Is it possible to determine whether the photons in a given ensemble are entangled to a third party? Suppose Alice sends Bob 2 ensembles with 1 billion photons each (or as many photons as you want). In one of the ensembles each of its photons are one of an entangled pair at parallel polarizations while Alice is holding their partners without interaction so they don't decohere.
Is it possible for Bob to create an experiment to determine which of the ensembles has entangled photons?
 A: No.
First ensemble
First consider the ensemble that is not entangled.
The way I read the question, we assume that each photon in the ensemble is in a random state and is not entangled with anything else.
Under those assumptions, the ensemble of photon polarizations can be thought of as an ensemble of arrows in 3D space with direction uniformly distributed over the unit sphere, and in this case, each photon's probability of spin up/down is 1/2 along any axis.
Second ensemble
Now consider the second distribution.
Each photon is entangled with a photon that Alice holds, and that pair is in the state
$$\frac{1}{\sqrt{2}} \left( \lvert \uparrow \uparrow \rangle + \lvert \downarrow \downarrow \rangle \right)$$
where in each ket, the first arrow refers to Alice's photon and the second arrow refers to Bob's photon.
We haven't specified what basis we're in here, and it doesn't matter.
Just imagine that we choose the basis for each pair to make that the correct representation of the state (which is possible because the question says that the photons are in parallel spin entangled states -- and never mind the sign).
Bob does not have access to Alice's photon, so from the point of view of any experiment Bob does, the state of his photon is a mixed state [1]
$$\frac{1}{2} \left( \left \lvert \uparrow \right \rangle \left \langle \uparrow \right \rvert + \left \lvert \downarrow \right \rangle \left \langle \downarrow \right \rvert \right)$$
which is exactly the same as a classical probability distribution with equal probability of spin up and spin down.
Therefore, no matter what we measure along, each photon's probability of spin up/down is 1/2.
Therefore, the ensembles are indistinguishable.
Discussion
It's quite interesting that the entangled ensemble is in some sense "more random" than that not-entangled one.
In the case of the not-entangled ensemble, each photon's polarization actually is pointing in a specific direction before the measurement.
If the directions of each photon's polarization are random and independent from one another (as we assumed) then there's no way to find out which way each photon's polarization was pointed.. but those polarizations do presumably exist... we just weren't given that information.
On the other hand, the photons in the second ensemble don't actually have a polarization direction.
The expectation value of the spin is zero along any axis.
We can show this as follows.
Let $\sigma$ be any one of the three Pauli operators.
Then
\begin{align}
\left \langle \sigma \right \rangle
&= \text{Trace} \left[ \frac{1}{2} \sigma \left( \left \lvert \uparrow \right \rangle \left \langle \uparrow \right \rvert + \left \lvert \downarrow \right \rangle\left \langle \downarrow \right \rvert \right)\right] \\
&=
\left \langle \uparrow \right \rvert \left[ \frac{1}{2} \sigma \left( \left \lvert \uparrow \right \rangle \left \langle \uparrow \right \rvert + \left \lvert \downarrow \right \rangle\left \langle \downarrow \right \rvert \right)\right] \left \lvert \uparrow \right \rangle
+
\left \langle \downarrow \right \rvert \left[ \frac{1}{2} \sigma \left( \left \lvert \uparrow \right \rangle \left \langle \uparrow \right \rvert + \left \lvert \downarrow \right \rangle\left \langle \downarrow \right \rvert \right)\right] \left \lvert \downarrow \right \rangle \\
&= \frac{1}{2} \left( \left \langle \uparrow \rvert \sigma \lvert \uparrow \right \rangle + \left \langle \downarrow \rvert \sigma \lvert \downarrow \right \rangle \right) \\
&= 0 \, .
\end{align}
You can check that the last line is zero by explicitly checking each of the three Pauli operators.
Since any spin direction operator can be expressed as a linear combination of the Pauli operators, we've proven that the expectation value of any spin operator is identically zero for the entangled photons.
[1]: Someone please link to an actually-good explanation of the difference between pure and mixed states, why we use the density matrix to represent mixed states, and why you use the partial trace to find the mixed state of a sub-part of an entangled quantum system.
A: Not unless the ensembles are different in some other way too. Nothing interesting happens to Alice's half of each entangled pair, so we can "trace over" that part of the system to obtain a density matrix describing the photons accessible to Bob. Any measurement Bob makes can depend only on this density matrix. But by assumption, this density matrix is the same as the density matrix describing the other ensemble, so there can be no measurement that distinguishes them.
