In this question Alice and Bob play a game on $K_{2014}$, Alice directing one edge, Bob directing $1$ to $1000$ edges with Alice trying to make a cycle. The proof that Alice can win depended on the fact that Bob's maximum number of edges was less than half the number of vertices. I believe that in fact if Bob's maximum number of edges on $K_n$ is $n-3$ or less Alice can win. Bob can win with $n-2$-after Alice's first move, he directs all the other edges leaving from the root of Alice's edge outward and continues in that vein. I think the proof that Alice can win will depend on showing Bob really doesn't want to play edges until he needs to, and then at the end he can't play quite enough.
For a more tractable example, take $K_8$ and allow Bob $5$ moves. As there are only $28$ edges, in theory the game could be done in five moves. But let Alice move $1\to 2$ If Bob plays five more edges leaving $1$, Alice plays the other edge into $1$ and wins. If Bob plays $3 \to 4$, Alice can play $2 \to 3$ and has a chain of three. Bob has to use three of his moves $1 \to 3, 1\to 4, 2 \to 4$ and doesn't want to play any more. Now Alice will win by $4 \to 5, 5 \to 6, 6 \to 7$ unless Bob leaves something open earlier. Bob's best first move is probably just $1 \to 3$, but even then Alice can take advantage by playing $4 \to 1$ and so on. It isn't a proof, but the more I play with it the more convinced I become.
Can we prove that Alice wins on $K_8$ with Bob having at most $5$ moves? That is easily accessible to exhaustive search. Can we prove it for $K_n$ with Bob having $n-3$ moves?