Abstract ${\rm F{\small IXED-}F{\small LOOD-}I{\small T}}$ and ${\rm F{\small REE-}F{\small LOOD-}I{\small T}}$ are combinatorial problems on graphs that generalize a very popular puzzle called Flood-It. Both problems consist of recoloring moves whose goal is to produce a monochromatic ("flooded") graph as quickly as possible. Their difference is that in ${\rm F{\small REE-}F{\small LOOD-}I{\small T}}$ the player has the additional freedom of choosing the vertex to play in each move. In this paper, we investigate how this freedom affects the complexity of the problem. It turns out that the freedom is bad in some sense. We show that some cases trivially solvable for ${\rm F{\small IXED-}F{\small LOOD-}I{\small T}}$ become intractable for ${\rm F{\small REE-}F{\small LOOD-}I{\small T}}$. We also show that some tractable cases for ${\rm F{\small IXED-}F{\small LOOD-}I{\small T}}$ are still tractable for ${\rm F{\small REE-}F{\small LOOD-}I{\small T}}$ but need considerably more involved arguments. We finally present some combinatorial properties connecting or separating the two problems. In particular, we show that the length of an optimal solution for ${\rm F{\small IXED-}F{\small LOOD-}I{\small T}}$ is always at most twice that of ${\rm F{\small REE-}F{\small LOOD-}I{\small T}}$, and this is tight.