Answer: the first player has a winning strategy.

Assume there are $n$ points. The first to play so that $n-2$ points each have at least one segment loses, because the other player simply joins the last two points and the game ends. But there are $\mathrm{N}=(\mathrm{n}-3)(\mathrm{n}-4) / 2$ possible plays amongst the first $\mathrm{n}-3$ points to get a segment. For $\mathrm{n}=1$ or $2 \bmod 4, \mathrm{~N}$ is odd and for $\mathrm{n}=0$ or 3 it is even. So the first player wins for $\mathrm{n}=1$ or $2 \bmod 4$ (and in particular for $\mathrm{n}=98$ ) and the second player for $\mathrm{n}=0$ or $3 \bmod 4$.