@RyderRude If $|\psi\rangle=\sum_{n=0}^{\infty} c_{n_i} |n_i\rangle$, where $|\psi\rangle$ is some state of a mode and the index "i" is used for different modes of the em. field, then:
$(a^{\dagger})^2|\psi\rangle=\sum_{n=0}^{\infty} c_{n_i}\sqrt{n_i}\sqrt{n_i+1} |n_i+2\rangle$. For $n_i \ge 0$ the lowerst fock state is $|2\rangle$. So when I say, there's 100% chance of NOT measuring 1 photon, what I wrote above is what I meant