This is a very interesting question, and to answer it I am going to consider the following example. Suppose, as you said, that we have a source that, somehow (there are different techniques which allow us to obtain entangled photons such as employing nonlinear cristals) generates a pair of photons that are in the following global state
\begin{equation}
|\Phi\rangle = \dfrac{1}{\sqrt{2}}\big[|0_A1_B\rangle + |1_A0_B\rangle\big],
\end{equation}
where the $0$ represents vertical polarization and $1$ horizontal polarization, for instance. This is an example of maximally entangled state, which belong to the basis of Bell states (see the book Quantum Information and Quantum COmputation by Nielsen and Chuang, for example). Each of the photons are sent to system $A$ (usually known as Alice) and system $B$ (usually known as Bob), which explains the presence of the subscripts that I have considered above. Also, I assume that Alice and Bob are far away from each other and that they can only communicate classically (via telephone, for example).
Now, as you said, Alice performs a random polarization measurement so with probability $p_A = \frac{1}{2}$ she can obtain either horizontal or vertical polarization. So, if she obtains after the measurement horizontal polarization, i.e., a $1$, the global state of the system collapses to
\begin{equation}
|\Phi'\rangle = |1_A 0_B\rangle.
\end{equation}
But, what about Bob? He doesn't know what the result of Alice measure was, so if we want to put ourselves in his position, we have to eliminate somehow Alice degrees of freedom. This is done in quantum information via the density matrix representation, which in our case is given by
\begin{equation}
\rho = |\Phi\rangle \langle \Phi |
= \dfrac{1}{2}
\Big[
|1_A0_B\rangle \langle 1_A 0_B| + |0_A1_B\rangle \langle 0_A1_B|
+ |1_A0_B\rangle\langle 0_A1_B| + |0_A1_B\rangle \langle 1_A 0_B|\Big],
\end{equation}
and taking the partial trace with respect to $A$, that is, to sum over Alice degrees of freedom, we get that Bob sees the state
\begin{equation}
\rho_B = \text{tr}_A \rho = \sum_{i=0,1} \langle i |\rho|i \rangle =
\dfrac{1}{2}|0\rangle \langle 0 | + |1\rangle \langle 1 | = \dfrac{\boldsymbol{1}}{2},
\end{equation}
where $\boldsymbol{1}$ represents the identity. Hence, no matter what Alice does on her state that, if Bob does not know what the result o her measurement was, every horizontal or vertical polarization measurement that he does on his state will have an outcome probability equal to $\frac{1}{2}$.
Nevertheless, if after the measurement Alice tells Bob her result (in our case horizontal polarization), then the density matrix of the global system will be
\begin{equation}
\rho = |1_A 0_B\rangle \langle 1_A 0_B|,
\end{equation}
and then Bob gets with probability $1$ vertical polarization.
Hope this answers your question satisfactorilly!
