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## What does this example say about the applicability of Bell's inequalities?

 Quote by billschnieder Every time a button is pressed, calculate calculate sin(t) where t is the time read off the internal clock. If sin(t) > 0, program coin "a" to be biased for heads (+1) and coin "c" to be biased for tails (-1). Then randomly pick two of the three coins. If coin "b" is one of the picks, program coin "b" to be biased for tails (-1) if the other pick is coin "a", otherwise program coin "b" to be biased for heads (+1)[corrected]. If sin(t) <= 0 reverse all the signs.
Bill,

This algorithm, by your own admission, produces -1 (perfect correlation) for every trial. No reason to put the other stuff in. You may as well say that a perfectly correlated pair comes out every time with a random orientation of +1-1 or -1+1.

So then the question becomes: does a perfectly correlated outcome violate a Bell Inequality? That is after all the criteria for the EPR state (elements of reality). Clearly the answer should be NO if this is a good analogy, because this case was considered specifically by Bell. He considered this as being simple.

But here the answer is actually YES (as you say), because we have not specified any angle settings. You get the same results at all angle settings, including 0 and 90 (which should be anti-correlated relative to each other).

So we conclude the following:

a) This analogy does not mimic a Bell setup as it produces predictions which are counter to any experiment. Ergo, it predicts a different result than QM, which matches experiment. Therefore the main result of Bell, stated below stands:

No physical theory of local Hidden Variables can ever reproduce all of the predictions of Quantum Mechanics.

b) It is not truly local, as the results are observer dependent in most cases (whenever b is one of the 3 selected, which occurs about 2/3 of the time). In other words, the outcome b is predicated on whether the observer gets a or c for the other coin. This is a contextual setup, which essentially will always violate the locality/separability condition. Note that b appears to be undefined when a and c are selected, though you could fix that and would need to if you claim to have a realistic example.

This analogy doesn't pass the test, no matter what you say about your Bell Inequality being violated.

 Quote by Delta Kilo ...... it's the same bloody thing!
No it is not. Bell's theorem relates the inequalities to QM. The inequalities themselves do not need QM to be derived. Get it? Just because the inequalities are relevant for the theorem does not mean the inequalities by themselves are not worthy of independent discussion without being side-tracked by reference to the theorem.

 1. Do you agree that your setup does not satisfy Bell's "vital assumption", as quoted earlier?
I disagree that the "vital assumption" quoted earlier is relevant to this thread. Simply look at the OP and explain why the proof of the inequality stated in the first part of the OP should not apply to the second part of the OP. Everything is contained in the openning post. Not need to waste time looking elsewhere.
 2. Do you agree that the derivation of Bell's inequality requires "viltal assumption" to be valid?
Again, look at the openning post where the inequality is proven by brute force calculation of all the possibilities. Is it your claim that the inequality is different from Bell's? Just because it was proven differently? So your so called "vital assumption" may be interesting for another thread not this one.

 3. Can you now make a great leap of logic and put 1. and 2. together and agree that Bells inequalities are inapplicable to your setup?
Again it is you who is failing to make the great leap of logic: We have the proof of the inequalities for three coins labelled "a", "b", "c", not angles. You do not argue that the inequality which is proven in the openning post is invalid. Neither do you argue that the inequality is not Bell's inequality. Then we have an example using three coins which violates the inequality just proven. The issue for discussion is: How come? Everything required for answering the question is contained in the openning post. You do not need any reference to QM predictions because there is none for the specific example with coins mentioned in the OP. We can go off trail discussing how "angles" are missing from the example, or how distant separation is missing, but nobody expects coins to be tossed at angles etc.

So again here are the issues for this thread.
1) Are the inequalities proven in the openning post valid for the situation described?
2) How come it is violated by the experiment described?
3) What does this mean about the applicability of inequalities of the type proven in the OP, to experiments of the type described in the OP?

Hi DrC,
Thanks for correcting the typo in your quote.
 Quote by DrChinese This algorithm, by your own admission, produces -1 (perfect correlation) for every trial.
Actually, it is perfect anti-correlation since the product of the two coins is always -1. Perfect correlation would be a product of +1 every time.

 So then the question becomes: does a perfectly correlated outcome violate a Bell Inequality? That is after all the criteria for the EPR state (elements of reality). Clearly the answer should be NO if this is a good analogy, because this case was considered specifically by Bell.
If the question was changed to "does a perfectly anti-correlated outcome violate a Bell Inequality?", the is YES. To verify replace each of the expectation values with a -1 and you get 3 on the LHS.

