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Imagine the following experimental setup, thinking semi-classically:

We have a source which produces a stream of photon pairs with random polarization, but with both photons in the pair having identical polarization. We block one of the streams, and direct the other one to towards a pair of polarizing filters (A,B) placed in series. The probability that the particles pass through the first polarizer is P(A) = 0.5, and is independent of orientation. According to Malus law, the probability that the particles pass through the second polarizer, after passing through the first one is given by P(AB) = P(A)P(B|A) = cos^2(a-b) where (a,b) are the angle settings for each polarizer. The reason we use P(AB) = P(A)P(B|A) instead of P(AB) = P(A)P(B) is because for the B filter, the particles under consideration are only the subset which passed through A. P(B|A) reads "for those particles which went through A, what proportion of them will now go through B."

Now instead of two polarizers in sequence, and a single stream of particles, let us change the setup a little. We separate out our polarizers and open up the other beam. This time, one stream of particle pairs goes to A and the other one goes to B. As expected, the probability of a particle going through A, P(A) = 0.5 and the probability of a particle going through B, P(B) is 0.5, in full agreement with the single stream case. Then, we decide to post-select from all the particles going through B, only those for which their twin sister particle also went through A. In other words, we are now calculating not P(B), but P(B|A). In this case P(B|A) reads "for those particles whose twin went through A, what proportion of them will now go through B". Given that the twins are identical, P(B|A) means exactly the same thing as it did in the single stream case. And it turns out we observe experimentally that in the second case P(AB) = P(A)P(B|A) = cos^2(a-b) as well!

The two experiments are conceptually the same. You may now notice that the second result is exactly the QM expectation value for "entangled" pairs. Also notice the importance of post-processing for the result of the second experiment.

So my question is: Where is the mystery?

Good question!

You will not be surprised to learn that in my opinion there is no mystery, even for an arbitrarily large sequence of measurements.

For my own GA version of the derivation of the Malus's law, please see page 10 of this paper: http://arxiv.org/pdf/0707.1333.pdf.

Minkwe and Joy,

With respect, but imho: The ONLY mystery to me is that you two have not corrected/clarified the OP. For, as I read it, it is erroneous or misleading.

Perhaps my problem (among others) is that I am misreading this: "but with both photons in the pair having identical polarisation"?

For I take that to mean that both photons are prepared identically linearly-polarised (in one random direction orthogonal to the line of flight). This (to me) would be equivalent to the outputs from a Source consisting of an Aspect/EPRB-style source of entangled photons being sandwiched between two aligned and coupled linear-polarisers which rotate randomly, stepwise, in unison.

minkwe wrote:Imagine the following experimental setup, thinking semi-classically:

We have a source which produces a stream of photon pairs with random polarization, but with both photons in the pair having identical polarization. We block one of the streams, and direct the other one to towards a pair of polarizing filters (A,B) placed in series. The probability that the particles pass through the first polarizer is P(A) = 0.5, and is independent of orientation. According to Malus law, the probability that the particles pass through the second polarizer, after passing through the first one is given by P(AB) = P(A)P(B|A) = cos^2(a-b) where (a,b) are the angle settings for each polarizer. The reason we use P(AB) = P(A)P(B|A) instead of P(AB) = P(A)P(B) is because for the B filter, the particles under consideration are only the subset which passed through A. P(B|A) reads "for those particles which went through A, what proportion of them will now go through B."

Now instead of two polarizers in sequence, and a single stream of particles, let us change the setup a little. We separate out our polarizers and open up the other beam. This time, one stream of particle pairs goes to A and the other one goes to B. As expected, the probability of a particle going through A, P(A) = 0.5 and the probability of a particle going through B, P(B) is 0.5, in full agreement with the single stream case. Then, we decide to post-select from all the particles going through B, only those for which their twin sister particle also went through A. In other words, we are now calculating not P(B), but P(B|A). In this case P(B|A) reads "for those particles whose twin went through A, what proportion of them will now go through B". Given that the twins are identical, P(B|A) means exactly the same thing as it did in the single stream case. And it turns out we observe experimentally that in the second case P(AB) = P(A)P(B|A) = cos^2(a-b) as well!

