## understanding bell's theorem: why hidden variables imply a linear relationship?

Understanding/checking Bell's theorem via scenario analysis

Theoretical simulation (of a case(s)) can be another way to understand/validate Bell's theorem.

So lets take a scenario:

If the entangled photons were oriented at say 0 degrees(see note below) and one polarizer (A) was at 0 degrees and the other (B) at 15 degrees.

What is the logic, and calculations, we use to calculate the match prediction per LHV hypothesis? and per QM/actual?

Note: per QM/QE - it is impossible to tell what degree the photons are oriented at because they are in an indeterminate state and as soon as we measure it, the entanglement breaks. That said.... lets assume hypothetically/theoretically.

 Quote by lugita15 The probabilities P(A), P(B), and P(C) depend on the results of both photons.
Then those probabilities are conditional probabilities. Besides, when you look deeply enought, all probabilities are conditional.
 And I already gave told you the definition of this probability, for both (b) and (c), back in post #224. Now you can definitely say that different probability spaces are being considered in (b) and (c). Indeed, there are three probability spaces in (b) for the three different factual scenarios. In each of these three probability spaces, only one of the three probabilities P(A), P(B), and P(C) is well-defined.
So what in your opinion is P(ABC) for scenario (b)? Unless P(ABC) is well defined, an expression involving P(A), P(B) and P(C) does not make any practical sense.

 And in (c), there is yet another probability space, a bigger one that includes all the particle pairs because in (c) the three statements A, B, and C are all simultaneously well-defined for all particle pairs, so that P(A), P(B), and P(C) are all well-defined in this probability space.
But P(ABC) is well defined in this scenario (c) contrary to the (b) scenario.

Le us represent the sets of photons as one set "w" for scenario (c), and three different sets ("x", "y","z") for scenario (b). The the probabilities for (c) are more accurately represented as

P(A|w), P(B|w), P(C|w). Surely we can combine these probabilities into the same expression and obtain the trivial inequality P(C|w) <= P(A|w) + P(B|w). This makes sense because we are dealing with the same space "w" (aka, the same context, condition, etc). There is no question here.

However the probabilities for scenario (b) are more accurately represented as P(A|x), P(B|y), P(C|z). You are saying that because for the specific physical scenario you described, P(A|x) = P(A|w) and P(B|y) = P(B|w) and P(C|z) = P(C|w), therefore P(C|z) <= P(A|x) + P(B|y) should be valid. But that is not very credible. The inequality is a deeper relationship between probabilities not just specific values of the probabilities. This is evident in the fact that P(ABC|w) is well defined and P(ABC|xyz) is not.

There is nothing new here, this same issue was addressed by Boole 150 years ago. The article by Itamar Pitowsky I cited earlier goes into great details about this. Violation of inequality simply means the data are not from the same population. How does no-conspiracy enable you to recombine separate incompatible samples into a single one? It can't, they are not called incompatible for no reason.[/QUOTE]

