Hardness of boolean functions
In the average case, you want the circuit to succeed in computing the function for a large portion of all possible inputs. Since a constant function always succeeds for at least half the inputs (the majority of $f$), the interesting case in where you can achieve advantage which is greater than $\frac{1}{2}$.
In the worst case, you want the circuit to succeed in computing the function for all inputs, so you want to say that $f$ is at least $H(n)$ worst case hard if any circuit of size $\le H(n)$ disagrees with $f$ on at least one input in $\{0,1\}^n$.