Exact tails and mean bias

For $E>0$,

Indeed, $(1-E/L)|0⟩⟨0|+(E/L)|L⟩⟨L|$, with $L>\max (E,N)$, has bias $E(1-N/L)\to E$. A mean-only constraint gives a probability bound $\mathrm{Pr}(N_h>N)\le E/(N+1)$, not a uniformly vanishing mean bias.