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Quote by billschnieder
 The probabilities P(A), P(B), and P(C) depend on the results of both photons.
Then those probabilities are conditional probabilities.
They don't seem like conditional probabilities to me. P(A) is not "the probability of mismatch when the polarizers are oriented at -30 and 0, given that the result of photon 1 was such-and-such and the result of photon 2 was so-and-so." Rather, it is just "the probability of mismatch when the polarizers are oriented at -30 and 0, given no information as to what the results of the two photons are".
 Besides, when you look deeply enought, all probabilities are conditional.
Well, I suppose that's trivially true. Any regular probability can be written as "the probability that such-and-such is the case, given nothing."
 So what in your opinion is P(ABC) for scenario (b)? Unless P(ABC) is well defined, an expression involving P(A), P(B) and P(C) does not make any practical sense.
Sorry, what does P(ABC) mean? In fact, what does ABC mean? Does it mean "A and B and C" or does it mean something else?
 Le us represent the sets of photons as one set "w" for scenario (c), and three different sets ("x", "y","z") for scenario (b). The the probabilities for (c) are more accurately represented as P(A|w), P(B|w), P(C|w).
I don't think this is the right notation. Usually a conditional probability is of the form "the probability that statement X is true, given statement Y." But you're putting a set of photons on the other side of the bar. I think you mean "The probability that A is true for the photons in w", not "The probability that A is true given w", which doesn't make sense to me. But anyway, this is a minor point, since it is just a notation.
 Surely we can combine these probabilities into the same expression and obtain the trivial inequality P(C|w) <= P(A|w) + P(B|w). This makes sense because we are dealing with the same space "w" (aka, the same context, condition, etc). There is no question here.
Good, at least we're agreed on that.
 However the probabilities for scenario (b) are more accurately represented as P(A|x), P(B|y), P(C|z). You are saying that because for the specific physical scenario you described, P(A|x) = P(A|w) and P(B|y) = P(B|w) and P(C|z) = P(C|w), therefore P(C|z) <= P(A|x) + P(B|y) should be valid.
Yes, that is exactly what I'm saying. If we know that p≤q+r, and we also know that p=p',q=q', and r=r', can't we conclude that p'≤q'+r'? How can you possibly disagree with that?
 But that is not very credible. The inequality is a deeper relationship between probabilities not just specific values of the probabilities.
This makes absolutely no sense to me. It's just an inequality. It just says that this number is less than or equal to this number plus this number. Then we're just using the substiution property of equality.

It seems so obvious to me. P(A|x)=.25, P(B|y)=.25, and P(C|z)=.75. Thus P(A|w)=.25, P(B|w)=.25, and P(C|w)=.75. And you agreed that P(C|w) ≤ P(A|w) + P(B|w). Then we reach the conclusion .75≤.25+.25. What do you think I'm doing wrong here?

 Quote by lugita15 And you agreed that P(C|w) ≤ P(A|w) + P(B|w). Then we reach the conclusion .75≤.25+.25.
The mismatch is 0.25 more than expected (by the additive law of probability)

What is the explanation for that per QM/QE?

Is it something like -

Since the two particles are entangled and act as one prior to interaction (with the polarizer):

the angle/orientation at both the polarizers will be "taken into account" and this causes the increase in mismatch.

side note: Do the calculation involve some sort of "joint probabilities" ?

 Quote by lugita15 Well, I suppose that's trivially true. Any regular probability can be written as "the probability that such-and-such is the case, given nothing."
"Given nothing", you have nothing. But you always have something by which to define you probability space otherwise any calculation is meaningless.
 Sorry, what does P(ABC) mean?
That is the joint probability for the events A, B, C.
 But you're putting a set of photons on the other side of the bar I think you mean "The probability that A is true for the photons in w", not "The probability that A is true given w", which doesn't make sense to me.
If A = "mismatch at -30 and 0". P(A|w) is the probability of mismatch at -30 and 0 for the probability space "w". If "w" was all we were ever talking about, it will make sense to simply leave it out and write P(A) but that will not mean P(A) is not a conditional probability. However your argument is not such a case; therefore to to be precise and avoid mistakes, it is recommended to write P(A|w).
 How can you possibly disagree with that?This makes absolutely no sense to me. It's just an inequality. It just says that this number is less than or equal to this number plus this number. Then we're just using the substiution property of equality.
I can disagree with that because I'm not limitting my thinking to an expression on a sheet of paper or on screen. I can disagree because I understand the difference between an empirical manipulation and an algebraic manipulation. You are focusing on the algebraic and ignoring the empirical. This is what I've been trying to tell you about Boole's work.

And if you attempt to tell me what the joint probability P(ABC) is for both scenario (b) and (c) then you may start having a paradigm shift to understand the issue I have with your argument. I suppose your no-conspiracy condition implies that they two are equal. But that is impossible because P(ABC) is undefined for scenario (b). Let me even make it easier: your expression P(C|w) <= P(A|w) + P(B|w) is actually derived from the equality P(C|w) = P(A|w) + P(B|w) - P(AB|w) so let us stay with the equality for a moment. Please use your so called "no-conspiracay condition" to write down the equivalent equality for scenario (b) with appropriate notation. As soon as you start writing it you will realize that after the second term on the RHS you are stuck, there is no P(AB|..) term because A and B are incompatible experiments. So contrary to appearances, your "no-conspiracy" inequality P(C|z) <= P(A|x) + P(B|y) is nonsense. By writing it as an inequality, you have hidden the error of the impossible P(AB|..) term.