 You get the same results at all angle settings, including 0 and 90 (which should be anti-correlated relative to each other).
Again as I explained to DK, focusing on angles just muddies the water and takes us off topic. There are no angles involved in the example. "a", "b", "c" are just coins. In case you are tempted to think angles are relevant for "Bell's inequalities" as opposed to "Bell's theorem", remember than "a", "b", "c" in the mathematical derivation performed by Bell are just symbols. The derivation will proceed in exactly the same way to obtain the inequalities, no matter what those symbols represented. The algebra used to derive the inequalities makes absolutely no use of the idea that values for "a", "b", "c" can be changed, which means the inequalities are valid even if "a", "b", "c" are fixed. In our case "a", "b", "c" are just labels for three coins. The fact that the inequalities are valid even for the case in which "a", "b", "c" represent three coins is proven in the openning post. Just in case this is not clear still. Imagine that in the actual EPR experiment, the angles were fixed to just those in which a violation occured, and Alice and Bob were not allowed to change them. Would you argue in that instance that the inequality is not valid because they were not allowed to change the angles? The simple fact that a violation occured whether the angles could be changed or not is significant. That is the scenario you should be comparing this example with, if any.

 So we conclude the following: a) This analogy does not mimic a Bell setup as it produces predictions which are counter to any experiment. Ergo, it predicts a different result than QM, which matches experiment. Therefore the main result of Bell, stated below stands:
And what may those predictions of QM for a coin-toss experiment be? You are comparing apples and oranges here. No claim is made that this example reproduces QM results. The issue is the inequalities, period.
 b) It is not truly local
Now that is an interesting claim. You think the example as described in the openning post is non-local? Please elaborate.

 In other words, the outcome b is predicated on whether the observer gets a or c for the other coin.
But the coins are produced that way by the box due to conservation of it's mechanism of functioning.

 This is a contextual setup, which essentially will always violate the locality/separability condition.
Really? At any one moment in time, each of the three coins has a definite outcome value. Prior to the tossing of the coins so it is not contextual. Unless you have a different meaning for the word.

 Note that b appears to be undefined when a and c are selected, though you could fix that and would need to if you claim to have a realistic example.
I did not tell you what b was in that case but of course you understand that it could be easily fixed and it will be realistic and still violate the inequality.

So let us see:

- Every outcome is predetermined
- Every outcome is known for certain before the toss
- Every coin has a definite state before the toss
- The box and coins operate in a local and realistic manner.
- The outcomes are non-contextual (not obeserver dependent) -- there is nothing the observers can do to change the results except refuse to toss.

Yet the inequality is violated. Why?

 Quote by lugita15 It's not that the Bell inequality will never be violated in such a case, it's that the inequality won't be derivable in the first place.
Really? Look again at the openning post. The very first part provides a proof of the inequality for the very example involving three coins labelled "a", "b", "c". So your claim that the inequalities are not derivable is just not true.
 This example that the OP gives is very good in showing the problems you're driven into when considering the results of mutually exclusive experiments, as done in the quantum case as well. On one hand, i can see that this classical situation isn't entirely similar to the quantum case, since we have no freedom on which coins we will measure. You give me two specific coins, you tell me 'measure them' and this way the inequality is violated 'by hand'. It's like putting the numbers in the inequality by hand. If i was able to choose however which of the 3 coins to measure, the inequality would not be violated.. On the other hand, it's not yet clear to me what happens in the quantum case since we use results of mutually exclusive experiments to violate the Bell inequality.. Ofcourse, in this case we have freedom of measurement choice, something that isn't present in the OP's example. But still, im not sure yet if the freedom of choice is enough to assure that including results of exclusive experiments is consistent and doesn't lead to absurd results. One first thought on this problem is to try to consider a classical experiment (using hidden varibles), where given freedom of measurement choice we can still violate Bell's inequality. If it turns out that this isn't possible, then we can safely conclude that there's something strange about quantum mechanics :). But if it is possible then we have a serious problem... My personal opinion is that it's not possible, but i'll give it a little more thought these days.

 Quote by billschnieder Is it your claim that the inequality is different from Bell's?
Yes. In your 'proof' the inequality applies to averages from the same series of $\{a_i,b_i,c_i\}$, where the average is defined as arithmetic mean of all elements in the series, even though you did not explicitly state it. It does not allow you to pick different subsets to calculate <ab>, <ac>, and <bc>. But of course, since you did not articulate your proof, you have missed this. If, instead of handwaving, you actually try to write out the steps, this is as far as you can get. Bell's inequality, on the other hand, applies to expectation values. It is a high time you learn the difference between the two.

