I'm taking a cryptography introduction course, and we're covering the definition of a secure PRF.
I understand the test goes as follows: A challenger picks a function $f$ such that
$f \leftarrow \left\{ \begin{array}{lr} b=1: k\leftarrow K, f\leftarrow \left(k,\cdot \right ) \\ b=0: Funs \left(X,Y \right ) \end{array} \right.$
Where $Funs(X,Y)$ is the set of all possible PRFs, The entire experiment is defined as $EXP\left(b \right )$, such that the output is the adversary's guess as to whether or not the challenger chose a function at random. A secure PRF is then defined as the conditions where
$\left | Pr\left[EXP\left(0 \right ) =1\right ] - Pr\left[EXP\left(1 \right ) = 1\right ] \right |$ is negligible. Or in other words, when the probability that the attacker's result is zero minus the probability that the attacker's result is one is negligible.
However, what if the adversary's detection method always returns one? This would imply that the cipher is not secure, but that can't be true.
What am I missing here?