The two experiments are conceptually the same. You may now notice that the second result is exactly the QM expectation value for "entangled" pairs. Also notice the importance of post-processing for the result of the second experiment.

So my question is: Where is the mystery?

1. We read, "Imagine the following experimental setup, thinking semi-classically:"

I would say, "Imagine the following experimental setup, thinking local-realistically:"

Reason: minor point, but, imho, local-realists have no need to think "semi-classically."

2. We read, "According to Malus law, the probability that the particles pass through the second polarizer, after passing through the first one is given by P(AB) = P(A)P(B|A) = cos^2(a-b) where (a,b) are the angle settings for each polariser."

But this is garbled, as I read it.

For me: Identifying this experiment as experiment S (= single-stream), I would say, "According to Malus law, the probability that the particles pass through the second polarizer, after passing through the first one is given by P(B|SA) = cos^2(a-b) where (a,b) are the angle settings for each polariser."

Alternatively: "According to Malus law, the probability that the particles pass through both polarisers, is given by P(AB|S) = P(A|S)P(B|SA) = (1/2)cos^2(a-b) where (a,b) are the angle settings for each polariser."

3. We read, "Given that the twins are identical, P(B|A) means exactly the same thing as it did in the single stream case. And it turns out we observe experimentally that in the second case P(AB) = P(A)P(B|A) = cos^2(a-b) as well!"

But this is garbled, as I read it.

For me: Identifying this experiment as experiment D (= double-stream), I would say, "Given that the twins are prepared as before, P(B|DA) means exactly the same thing as P(B|SA) did in the single stream case. And it turns out we observe experimentally that in the second case P(AB|D) = P(A|D)P(B|DA) = 1/8 + (1/4)cos^2(a-b)!

Thus the mystery for me now is this: What am I missing?

<EDIT> Of course, if you are referring to the apparent "Malus Law" that emerges from the analysis of Bell-tests, that is another matter: though there is still no mystery. In my terms, that law is better labelled as "Malus Law Generalised" (MLG). Moreover, it also accords with local-realism.

To be clear, if this is your aim: I do not see how the wording of the OP (with its "identically polarised photons") explicates or engages with the entangled case and MLG.

PS: Though QM's collapse postulate is [maybe] open to an interpretation [via an "AS IF"] that is a little like what you are [maybe] attempting to describe?

E & OE.

Xray

Hi Xray,
It's been a while I completely garbled up the point I wanted to make didn't I . So let me try again. For experiments S and D, without any reference to what the specific results are. Let us say only that we are using the exact same source for both cases, no reference to the detailed nature of the particles, or "polarization", we simply say both particles in a pair are identical in every sense relevant for their subsequent behavior, fully local realistic.

Let us then also say that we are in both cases interested in the joint probability P(AB|X) , X being S or D. The settings at A and B are arbitrary and not really relevant for the analysis. In the first case (S)

P(AB|S) = P(A|S)P(B|AS) NOT P(AB|S) = P(A|S)P(B|S)

In the second case (D)

P(AB|D) = P(A|D)P(B|AD) NOT P(AB|D) = P(A|D)P(B|D)

Obviously, both cases are fully local realistic, and we recognize that the joint probabilities can not be factored. Specifically, for experiment D, we can separate the A and B beyond any light-cone influences and still the analysis above applies just the same. It might be helpful to think about how P(AB|D) might actually be measured and calculated in an experiment.

Why is anyone mystified that P(AB|D) =/= P(A|D)P(B|D) in both cases.

minkwe wrote:...Why is anyone mystified that P(AB|D) =/= P(A|D)P(B|D) in both cases...

