The change of probability

Two photons income a 50:50 beam splitter from a and b respectly. There are four results. . If the two photons are identical in their physical properties and indistinguishable, we cannot distinguish between the output states of possibilities 2 and 3, and the probability of possibilities 2 and 3 is 50%.
If they arrive at the splitter simultaneously, as demonstrated in reference 1
\begin{equation}
\begin{aligned}
\vert \psi_f \rangle&=\frac{1}{2\sqrt{2}}[(\alpha \gamma + \beta \delta)(\vert H\rangle_1 |H\rangle_2 + |V\rangle_1 |V\rangle_2 )\cdot i(|c\rangle_1 |c\rangle_2 + |d\rangle_1 |d\rangle_2)\\
&\phantom{=\frac{1}{2\sqrt{2}}}+(\alpha \gamma - \beta \delta)(\vert H\rangle_1 |H\rangle_2 - |V\rangle_1 |V\rangle_2 )\cdot i(|c\rangle_1 |c\rangle_2 + |d\rangle_1 |d\rangle_2)\\
&\phantom{=\frac{1}{2\sqrt{2}}}+(\alpha \delta + \beta \gamma)(\vert H\rangle_1 |V\rangle_2 + |V\rangle_1 |H\rangle_2 )\cdot i(|c\rangle_1 |c\rangle_2 + |d\rangle_1 |d\rangle_2)\\
&\phantom{=\frac{1}{2\sqrt{2}}}+(\alpha \delta - \beta \gamma)(\vert H\rangle_1 |V\rangle_2 - |H\rangle_1 |V\rangle_2 )\cdot (|d\rangle_1 |c\rangle_2 - |c\rangle_1 |d\rangle_2)]
\end{aligned}
\label{phif}
\end{equation}
the probability of 2 and 3 changes from 50% to \(\frac{1}{8}|\alpha\delta-\beta\gamma|^2\)
In section 3.2.2 of reference 1, which explains the concept of quantum teleportation, the probability of 2 and 3 changes to 25%. Because photon 2 and 3 are entangled and by rewriting the state of photon 1,2 and 3 it is easy to find that the probability of \(|\Psi^-\rangle_{12}\) is 25%. And we can find in eq(1)) that \(|\Psi^-\rangle_{12}\)means probability of 2 and 3 happens.

Reference

1. Quantum Teleportation and Multi-photon Entanglement, Jian-Wei Pan