournament has $2^n$ players with distinct strengths, and the stronger player always wins a game. The draw is completely random. What is the probability that the two strongest players face each other in the final round?

$2^{n-1}/(2^n-1)$. The top two can only lose to each other, so they meet in the final if and only if they are drawn into different halves. Fixing the top player's position, the second player lands in the other half in $2^{n-1}$ of the $2^n-1$ remaining slots.