Perhaps because, in agreement with what I think Xray is saying regarding your post #1 scenario, the two cases are entirely different. For single stream case, reading past the second polarizer is directly influenced by the result past the first polarizer. Malus's law follows as applicable both classically (product of individual transmission coefficients) and quantum mechanically (sequential probabilities). In the two streams scenario, usual classical expectation is just proportional to the sum of individual transmission coefficients past each polarizer separately. No influence of one polarizer result should directly effect the other. The sole joint property being strict sharing of polarization of both streams. Hence applying Malus's law as in first case (product formula) is just inapplicable classically. That it does continue to apply in QM case is the mystery that JC and one or more other members here claim to explain via GA or some such.

Q-reeus wrote:

minkwe wrote:...Why is anyone mystified that P(AB|D) =/= P(A|D)P(B|D) in both cases...

Perhaps because, in agreement with what I think Xray is saying regarding your post #1 scenario, the two cases are entirely different. ...

I already admitted my post #1 was quite poorly written. The point as I've subsequently clarified is the fact that in both situations, the expression for the joint probability has the same form P(AB|X) = P(A|X)P(B|AX) =/= P(A|X)P(B|X).

Now are you saying P(AB|S) = P(A|S)P(B|AS) =/= P(A|S)P(B|S) is accurate but P(AB|D) = P(A|D)P(B|AD) =/= P(A|D)P(B|D) is not, if A and B are separated beyond their light cones?

Again, remember the issue is not to claim that the two experiments are the same but that the joint probabilities have the exact same form, NOT simply a product of separate one-sided probabilities, even in the D case.

minkwe wrote:I already admitted my post #1 was quite poorly written. The point as I've subsequently clarified is the fact that in both situations, the expression for the joint probability has the same form P(AB|X) = P(A|X)P(B|AX) =/= P(A|X)P(B|X).

Now are you saying P(AB|S) = P(A|S)P(B|AS) =/= P(A|S)P(B|S) is accurate but P(AB|D) = P(A|D)P(B|AD) =/= P(A|D)P(B|D) is not, if A and B are separated beyond their light cones?

Regardless of any degree of spatial separation (I don't accept notion of 'communication' going on, superluminal or not), how can you expect an equivalent outcome between two physically quite different arrangements? A joint probability as product in single stream sequential filters case is physically sensible but I cannot fathom your logic in applying that same formula to twin stream separate filters case. Where classically no interaction term is involved and one simply has a net result from two independent (polarizer angle settings) stream/filter interactions. What is the physical meaning of multiplying those individual outcomes? And I take it your probabilities are not averages over all angles but apply to any given instance of polarization & angle settings.

Q-reeus wrote:how can you expect an equivalent outcome between two physically quite different arrangements? A joint probability as product in single stream sequential filters case is physically sensible but I cannot fathom your logic in applying that same formula to twin stream separate filters case.

Again I'm not saying the outcome is exactly the same. Please focus on the mathematical form of the expressions. The actual results are irrelevant to the point I'm making.

So you are saying you accept P(AB|S) = P(A|S)P(B|AS) =/= P(A|S)P(B|S) as correct. But you refuse to accept P(AB|D) = P(A|D)P(B|AD) =/= P(A|D)P(B|D)? Is that a correct representation of your point?

Where classically no interaction term is involved and one simply has a net result from two independent (polarizer angle settings) stream/filter interactions. What is the physical meaning of multiplying those individual outcomes? And I take it your probabilities are not averages over all angles but apply to any given instance of polarization & angle settings.

We will get to physical meaning in a moment. First let us be clear that you are saying my analysis of the D experiment is wrong?

minkwe wrote:Again I'm not saying the outcome is exactly the same. Please focus on the mathematical form of the expressions. The actual results are irrelevant to the point I'm making.

So you are saying you accept P(AB|S) = P(A|S)P(B|AS) =/= P(A|S)P(B|S) as correct.