And that means you are not allowed, ever, to use P(A|x), P(B|y) and P(C|z) in the same expression the way you do. You can only do that if the three are defined over the exact same probability space where the joint probabilities are well defined. Since you insist on doing the forbidden, don't blame anything else for violations. Going back to the inequality for a bit, if you insist on using the inquality and obtain a violation for scenario (b), such a violation means P(AB|..) is negative which is impossible! But of cause that is because you insisted on using an impossible probability P(AB|..) to start with so "garbage-in", "garbage out".

This is what Boole published 150 years ago. Please read the article I cited earlier.

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Quote by billschnieder
 Sorry, what does P(ABC) mean?
That is the joint probability for the events A, B, C.
I assume you mean P(A and B and C), more compactly written as P(A & B & C).
 If A = "mismatch at -30 and 0". P(A|w) is the probability of mismatch at -30 and 0 for the probability space "w". If "w" was all we were ever talking about, it will make sense to simply leave it out and write P(A) but that will not mean P(A) is not a conditional probability. However your argument is not such a case; therefore to to be precise and avoid mistakes, it is recommended to write P(A|w).
It's fine if you want to use that notation, but my only quibble was that usually you put a statement on each side of the bar, you don't put a probability space.
 I can disagree with that because I'm not limitting my thinking to an expression on a sheet of paper or on screen. I can disagree because I understand the difference between an empirical manipulation and an algebraic manipulation. You are focusing on the algebraic and ignoring the empirical.
Well, we are talking about numbers, so naturally algebra is where I turn.
 And if you attempt to tell me what the joint probability P(ABC) is for both scenario (b) and (c) then you may start having a paradigm shift to understand the issue I have with your argument.
Sorry, I'm not having any paradigm shift. Of course A is only a meaningful statement in x, B is only a meaningful statement in y, C is only a meaningful statement in z, and all three statements are meaningful in w. So of course "A & B & C" is a meaningless statement in any of the three probability spaces in (b), and thus P(A & B & C) is not well-defined in any of them.
 I suppose your no-conspiracy condition implies that they two are equal.
No, it doesn't. It's a conspiracy if the probability of mismatch you would get if you measured at -30 and 0 depends on whether you actually measure at -30 and 0, because the particles don't "know" whether they're going to be measured at -30 and 0. But it's not a conspiracy if C, and thus P(C), is well-defined for w but not for x. That just means that (c) allows for counterfactual reasoning and (b) doesn't.
 Let me even make it easier: your expression P(C|w) <= P(A|w) + P(B|w) is actually derived from the equality P(C|w) = P(A|w) + P(B|w) - P(AB|w) so let us stay with the equality for a moment.
Actually, you forgot a factor of 2. The correct equation is P(C|w)=P(A XOR B|w)=P(A|w)+P(B|w)-2P(A & B|w). That's because if both A and B are true, then C is false.
 Please use your so called "no-conspiracay condition" to write down the equivalent equality for scenario (b) with appropriate notation. As soon as you start writing it you will realize that after the second term on the RHS you are stuck, there is no P(AB|..) term because A and B are incompatible experiments.
Yes, the statement "A & B" is meaningless in any of the three probability spaces in (b), and thus P(A & B) is not well-defined in any of three probability spaces in (b), the equation doesn't make any sense using probabilities in (b).
 So contrary to appearances, your "no-conspiracy" inequality P(C|z) <= P(A|x) + P(B|y) is nonsense. By writing it as an inequality, you have hidden the error of the impossible P(AB|..) term.
Look, I see no contradiction in saying the following:
1. The equation is meaningless in (b).
2. The equation is meaningful and correct in (c).
3. The equation in (c) implies the inequality in (c).
4. The inequality in (c) implies the inequality in (b)
5. The inequality is meaningful and correct in (b).

