Frog on a ring of 100 pads reaching the far pad last

A frog sits on pad $0$ of a ring of $100$ lily pads labelled $0$ to $99$ in order around the circle. Each hop takes it to one of the two neighbouring pads, each with probability $\tfrac12$. What is the probability that by the time it first lands on pad $50$, it has already visited all $98$ other pads?

$\tfrac{1}{99}$. Pad $50$ is last exactly when, after first reaching one of its neighbours (say $49$), the frog goes all the way round to $51$, which is $98$ steps away, before stepping onto $50$, which is $1$ step away. Gambler's ruin gives $\tfrac{1}{1+98}=\tfrac1{99}$. (On a cycle, every pad other than the start is equally likely to be the last one visited.)