Yes - assuming it refers to general case of an arbitrary given instance of polarizers having some relative angle settings. Obviously there are particular angular settings when equality applies.

But you refuse to accept P(AB|D) = P(A|D)P(B|AD) =/= P(A|D)P(B|D)? Is that a correct representation of your point?

Far as I can see both joint probabilities as products have no meaningful physical content given the classical scenario, hence the inequality is also meaningless.

We will get to physical meaning in a moment. First let us be clear that you are saying my analysis of the D experiment is wrong?

As per above, yes. Then again might have completely misunderstood your argument.

Q-reeus wrote:

But you refuse to accept P(AB|D) = P(A|D)P(B|AD) =/= P(A|D)P(B|D)? Is that a correct representation of your point?

Far as I can see both joint probabilities as products have no meaningful physical content given the classical scenario, hence the inequality is also meaningless.

I don't follow your gripe about "physical content". Remember, that P(AB|D) means the probability that the one particle of a pair passes through A and it's twin sibling also passes through B for the D experiment. We do not care what settings or arrangements are in place at A or B. It is completely arbitrary, there may or may not be any settings at all. Are you saying P(AB|D) = P(A|D)P(B|AD) is not a correct analysis of this experiment? If so what is your analysis?

minkwe wrote:I don't follow your gripe about "physical content". Remember, that P(AB|D) means the probability that the one particle of a pair passes through A and it's twin sibling also passes through B for the D experiment. We do not care what settings or arrangements are in place at A or B. It is completely arbitrary, there may or may not be any settings at all. Are you saying P(AB|D) = P(A|D)P(B|AD) is not a correct analysis of this experiment? If so what is your analysis?

Think about it. Take a specific instance of you OP setup: horizontally polarized stream/beam, with absorbing polarizer filter A oriented at +45 degrees, filter B at -45 degrees. What is your joint probability for S case? What is it for D case? Do you see a problem here?
Must go - back much later.

One step at a time. Since you think P(AB|D) = P(A|D)P(B|AD) is not correct, please could you write down what you think is the correct expression for

P(AB|D) = ?

Again: don't be misled by the OP, as I've already explained many times:

minkwe wrote:So let me try again. For experiments S and D, without any reference to what the specific results are. Let us say only that we are using the exact same source for both cases, no reference to the detailed nature of the particles, or "polarization", we simply say both particles in a pair are identical in every sense relevant for their subsequent behavior, fully local realistic.

Let us then also say that we are in both cases interested in the joint probability P(AB|X) , X being S or D. The settings at A and B are arbitrary and not really relevant for the analysis.

minkwe wrote: ...

Why is anyone mystified that P(AB|D) =/= P(A|D)P(B|D) in both cases.

Dear minkwe and Q-eerus,

Let me first fix minkwe's nice question and then address your dispute:

1. Let A and B be correlated events under condition X. Then, per (little-known) Watson's Law*, a law supported by many experiments and (so far) never refuted:

P(AB|X) = P(A|X)P(B|XA) = P(B|X)P(A|XB). (1)

2. So, per Minkwe's Question: Why is anyone mystified that in both cases (ie, when X = S and X = D):

P(AB|X) ≠ P(A|Χ)P(B|Χ)? (2)

3. Response: Since A and B are correlated events under both X = S and X = D, we can invoke Watson's Law. (Noting, in passing, that the S and D experiments will further confirm the validity, here, of that law.)

4. So why is anyone mystified by (2) -- a TRUISM (per Watson's Law) under minkwe's specification?

I trace that wide-spread mystification amongst physicists to Bell's later introduction of a new error into his already (1964-style) defective analysis.

(A mystification/error that was famously and publicly evident at about the same time as Bell wrote; see http://en.wikipedia.org/wiki/Roy_Meadow : Watson's law applying to P(Bell's-error, Meadow's-error|Z); Z real!)