 Quote by billschnieder Just because it was proven differently?
Of course. Applicability of any formula depends on the underlying assumptions.
 Quote by billschnieder 1) Are the inequalities proven in the openning post valid for the situation described?
No.
 Quote by billschnieder 2) How come it is violated by the experiment described?
Because your so-called 'proof' does not hold water.
 Quote by billschnieder 3) What does this mean about the applicability of inequalities of the type proven in the OP, to experiments of the type described in the OP?
That you have to learn the difference between average and expectation value. Didn't I tell you that about a month ago?

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 Quote by billschnieder 1. Actually, it is perfect anti-correlation since the product of the two coins is always -1. Perfect correlation would be a product of +1 every time... If the question was changed to "does a perfectly anti-correlated outcome violate a Bell Inequality?", the is YES. To verify replace each of the expectation values with a -1 and you get 3 on the LHS. 2. Again as I explained to DK, focusing on angles just muddies the water and takes us off topic. There are no angles involved in the example. "a", "b", "c" are just coins. In case you are tempted to think angles are relevant for "Bell's inequalities" as opposed to "Bell's theorem", remember than "a", "b", "c" in the mathematical derivation performed by Bell are just symbols... 3. Now that is an interesting claim. You think the example as described in the openning post is non-local? Please elaborate... - The outcomes are non-contextual (not obeserver dependent) -- there is nothing the observers can do to change the results except refuse to toss. 4. Yet the inequality is violated. Why?
1. I know that in your particular example, perfect correlation is a -1 result. So I am not questioning the arithmetic.

2. Well, this does matter if you are setting up a comparison to Bell. There must be 3 angles, usually labelled a/b/c or A/B/C, so that Alice can pick 1 and Bob can pick 1. The selection is done independently by each. This is necessary so we can have condition that the result will be local.

I am not questioning that you have the right to specify the outcomes in advance; in fact to be realistic, you MUST do this for all 3. Your example meets this criteria sufficiently.

3. Here is a key issue: you change the results according to which 2 of a/b/c are selected. This violates one of the Bell conditions which is necessary to get the inequality. You cannot bypass this and expect to convince anyone that this is a local realistic simulation. The rule is: you must specify the possible results independent of what attribute (a/b/c) the observers Alice & Bob freely choose to examine.

4. The root rule is not the CHSH inequality. The root requirements are:
i) the possibility of any permutation must be within the range of 0 to 1 (realism);
ii) the outcome cannot change based on what the observers choose to look at (locality or separability or whatever you want to call it).

In your example, i) is satisfied but ii) is not. Note that QM predicts values outside the range of i). You can see that negative probabilities are such a prediction at:

http://drchinese.com/David/Bell_Theo...babilities.htm

To convince yourself that your example, if properly respecting ii), would not violate any type of Bell inequality, just write down a set of data points for your a/b/c (we have had this discussion previously of course). To make things work out for you, I will make it simple: I will always select a for one of the two, and will randomly pick between b and c. That way, you can respect ii) (since the outcome b does not change). Below are the only 2 permutations:

a b c
+ - -
- + +

This has <ab>=-1, <ac>=-1, <bc>=+1, and the inequality is not violated (as it was when requirement i was violated). QED.