Why do you see a contradiction in this? It's true that the equation doesn't apply to (b), but that is irrelevant. The reason the inequality applies to (b) is not because the equation applies to (b), but rather because the inequality applies to (c), which is ultimately because the equation applies to (c).
 And that means you are not allowed, ever, to use P(A|x), P(B|y) and P(C|z) in the same expression the way you do. You can only do that if the three are defined over the exact same probability space where the joint probabilities are well defined.
But they're just numbers. Why am I not allowed to use numbers in whatever expression I want to?
 Since you insist on doing the forbidden, don't blame anything else for violations.
But how in the world is it forbidden?
 Going back to the inequality for a bit, if you insist on using the inquality and obtain a violation for scenario (b), such a violation means P(AB|..) is negative which is impossible! But of cause that is because you insisted on using an impossible probability P(AB|..) to start with so "garbage-in", "garbage out".
Yes, a violation of the inequality in (b) implies a violation of the inequality in (c), which implies that P(A & B|w) is negative, which is impossible. But I am never invoking the ill-defined probabilities P(A & B|x), P(A & B|y), or P(A & B|z), so I don't see any garbage-in, garbage-out here.

Your argument is that no-conspirary allows you to conclude that the inequality P(C|w) <= P(A|w) + P(B|w) which is valid for scenario (c) must be valid for scenario (b), ie P(C|z) <= P(A|x) + P(B|y) because according to you the no-conspiracy implies that P(C|z) = P(C|w) and P(A|x)=P(A|w) and P(B|w)= P(B|y). This is your argument.
My argument is is the following:
The inequality P(C|w) <= P(A|w) + P(B|w) is obtained from P(C|w) = P(A|w) + P(B|w) - 2P(AB|w). So if you are claiming that the inequality should apply to scenario (b) it means the equality should also apply. You can not take the inequality and reject the equality, the inequality is just a different view into the what is already there in the equality.
 Yes, the statement "A & B" is meaningless in any of the three probability spaces in (b), and thus P(A & B) is not well-defined in any of three probability spaces in (b), the equation doesn't make any sense using probabilities in (b).
You agree that the *equality* does not make sense for scenario (b) yet you insist on using the inequality. You claim that 3 of the 4 terms in the equality are the same for both (b) and (c) due to your so called "no-conspiracy condition". However, you conveniently hide the 4th term by removing it and changing the equality sign to an inequality. By doing that, you are assuming that the meaningless undefined probability which does not exist has a value. Yet you do not see the problem. This is why I asked you earlier to derive the inequality directly from scenario (b) using any assumptions you like without going through (c). But you didn't and can't.
What more do you want me to say if you refuse to see the fallacy.
 Look, I see no contradiction in saying the following: 1. The equation is meaningless in (b). 2. The equation is meaningful and correct in (c). 3. The equation in (c) implies the inequality in (c). 4. The inequality in (c) implies the inequality in (b) 5. The inequality is meaningful and correct in (b).
 Why do you see a contradiction in this?
Because there is one.
 It's true that the equation doesn't apply to (b), but that is irrelevant.
It is very relevant as I have explained. The inequality does not exist in the ether. You can not have a valid inequality if the equality is invalid. This much is obvious from you inability to derive the inequality directly under scenario (b) without invoking scenario (c) at all.
 The reason the inequality applies to (b) is not because the equation applies to (b), but rather because the inequality applies to (c), which is ultimately because the equation applies to (c).
This is precisely the fallacy you continue to make. You can not put P(A|x) and P(B|y) and P(C|z) together in a single expression and calculate anything empirically meaningful. There is no alternate logic, or alternate probability theory which allows you do do that. Yet you insist on doing just that.
 But they're just numbers. Why am I not allowed to use numbers in whatever expression I want to?
You are free to use them in whatever expression you like. But don't fool yourself to think the result will be meaningful in any way. This is what Boole, the father of Boolean logic, showed 150 years ago, which I've been encouraging you to read.
 Yes, a violation of the inequality in (b) implies a violation of the inequality in (c), which implies that P(A & B|w) is negative, which is impossible. But I am never invoking the ill-defined probabilities P(A & B|x), P(A & B|y), or P(A & B|z), so I don't see any garbage-in, garbage-out here.
As soon as you write down your inequality and associate it with scenario (b), you have hidden those terms under a rug. They are embedded inside the inequality sign. So contrary to your claim that you are not using them, you are infact using them. I take back my suggestion that anyting was "forbidden". You are ofcourse free to do what you want but there is nothing more I can say. I have better things to do with my time.