5. Bell's equivalent error was his re-interpretation of "local causality" thus: If Y is a complete specification of the conditions relating to events A and B, then

P(AB|Y) = P(A|Y)P(B|Y). (3)

6. Bell's use of (3) can be seen in equation (11) of his "Bertlmann's socks" essay (available online: https://cds.cern.ch/record/142461/files/198009299.pdf ).

7. But (3) is ALWAYS false if events A and B are correlated.

8. Even if A and B are limited to a common degree of freedom equalling one part in infinity (to define a minimal correlation) and we fully specify that part (to satisfy Bell's condition Y): (3) is provably false; both mathematically and experimentally.

9. One example of the maths and experiment referred to in #8 follows from the application of Malus Law to the analysis of the experiment foreshadowed in viewtopic.php?f=6&t=101#p3643

* Akin to Bayes theorem, Watson's Law differs in that it is not directly related to "current and prior beliefs", etc, typically associated with Bayes: http://en.wikipedia.org/wiki/Bayes'_theorem

With E & OE, please excuse my haste,

Xray

minkwe wrote:One step at a time. Since you think P(AB|D) = P(A|D)P(B|AD) is not correct, please could you write down what you think is the correct expression for
P(AB|D) = ?

P(AB|D) = 0.5(P(A|D) + P(B|D))
That gives the normalized net count past both filters wrt net source count. My idea of joint probability for that strictly classical situation. Applying above sum formula to specific example I gave, one has 0.5(0.5 + 0.5) = 0.5. If averaged over all stream/beam polarizations, we have the form 0.5(Cos^2(theta) + Sin^2(theta)) = 0.5 (theta being the relative angle between stream/beam polarization vector and any one given polarization filter). So average of net transmission probability past both filters is 0.5 independent of beam polarization angle.
As I wrote before, the product formula P(A|S)P(B|AS) is applicable to the very different S case, and for the specific example I gave, it comes to 0 for any polarization of stream/beam. Hence the average over all stream/beam polarizations is 0.
Independent filter interactions - D case, vs tandem cascade - S case. Chalk and cheese.

Now explain your notion of joint probability for D case, what it actually means (avoiding circular definition), how it comes from a product formula P(A|D)P(B|AD), and what it yields for specific example I gave - both for fixed beam polarization, and when averaged.

As to what obtains in S vs D cases for further averaging over all possible combinations of polarizer filter angle settings, I do not really care.

Xray wrote:

minkwe wrote: ...

Why is anyone mystified that P(AB|D) =/= P(A|D)P(B|D) in both cases.

Dear minkwe and Q-eerus,

Let me first fix minkwe's nice question and then address your dispute:

1. Let A and B be correlated events under condition X. Then, per (little-known) Watson's Law*, a law supported by many experiments and (so far) never refuted:

P(AB|X) = P(A|X)P(B|XA) = P(B|X)P(A|XB). (1)

Commutativity thus reciprocity? Assuming a product formula universally applies and is meaningful, (1) seems trivial given all P's are reals - right?

2. So, per Minkwe's Question: Why is anyone mystified that in both cases (ie, when X = S and X = D):

P(AB|X) ≠ P(A|Χ)P(B|Χ)? (2)

3. Response: Since A and B are correlated events under both X = S and X = D, we can invoke Watson's Law. (Noting, in passing, that the S and D experiments will further confirm the validity, here, of that law.)

4. So why is anyone mystified by (2) -- a TRUISM (per Watson's Law) under minkwe's specification?

I trace that wide-spread mystification amongst physicists to Bell's later introduction of a new error into his already (1964-style) defective analysis.

(A mystification/error that was famously and publicly evident at about the same time as Bell wrote; see http://en.wikipedia.org/wiki/Roy_Meadow : Watson's law applying to P(Bell's-error, Meadow's-error|Z); Z real!)

