Why is quantum entanglement associated with increase rather then decrease of a system's entropy? As far as i understand quantum entanglement (and I'm far from being an expert on it), it is a phenomena involving two or more subatomic particles that reflects itself in immediate correlation between the measurable physical properties of the particles. For example, when a photon decays into electron-positron pair, the conservation of angular momentum implies that the spins of the two particles will be opposite (so the spins sum will be zero) and therefore there will be (anti) correlation between the spins.
From statistical mechanics point of view, the existence of entangled parts of a system of particles naturally lowers (this is how I initially thought) the number of possible microstates that correspond to a given macrostate, because the entangled parts are correlated and therefore the properties of entangled pairs can't assume arbitrary values. For example, in a spin anti-correlation of two electrons, only two out of four microscopic states are possible: up-down and down-up, instead of up-up, up-down, down-up,and down-down. Therefore, an entangled system entropy is supposed to be slightly lower then a non-entangled system entropy at similar conditions (identical temperature, number and type of particles, etc).
However, i read the subsection on entropy in the wikipedia article on quantum entanglement, and although i didn't understand much of the mathematical formulation there (which involves von Neumann entropy), i did notice that it arrives at opposite conclusion to the one I wrote:

The reversibility of a process is associated with the resulting entropy change, i.e., a process is reversible if, and only if, it leaves the entropy of the system invariant. Therefore, the march of the arrow of time towards thermodynamic equilibrium is simply the growing spread of quantum entanglement. This provides a connection between quantum information theory and thermodynamics.

This says that increase in entanglement actually increases in entropy. So what am I misunderstanding here? why is my statistical interpretation of entanglement wrong?
 A: I am not an expert on this, but let me give this question a try. Here is an argument.
Suppose that your system is composed of two particles in an entangled state:
$$|\psi\rangle=\frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle +|\downarrow\uparrow\rangle$$
this is a pure state, so under von Neumman definition, it has entropy it has zero entropy.
Now imagine that you can only measure one of the particles and you do not care about the second one. That means that you have to write the density matrix of your systems, i.e.
$$\rho=|\psi\rangle\langle\psi|=\frac{1}{2} (|\uparrow\downarrow\rangle\langle|\uparrow\downarrow|+|\downarrow\uparrow\rangle\langle|\uparrow\downarrow|+|\uparrow\downarrow\rangle\langle|\downarrow\uparrow|+|\downarrow\uparrow\rangle\langle|\downarrow\uparrow|) $$
and trace the effects of the other one (let's say the particle on the right), we have that your subsystem is described by
$$\rho_1=\operatorname{tr}_2(\rho)=\frac{1}{2}(|\downarrow\rangle\langle{\downarrow}|+|\uparrow\rangle\langle\uparrow| )=\frac{1}{2}\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$
which is now a mixed state (cannot be written as a linear superposition of single states). You can check by noticing that $\operatorname{tr}(\rho_1^2)=1/2<1$. Now you calculate von Neumann's entropy of the subsystem:
$$ S_1 = -\sum_j \eta_j \ln \eta_j$$
where $\eta_j$ are the eigenvalues of $\rho_1$. You find that $S_1=\log(2)>0$. The entropy of your subsystem is non-zero! (and positive by definition).
What all that means
If you can only account for a single particle (your particle is NOT in superposition of $\uparrow$ and $\downarrow$), for you the particle is in a state that is equivalent to a classical indeterminacy between either $\uparrow$ or $\downarrow$. It is the analogous of flipping a coin and hiding it under your hand. You cannot know if the coin is heads or tails (you could try to gather some information from sound, vibrations and air molecules but that would be impossibly hard).
Conclusion for larger systems
The argument above indicates that if you are describing a closed system, there is no problem. But if you start to describe a system that can interact with the environment, then it can get entangled with states that are outside the subsystem. As you cannot account for all interactions, you effectively see an increase of entropy the longer your systems interacts with the outside world due to entanglement.
Alternative picture
Here is another try. Entropy is related to irreversibility. What is harder to recreate, an entangled state of $N$ particles entangled like this:
$$|\psi_N\rangle= \frac{1}{\sqrt 2} (|\uparrow\uparrow\uparrow\cdots\uparrow\rangle +|\downarrow\downarrow\downarrow\cdots\downarrow\rangle) $$
or a system of $N$ particles that is not entangled and it is just a product of superposed states?
$$|\psi_N'\rangle= \frac{1}{2^{N/2}}(|\uparrow\rangle+|\downarrow\rangle)(|\uparrow\rangle+|\downarrow\rangle)\otimes\cdots\otimes(|\uparrow\rangle+|\downarrow\rangle) $$
Consider what happens if some noise in the first system collapses the first particle. All the particles in the system collapse to an unknown state and you lose the whole state. You will need to manipulate all particles again to comeback to the initial state.
In the second case, if the first particle collapses. You still have all other particles in a quantum superposition (you still conserve all the information of the other N-1 particles). You can just revert it back to the initial state by putting the first particle back in a superposition.
A: Because it is chaotic Markov chain related process; it bounds two particles and there are more possible ways of binding/tangling two particles, but there is one and only one way of having them not bound to each other. Entropy increases as the result, because the state of the particles being tangled is more messy and chaotic.
