Second seed surviving ten rounds of winner-stays

Fifty players numbered $1$–$50$ take turns at one table under winner-stays rules: the higher number always wins a match, the first match is between two randomly chosen players, and each later match pits the reigning winner against a random player who has not yet played. The loser of each match leaves; $49$ matches are played in all. What is the probability that player $49$ has not been eliminated once $10$ matches are complete?

$234/245$. Entry order is a random permutation, and $10$ matches involve the first $11$ entrants. Player $49$ survives if it is not among them ($\tfrac{39}{50}$), or if it is but player $50$ is not ($\tfrac{11}{50}\cdot\tfrac{39}{49}$). The total is $\tfrac{1911+429}{2450}=\tfrac{234}{245}$.