5. Bell's equivalent error was his re-interpretation of "local causality" thus: If Y is a complete specification of the conditions relating to events A and B, then

P(AB|Y) = P(A|Y)P(B|Y). (3)

6. Bell's use of (3) can be seen in equation (11) of his "Bertlmann's socks" essay (available online: https://cds.cern.ch/record/142461/files/198009299.pdf ).

7. But (3) is ALWAYS false if events A and B are correlated.

8. Even if A and B are limited to a common degree of freedom equalling one part in infinity (to define a minimal correlation) and we fully specify that part (to satisfy Bell's condition Y): (3) is provably false; both mathematically and experimentally.

9. One example of the maths and experiment referred to in #8 follows from the application of Malus Law to the analysis of the experiment foreshadowed in viewtopic.php?f=6&t=101#p3643

* Akin to Bayes theorem, Watson's Law differs in that it is not directly related to "current and prior beliefs", etc, typically associated with Bayes: http://en.wikipedia.org/wiki/Bayes'_theorem

With E & OE, please excuse my haste,

Xray

Hi Xray. Regret ever wading into this swampland, having never really studied Bell's theorem beyond a cursory inspection. Seems certain accepted notions of 'correlations' and 'joint probability' are quite foreign to my thinking. Back to the sidelines for me!

Q-reeus wrote:

minkwe wrote:One step at a time. Since you think P(AB|D) = P(A|D)P(B|AD) is not correct, please could you write down what you think is the correct expression for
P(AB|D) = ?

P(AB|D) = 0.5(P(A|D) + P(B|D))

Huh? I'm curious were you got this expression for joint probability. Ever heard if the chain rule?

minkwe wrote:

Q-reeus wrote:

minkwe wrote:One step at a time. Since you think P(AB|D) = P(A|D)P(B|AD) is not correct, please could you write down what you think is the correct expression for
P(AB|D) = ?

P(AB|D) = 0.5(P(A|D) + P(B|D))

Huh? I'm curious were you got this expression for joint probability. Ever heard if the chain rule?

Just looked up https://en.wikipedia.org/wiki/Joint_pro ... stribution
There are various definitions given for joint probability. You gave your definition of what it meant for S case back in 1st para of post #1 - fraction of initial stream that makes it past 2nd polarizer. That's the net fractional count past both polarizers. I have given you the expression that gives the equivalent value for D case. You disagree? Further, how about you answering my questions already given earlier, instead of culling them out completely from your quotes. Otherwise, I have no further interest in this thread.

Q-reeus wrote:There are various definitions given for joint probability.

Which one are you using? I already explained what P(AB|D) means:

minkwe wrote:Remember, that P(AB|D) means the probability that the one particle of a pair passes through A and it's twin sibling also passes through B for the D experiment. We do not care what settings or arrangements are in place at A or B. It is completely arbitrary, there may or may not be any settings at all. Are you saying P(AB|D) = P(A|D)P(B|AD) is not a correct analysis of this experiment? If so what is your analysis?

Q-reeus wrote:I have given you the expression that gives the equivalent value for D case.

Which is blatantly wrong as can be easily verified. Say in an experiment D, involving N photon pairs 1/2 of the "A-stream" photons pass through device A. 1/2 of the "B-stream" photons pass through device B. But of the N/2 particles who pass through A, only 1/2 of their twins make it through B and vice versa. So we have P(A|D) =1/2, P(B|D) = 1/2. P(B|AD) = 1/4, P(A|BD) = 1/4. The joint probability P(AB|D) is simply asking what relative frequency of the original N photon pairs make it through both A and B.

You have P(AB|D) = 0.5 ( P(A|D) + P(B|D)) = 0.5 (0.5 + 0.5) = 0.5
I have P(AB|D) = P(A|D)P(B|AD) = P(B|D)P(A|BD) = 0.5*0.5 = 0.25.

Do you still think your formula is correct?

Further, how about you answering my questions already given earlier

What question do you feel is unanswered?

minkwe wrote:

Q-reeus wrote:There are various definitions given for joint probability.