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 Quote by billschnieder Your argument is that no-conspirary allows you to conclude that the inequality P(C|w) <= P(A|w) + P(B|w) which is valid for scenario (c) must be valid for scenario (b), ie P(C|z) <= P(A|x) + P(B|y) because according to you the no-conspiracy implies that P(C|z) = P(C|w) and P(A|x)=P(A|w) and P(B|w)= P(B|y). This is your argument.
Yes, it is.
 My argument is is the following: The inequality P(C|w) <= P(A|w) + P(B|w) is obtained from P(C|w) = P(A|w) + P(B|w) - 2P(AB|w). So if you are claiming that the inequality should apply to scenario (b) it means the equality should also apply. You can not take the inequality and reject the equality, the inequality is just a different view into the what is already there in the equality.
It is true that the equation in (c) implies the inequality in (c). But the inequality is its own statement, even if it's implied by another statement. Let me write things more explicitly so we can key in on our disagreement:
1. (P(A|w)+P(B|w)-P(C|w))/2=P(A & B|w)
2. P(A & B|w)≥0
3. (P(A|w)+P(B|w)-P(C|w))/2≥0
4. (P(A|x)+P(B|y)-P(C|z))/2≥0
Now I'm open in my "hiding" of a term. What I'm saying is that 1 and 2 imply 3, and even though 1 and 2 have no analogue for (b), it is nevertheless true that 3 implies 4.
 However, you conveniently hide the 4th term by removing it and changing the equality sign to an inequality. By doing that, you are assuming that the meaningless undefined probability which does not exist has a value. Yet you do not see the problem.
No, I'm not assuming that P(A & B) is defined in any of the three probability spaces in (b). The equation implies the inequality, but the inequality need not imply the equation.
 This is why I asked you earlier to derive the inequality directly from scenario (b) using any assumptions you like without going through (c). But you didn't and can't.
But there's a fundamental reason for that: counterfactual definiteness is a crucial assumption for Bell's theorem. If you were the kind of person who rejected counterfactual reasoning, in other words rejected reasoning in (c) as invalid, then this argument can't persuade you. If you restrict yourself to reasoning in (b), i.e. if you don't believe in counterfactual definiteness, then you can only carry out the argument for EPR, not Bell.
 Look, I see no contradiction in saying the following: 1. The equation is meaningless in (b). 2. The equation is meaningful and correct in (c). 3. The equation in (c) implies the inequality in (c). 4. The inequality in (c) implies the inequality in (b) 5. The inequality is meaningful and correct in (b).
What is the logical contradiction you see?
 This much is obvious from you inability to derive the inequality directly under scenario (b) without invoking scenario (c) at all.
I explained my inability to do so.
 This is precisely the fallacy you continue to make. You can not put P(A|x) and P(B|y) and P(C|z) together in a single expression and calculate anything empirically meaningful. There is no alternate logic, or alternate probability theory which allows you do do that. Yet you insist on doing just that.
OK, I still don't understand your point, since I'm just doing valid operations with numbers, but what if I side-stepped the issue and reasoned as follows:
1. P(C|w)≤P(A|w)+P(B|w)
2. P(A|x)=.25, P(B|y)=.25, P(C|z)=.75
3. P(A|x)=P(A|w), P(B|y)=P(B|w), P(C|z)=P(C|w)
4. P(A|w)=.25, P(B|w)=.25, P(C|w)=.75
5. .75 ≤ .25 + .25

What step do you think I'm going wrong in?

