Drawings on a circle game winner? Alice and Bob play ChordBash, which is played as follows:

*

*There is a circle with $n > 4$ points on it.

*A move consists of drawing a chord between two points.

*Every move after the first must be a chord that intersects every other chord (common endpoints count as intersections).

*The game stops once there are no possible legal moves left for a player, the last player to draw a chord wins.

For example, if we have the points 1 to 6 around a circle, in that order, and Alice choses 1-3, Bob can choose 1-2, 1-4, 1-5, 1-6, 2-3, 2-4, 2-5, 2-6, 3-4, 3-5, or 3-6. Assuming Alice goes first and that they alternate making moves, for which $n$ is Alice guaranteed win?
 A: If n is odd, say labeled $1,2,3,\dots, 2k+1$ around the circle, then Alice first connects $1$ and $2k+1$. Bob's move must connect either $1$ or $2k+1$ to another point, say $i$. Alice then proceeds to close the triangle, $1, i, 2k + 1$. At this point, any chord must cross all three sides of the triangle, which is only possible if the remaining points $< i$ are connected to $2k+1$ and the remaining points $> i$ are connected to $1$. Since $n$ is odd, there are an even number of moves remaining and it's Bob's turn, so Alice wins.
If n is even, then label the points $1,2, \dots, 2k$ around the circle so that Alice's first move connects $1$ to $i \neq 2k$. Bob then connects $1$ to $2k$. No matter where Alice goes, it's impossible to stop Bob from closing a $1, j, 2k$ triangle, at which point we are back in the same triangle situation, except it's Alice's turn with an even number of moves remaining, so Bob wins.
Edited: Unless I've missed something in the even case:
Alice connects $1$ to $i \neq 2k$, Bob connects $1$ to $2k$. Then Alice can make one of four types of move, each of which ends with the triangle case, Alice's turn, and an even $2k-4$ legal moves remaining:

*

*Connect $i$ to $2k$, after which Bob is free to make any legal move.

*Connect a chord containing a point $> i$, say $j$. Then it must be the chord from $1$ to $j$, since it must cross both $1,2k$ and $1,i$. Bob connects $i$ to $2k$.

*Connect a chord containing a point $< i$, say $j$, to $2k$. Bob connects $i$ to $2k$.

*Connect a chord containing a point $< i$, say $j$, to $1$. Then Bob connects $j$ to $2k$.