Which one are you using? I already explained what P(AB|D) means:

minkwe wrote:Remember, that P(AB|D) means the probability that the one particle of a pair passes through A and it's twin sibling also passes through B for the D experiment. We do not care what settings or arrangements are in place at A or B. It is completely arbitrary, there may or may not be any settings at all. Are you saying P(AB|D) = P(A|D)P(B|AD) is not a correct analysis of this experiment? If so what is your analysis?

Q-reeus wrote:I have given you the expression that gives the equivalent value for D case.

Which is blatantly wrong as can be easily verified. Say in an experiment D, involving N photon pairs 1/2 of the "A-stream" photons pass through device A. 1/2 of the "B-stream" photons pass through device B. But of the N/2 particles who pass through A, only 1/2 of their twins make it through B and vice versa. So we have P(A|D) =1/2, P(B|D) = 1/2. P(B|AD) = 1/4, P(A|BD) = 1/4. The joint probability P(AB|D) is simply asking what relative frequency of the original N photon pairs make it through both A and B.

You have P(AB|D) = 0.5 ( P(A|D) + P(B|D)) = 0.5 (0.5 + 0.5) = 0.5
I have P(AB|D) = P(A|D)P(B|AD) = P(B|D)P(A|BD) = 0.5*0.5 = 0.25.

Do you still think your formula is correct?

Reproducing your OP first para:

We have a source which produces a stream of photon pairs with random polarization, but with both photons in the pair having identical polarization. We block one of the streams, and direct the other one to towards a pair of polarizing filters (A,B) placed in series. The probability that the particles pass through the first polarizer is P(A) = 0.5, and is independent of orientation. According to Malus law, the probability that the particles pass through the second polarizer, after passing through the first one is given by P(AB) = P(A)P(B|A) = cos^2(a-b) where (a,b) are the angle settings for each polarizer. The reason we use P(AB) = P(A)P(B|A) instead of P(AB) = P(A)P(B) is because for the B filter, the particles under consideration are only the subset which passed through A. P(B|A) reads "for those particles which went through A, what proportion of them will now go through B."

What's in red is what I took and used as your definition. It's what I stated last time. And as defined there that formula has zero meaning for D case, but nevertheless an appropriate redefined formula gives the equivalent (fraction past both polarizer filters), and it's the mean of sum formula I gave.

What question do you feel is unanswered?

You can read what they were, but don't bother. What I hadn't noticed in your first para was the changed definition after that in red. So basically there was an unannounced switching to a new definition, and that's the one ran with thereafter. Whatever. As I wrote, product formula (defined by red text quote) yields a physically meaningful result (net transmission probability) in S case but not in D case.

Q-reeus, you keep going back to post #1, despite the fact that there were serious errors in it which I've since admitted and rectified long before you started posting in this thread, there was no such claimed "unannounced switching":

minkwe wrote:I completely garbled up the point I wanted to make didn't I . So let me try again. For experiments S and D, without any reference to what the specific results are. Let us say only that we are using the exact same source for both cases, no reference to the detailed nature of the particles, or "polarization", we simply say both particles in a pair are identical in every sense relevant for their subsequent behavior, fully local realistic.

Let us then also say that we are in both cases interested in the joint probability P(AB|X) , X being S or D. The settings at A and B are arbitrary and not really relevant for the analysis. In the first case (S)

P(AB|S) = P(A|S)P(B|AS) NOT P(AB|S) = P(A|S)P(B|S)

In the second case (D)

P(AB|D) = P(A|D)P(B|AD) NOT P(AB|D) = P(A|D)P(B|D)

Obviously, both cases are fully local realistic, and we recognize that the joint probabilities can not be factored. Specifically, for experiment D, we can separate the A and B beyond any light-cone influences and still the analysis above applies just the same. It might be helpful to think about how P(AB|D) might actually be measured and calculated in an experiment.

Why is anyone mystified that P(AB|D) =/= P(A|D)P(B|D) in both cases.