What if we test three photons instead of two in Bell's paradox? I am not a physicist and do not know anything about quantum mechanics (except that it can be formulated using Hilbert spaces), but watching a vulgarization video about Bell's paradox, I had a question that I do not know how to answer. Please excuse me if what I say is nonsensical.
In the video, Bell's paradox is formulated as follows. We have an entangled pair of photons with superposed state $\frac{1}{\sqrt{2}}(\left|H\right>⊗\left|H\right>+\left|V\right>⊗\left|V\right>)$ where $H$ is horizontal polarization and $V$ vertical polarization. I do not really know what polarization means, so maybe it makes no sense at all. To me it is only an element of some finite dimensional Hilbert space. We have three filters $A$, $B$ and $C$ with respective angles 0°, 22.5° and 45°. For each pair of filters, we can test the two photons with these filters and we get some probability distribution on the four possible outcomes.
As I understand the paradox, there is no probability distribution on the space of eight possible outcomes for each of the three filters ($2^3 = 8$) such that it is coherent with the probabilities given by testing only with two filters. Nonetheless, if we take three entangled photons instead of two, and test these photons with filters $A$, $B$ and $C$, we should get a probability distribution on the space of eight possible outcomes. So I was wondering what is this probability distribution, if we suppose the state of the three entangled photons is $\frac{1}{\sqrt{2}}(\left|HHH\right> + \left|VVV\right>)$.
How can we compute this probability distribution (if we can stay at the level of finite dimensional Hilbert spaces) and what is it? How does this probability distribution relate to the three probability distributions associated to each pair of filters?
I invented some calculations that worked with two photons (it gives the right probabilities) but I suspect they make no sense at all, so I'm not including them to shorten my post. If I try to apply it to three photons, it gives things that cannot be true because they are not rotation-invariant.
Moreover, I cannot imagine that if we run the experience with three photons, the probabilities can be different depending on if we keep the result from the last photon or not. What would be the "boundary" between the experience with two photons and the experience with one more photon? (Imagine the experimenters did not realize one more photon was emitted and tested by error, or something like that.)
I hope there is not too many conceptual misunderstandings in what I wrote. If so, what are they?
 A: Here is the recipe for doing your calculation.
First, I assume you mean an initial state of $\sigma=(1/\sqrt2)(HHH+VVV)$.
The possible outcomes for going through the zero-degree filter are $H$ and $V$.
The possible outcomes for going through the $\pi/8$ filter are $X=\cos(\pi/8)H+\sin(\pi/8)V$ and $Y=\cos(5\pi/8)H+\sin(5\pi/8)V$.
The possible outcomes for going through the $\pi/4$ filter are $Z=(\cos(\pi/4)H+\sin(\pi/4)V$ and $W=\cos(\pi/4)H+\sin(3\pi/4)V$.
Therefore the eight possible outcomes for your experiment are $HXZ,HXW,HYZ,HYW,VXZ,VXW,VYZ,VYW$.  Call these $T_1,\ldots T_8$.
To interpret these, use the distributive law so  that $HXZ=H\Big(\cos(\pi/8)H+\sin(\pi/8)V\Big)\Big(\cos(\pi/4)H+\sin(\pi/4)V\Big)$ is a sum of four nonzero terms, of which the first is $(\cos(\pi/8))(\cos(\pi/4))HHH$.
Now take your initial state $(1/\sqrt3)(HHH+VVV)$ and write it (uniquely) in the form $\alpha_1 T_1+\ldots \alpha_8T_8$.
Then the probability that you'll observe the outcome $T_i$ is $|\alpha_i|^2$.
A: If you can use three-photon states, you can do even better than Bell's inequality, you can prepare a Greenberger–Horne–Zeilinger (GHZ) state which allows you to make an entirely deterministic test of hidden variable and quantum mechanical theories, unlike the Bell inequality which relies on relative probabilities. In other words, you get an absolutely true or false experiment about the validity of quantum theory by using the GHZ state, if you do it correctly.
This GHZ state is written as
$$\vert GHZ \rangle = \frac{\vert 000\rangle +\vert 111\rangle}{\sqrt{2}}$$
You can read about the GHZ experiment at its wiki page here:
https://en.m.wikipedia.org/wiki/GHZ_experiment
