The analyses
in two recent MO questions
("recent" with respect to the original posting in 2011),
"Rolling a random walk on a sphere"
and
"Maneuvering with limited moves on $S^2$,"
suggest a Rolling-Ball Game, as follows.
A unit-radius ball sits on a grid point of
a $\delta \times \delta$ regular grid in the plane,
with $\delta \neq \pi/2$.
Player 1 (Blue) rolls the ball to an adjacent
grid point, and the track of the ball-plane contact point
is drawn on the ball's surface.
Player 2 (Red) rolls to an adjacent grid point.
The two players alternate until each possible
next move would cause the trace-path to touch itself,
at which stage the player who last moved wins.
In the following example, Red wins, as Blue cannot
move without the path self-intersecting.
Q1.
What is the shortest possible game, assuming the players cooperate
to end it as quickly as possible?
For $\delta=\pi/4$, the above example suggests 6, but
this min depends on $\delta$. It seems smaller $\delta$ need 8 moves to create
a cul-de-sac?
Q2.
What is the longest possible game, assuming the players cooperate
to extend it as much as possible?
Q3.
Is there any reasonable strategy if the players are truly competing
(as opposed to cooperating)?
Addendum.
We must have $\delta < 2 \pi$ to have
even one legal move,
and the first player wins immediately with one move for $\pi \le \delta < 2\pi$ (left below).
The right image just shows a non-intersecting path of no particular significance for $\delta=\pi/8$.
Best Answer
The game is, indeed, finite though the reason I see is slightly different. I claim that if $L$ is a sufficiently long piece of a fixed shape and $R$ is a rotation sufficiently close to the identity, then $RL$ intersects $L$. This is a combination of three observations:
1) If the angle $a$ of rotation is small enough, then the "far" intersections are ruled out automatically, i.e., if $E'$ and $E''$ are two edges and $RE'$ intersects $RE''$, then $E'$ and $E''$ are either the same or adjacent. Also, if $L$ and $RL$ do not intersect, then topologically everything is trivial: $L$ just "shifts to one side".
2) There are very many corners on $L$ and each corner sweeps the area about $\delta a$ under the rotation $R$, so the area between $L$ and $RL$ is about $Ca$ with large $C$.
3) The angle deficiency at each endpoint (if we close the curve somehow) is at most $a$.
The ending is obvious: there are finitely many possible shapes of fixed large length, each piece can have not too many copies (otherwise there are two pieces differing by a small rotation), so the total length is bounded.