Alice and Bob playing on a circle I want to solve this problem:

Let there be $n \ge 2$ points around a circle. Alice and Bob play a game on the circle. They take moves in turn with Alice beginning. At each move:

*

*Alice takes one point that has not been colored before and colors it red.


*Bob takes one point that has not been colored before and colors it blue.
When all $n$ points have been colored:

*

*Alice finds the maximum number of consecutive red points on the circle and call this $R$.


*Bob finds the maximum number of consecutive blue points on the circle and call this $B$.
If $R \gt B$, Alice wins. If $B \gt R$, Bob wins. If $R = B$, no one wins. Does any of the players have a winning strategy?

We still seem not to know for which odd $n$ Alice has a winning strategy. She does for $n=3$, and it seems also for $n=5$. But in general, I'm not sure. Could someone help?
 A: For even $n$, Bob can force a tie, as follows.  Divide the circle in half in any way, and then mirror Alice's moves on the other side.
For example, $1-2-3-4-5-6-7-8(-1)$ can be split into $1-2-3-4-$ and $5-6-7-8-$.  Then $1$ is mirrored to $5$, $2$ to $6$, $3$ to $7$, and $4$ to $8$.  Initially, all pairs are vacant; every time Alice moves into one element of a pair, Bob moves into the other.  If Alice moves into $1$, Bob moves into $5$, etc.  At the end of the game, the two haves will be opposite, so Bob's longest consecutive string is equal to Alice's.
In particular, this means that Alice cannot force a win. 
A: For even $n$, we can see that Alice can force a tie in a similar way to how @vadim123 shows that Bob can force a tie:
She divides the circle in half in any way (not necessarily the same way as Bob would).
She starts by picking any point (it does not matter which).
Then, for every move Bob makes, she looks at the opposite point, which is either


*

*Empty, in which case she should play that point.

*Not empty. By construction, it would be her point. She can again pick a random point.


By above's algorithm, Alice is guaranteed to have an exact mirror of Bob's points. Hence: a tie.
This shows that neither Alice nor Bob can have a winning strategy (for even $n$).
