Suppose that a dealer possesses a finite group $G$ of order $n$ with its elements enumerated $g_{1}, g_{2}, \dots, g_{n}$ and presents you with an empty Cayley table and informs you of the order of the group and that $g_{1}$ is the identity element.
He then ''deals'' you $k$ entries of the Cayley table (something resembling the example below) and you win if you can fill out the entire table. Is it known whether there exists some $m \in \mathbb{N}$ such that if $k \geq m$, then you can win the game for the particular game on $G$ or, conversely, if $k \leq m$ then you are guaranteed to lose on $G$? So, for an easy example, if you are dealt the following:
$$ \begin{array}{c|c c c c c} & g_{1} & g_{2} & g_{3} & g_{4} & g_{5} \\ \hline g_{1} & g_{1} & & & & \\ g_{2} & & g_{3} & & & g_{1}\\ g_{3} & & g_{4} & g_{5} &g_{1} &\\ g_{4} & & g_{5}& &g_{2} &g_{3} \\ g_{5} & & & & &\\ \end{array} $$ then you can win the game (in fact this is one of the first exercises about groups in Gallian's Contemporary Abstract Algebra).
Moreover, what if the dealer doesn't inform you of the identity element? (Note: I posted a similar question yesterday, but it was posed very poorly and I ended up deleting it.)