To properly describe the polarization state of a photon, it is useful to introduce some concepts of quantum optics. The polarization is in fact one of the possible degrees of freedom of a photon. As a matter of convenience, we can assume as a base for the polarization state of the photons the one formed by two orthogonal axis (e.g. an horizontal axis and a vertical axis). If the polarization of the photon we are describing is horizontal (from left to right), the wavefunction of the photon, only accounting for its polarization state, will be written as $|H\rangle$; on the other hand, if the polarization of the photon is vertical (from bottom to top), the associated state will be $|V\rangle$. The generic wavefunction, with respect to the polarization degree of freedom, of a photon will be written as
$$ |\Psi\rangle = a |H\rangle + b |V\rangle,\quad \langle\Psi|\Psi\rangle = 1 $$
Changing our representation of these states, we can identify the horizontal and vertical polarizations as the vectors
$$ |H\rangle \equiv \begin{bmatrix} 1 \\ 0 \end{bmatrix},\quad |V\rangle \equiv \begin{bmatrix} 0\\1\end{bmatrix}. $$
In this case, the previously considered wavefunction will be
$$ |\Psi\rangle \equiv a\begin{bmatrix} 1\\0\end{bmatrix} + b\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} a \\ b\end{bmatrix} .$$
We can now assume that this photon is interacting with a polarizing beam splitter (PBS) the will split radiation depending on its polarization state. If the optical axis of this polarizer is along the main diagonal (from bottom left to top right), then only photons with the associated polarization state
$$ |D\rangle \equiv \frac{1}{\sqrt{2}}\begin{bmatrix} 1\\1\end{bmatrix}$$
will be detected after passing through the PBS with a unitary probability, while photons with the orthogonal polarization state (along the anti-diagonal)
$$ |A\rangle \equiv \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1\end{bmatrix} $$
will be reflected by the PBS with unitary probability.
We want now to understand what will happen to a photon in the said generic polarization state $|\Psi\rangle$ when it is impinging on such a PBS. To do this, from a mathematical point of view, we need to project the generic wavefunction (associated to the polarization degree of freedom) on the basis associated to the PBS. In our case, with a few mathematical passages it is possible to show that
$$ |\Psi\rangle \equiv a\begin{bmatrix} 1 \\ 0 \end{bmatrix} + b \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \frac{a+b}{\sqrt{2}} \begin{bmatrix} 1\\ 1\end{bmatrix} + \frac{a-b}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} .$$
This means that the probability to detect the photon after the PBS is equal to the probability of measuring the photon on the $|D\rangle$ polarization state, that can thus be written as
$$ \mathbb{P}_D = \left| \frac{a+b}{\sqrt{2}}\right|^2 .$$
To sum up, we can apply this calculation to your case. Assuming a photon in a vertical polarization state, in this case $a=0$ and $b=1$. This means that we have a probability of 0.5 of measuring the photon after the PBS with diagonal polarization, while there is a 0.5 probability that the photon is reflected by the PBS (and that it has anti-diagonal polarization).
As already stated by garyp, there is not any meaning in asking what will happen "half of the time", when you are considering single photons: in every case, the photon is either fully transmitted or reflected (in the case of the PBS), with a quantum probability associated to these two outcomes. It is only repeating the experiment many times (or having several photons with the same polarization state impinging at the same time on the PBS) that you see 50% of the photons transmitted and 50% reflected.
