A polarizer does not work by transmitting one very specific polarization angle and blocking all other - even only very slightly different - polarizations completely.
Instead a polarizer obeys Malus' law: If the angle between the axis of polarization of the incident light and the polarizer's polarization angle is $\theta$, then the fraction of transmitted intensity to incident intensity is $\cos^2(\theta)$. Or, equivalently speaking in terms of photons, the probability that a photon with definite polarization at such an angle is transmitted and not absorbed is $\cos^2(\theta)$.
The reason for this is that any incident polarization can always be seen as a linear superposition of the polarizer's polarization and the one orthogonal to it. The orthogonal polarization is completely blocked and the parallel polarization is completely transmitted - this is the definition of what a perfect polarizer does. Now, an incident electric field vector $\vec E$ at angle $\theta$ is
$$ \vec E_0 = \cos(\theta) E_{||} + \sin(\theta)E_{\perp}$$
in terms of the field in the parallel direction $E_{||}$ and the field in the orthogonal direction $E_\perp$. After the polarizer, all that is left is $\vec E_1 = \cos(\theta) E_{||}$. Therefore, the ratio of the intensities is
$$ \frac{I_1}{I_0} \propto \frac{(\vec E_1)^2}{(\vec E_0)^2} = \cos^2(\theta),$$
which shows Malus' law.
If the polarization angle of the the incident radiation is uniformly randomly distributed among the possible angles, then the average intensity will be
$$ \frac{1}{2\pi}\int^{2\pi}_0\cos^2(\theta)\mathrm{d}\theta = \frac{1}{2},$$
i.e. "totally unpolarized" photons have, on average, an equal chance to be absorbed by or transmitted through the polarizer.