 Quote by lugita15 1. P(C|w)≤P(A|w)+P(B|w) 2. P(A|x)=.25, P(B|y)=.25, P(C|z)=.75 3. P(A|x)=P(A|w), P(B|y)=P(B|w), P(C|z)=P(C|w) 4. P(A|w)=.25, P(B|w)=.25, P(C|w)=.75 5. .75 ≤ .25 + .25 What step do you think I'm going wrong in?
1. P(C|w)=P(A|w)+P(B|w) - 2P(AB|w)
2. P(A|x)=.25, P(B|y)=.25, P(C|z)=.75, P(AB|x?y?)=????????
3. P(A|x)=P(A|w), P(B|y)=P(B|w), P(C|z)=P(C|w), P(AB|x?y?)=?????????
4. P(A|w)=.25, P(B|w)=.25, P(C|w)=.75, P(AB|w)=??
5. .75 = .25 + .25 + ??????????????????

Steps (1), (2), (3), (4) and (5). You are hiding the dirt under the inequality "rug". Enough said. If you still do not understand my argument that's your problem.
 A short note that: Bell's theorem, though fairly convincing, is not, as many of you are aware, the only "proof" of non-local action Other experiments, for example those listed below, also provide evidence of quantum entanglement - Two-photon interference - where interference is demonstrated (at a distance) without the entangled photons meeting at the beam-splitter which is not replicable via two (un-entangled) photons - Quantum entangled swapping - where photons are entangled (via their "twins") without ever meeting - improvements in Bell's test detection efficiency

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 Quote by billschnieder You are hiding the dirt under the inequality "rug".
OK, let me be more explicit in my logic and not use inequalities at all.
1. P(C|w)=P(A|w)+P(B|w)-2P(A & B)|w)
2. P(A|x)=.25, P(B|y)=.25, P(C|z)=.75
3. P(A|x)=P(A|w), P(B|y)=P(B|w), P(C|z)=P(C|w) (From no-conspiracy.)
4. P(A|w)=.25, P(B|w)=.25, P(C|w)=.75 (From 2 and 3)
5. .75 = .25 + .25 -2P(A & B|w) (From 1 and 4)
6. P(A & B|w) = -.125 (From 5)

Now what step is in error? I am no longer "hiding" ill-defined probabilities like P(A & B|x) behind inequality signs.

 Quote by lugita15 OK, let me be more explicit in my logic and not use inequalities at all. 1. P(C|w)=P(A|w)+P(B|w)-2P(A & B|w) 2. P(A|x)=.25, P(B|y)=.25, P(C|z)=.75 3. P(A|x)=P(A|w), P(B|y)=P(B|w), P(C|z)=P(C|w) (From no-conspiracy.) 4. P(A|w)=.25, P(B|w)=.25, P(C|w)=.75 (From 2 and 3) 5. .75 = .25 + .25 -2P(A & B|w) (From 1 and 4) 6. P(A & B|w) = -.125 (From 5) Now what step is in error? I am no longer "hiding" ill-defined probabilities like P(A & B|x) behind inequality signs.
Your argument is the same and the criticism is the same. P(AB|w) disappears in Step (2) and reappears in step (5). What happened to it in steps (2), (3) and (4)?????? In step (5) you have effectively the frankenstein equation P(C|z) = P(A|x) + P(B|y) - 2P(AB|w) from which you conclude in step (6) that 2P(AB|w) is negative! Why do you leave out the P(AB|..) terms from (2), (3) and (4)? Because you know that by including them, it reveals your error. Your equation (1) contains the term 2P(AB|w). How come your no-conspiracy condition can make a statement about P(C|w), P(A|w), and P(B|w) but conveniently ignores P(AB|w)????? Isn't it curious that your "no-conspiracy" condition "just happens" to be silent about P(AB|w), the same probability on which your conclusion (6) hinges????? That is the conspiracy of your "no-conspiracy".

How many different ways do I have to explain this?

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