 Quote by DrChinese 2. Well, this does matter if you are setting up a comparison to Bell. There must be 3 angles, usually labelled a/b/c or A/B/C, so that Alice can pick 1 and Bob can pick 1. The selection is done independently by each. This is necessary so we can have condition that the result will be local.
Again I ask you, if Alice and Bob are given their devices already fixed to a given angle, will you then argue that violation of the inequality is not significant? If not why do you think that line of argument is relevant here?
 3. Here is a key issue: you change the results according to which 2 of a/b/c are selected.
You forget that the experimenters do not select which coins they get. Whatever they get, that is what they measure. So it is wrong to say the results are "changed" based on what is selected. The coins they get is the coins they get. It is not up to them to "select" any thing so the results are not changed based on the situation. The results for each toss are predetermined, not contextual.
 This violates one of the Bell conditions which is necessary to get the inequality.
This is the crux of the matter. Could you elaborate what condition exactly which is violated. Please do not tell me about angles because the proof in the opening post does not use any angles yet it is a valid proof. So what condition required to obtain the proof in the OP is violated by the experiment in the OP.
 You cannot bypass this and expect to convince anyone that this is a local realistic simulation.
This is common sense. You do not have to look at any inequality to see that the experiment is local and realistic. There is no spooky action at a distance, there is no instantaneous influence. So it is surprising that you continue to claim that the experiment is non-local. In fact this experiment can be performed by humans. Just place a human in the box give him a stop-watch and a hand calculator, and give him 3 coins labelled "a", "b", "c". It is clearly obvious that everything is local and realistic and local and you know it.
 The rule is: you must specify the possible results independent of what attribute (a/b/c) the observers Alice & Bob freely choose to examine. 4. The root rule is not the CHSH inequality. The root requirements are: i) the possibility of any permutation must be within the range of 0 to 1 (realism); ii) the outcome cannot change based on what the observers choose to look at (locality or separability or whatever you want to call it).
Again, there is no free chosing of coins in the example in the OP. Yet the inequality is proven (see the first part of the OP). We calculate the LHS for every possible combination and observe that none of them ever violates the inequality. That is sufficient proof that the inequality is valid for three coins labelled "a", "b", "c", freedom of choice is just a red-herring in this example. Secondly as already explained, the outcomes are predetermined and the observers have no say in choosing what they look at, so it is wrong to suggest that the outcome in this example changes based on what observers choose to look at. The predetermined outcome in this example simply changes with time and that is sufficient to violate the inequality maximally.
 In your example, i) is satisfied but ii) is not. Note that QM predicts values outside the range of i). You can see that negative probabilities are such a prediction at: http://drchinese.com/David/Bell_Theo...babilities.htm
Negative probabilities do not make sense, unless you redefine what is mean by "probability". But that is fodder for a different thread.
 To convince yourself that your example, if properly respecting ii), would not violate any type of Bell inequality, just write down a set of data points ...
No need to, I have already proven the inequality in the first part of the OP by calculating for every possibility. Yet we have here an obviously local and realistic macroscopic experiment which violates the proven inequalities for a specific reason which has nothing to do with locality or realism or freedom of choice. Why.

 Quote by Delta Kilo Yes. In your 'proof' the inequality applies to averages from the same series of $\{a_i,b_i,c_i\}$, where the average is defined as arithmetic mean of all elements in the series, even though you did not explicitly state it. It does not allow you to pick different subsets to calculate , , and . But of course, since you did not articulate your proof, you have missed this. If, instead of handwaving, you actually try to write out the steps, this is as far as you can get.
Interesting you throw that accusation given that you do exactly the same thing when analyzing the EPR experiment. Note: this example was designed to reveal precisely this kind of misunderstanding. Now it is you making the argument which you have repeatedly rejected without understanding. I'm happy you are getting the point, which you have repeatedly refused to understand when I made it. Now you are forced to make the argument yourself when faced with an obviously local and realistic example which violates the inequalities.
 Bell's inequality, on the other hand, applies to expectation values. It is a high time you learn the difference between the two.
Really DK? I thought you knew better than to make such a ridiculous claim:
 Quote by Wikipedia The expected value may be intuitively understood by the law of large numbers: the expected value, when it exists, is almost surely the limit of the sample mean as sample size grows to infinity. More informally, it can be interpreted as the long-run average of the results of many independent repetitions of an experiment (e.g. a dice roll). ... In quantum mechanics, the expectation value is the predicted mean value of the result (measurement) of an experiment.
Now even a cave man can understand that <ab> is the expectation value of paired product of outcomes a*b for a pair of dice thrown a very large number of times, just like in the OP example. You need to learn about expectation values, einstein. I did not hear you complaining that Aspect, or Weihs, etc calculated averages from their experiments not expectation values. Such a suggestion will be laughed at, or rather cried at.

 Quote by JK423 This example that the OP gives is very good in showing the problems you're driven into when considering the results of mutually exclusive experiments, as done in the quantum case as well.
Thanks. I'm happy the point got across.

 But still, im not sure yet if the freedom of choice is enough to assure that including results of exclusive experiments is consistent and doesn't lead to absurd results. One first thought on this problem is to try to consider a classical experiment (using hidden varibles), where given freedom of measurement choice we can still violate Bell's inequality. If it turns out that this isn't possible, then we can safely conclude that there's something strange about quantum mechanics :). But if it is possible then we have a serious problem...
This has been done by the group of Hans De Raedt. See http://arxiv.org/abs/1112.2629 However, many people still do not understand their result and simply wave it off. The OP was meant as way to re-cast their results in a manner that is simple enough to highlight the just the major issues while eliminating the peripheral stuff which often obstructs understanding of the major issues.
 My personal opinion is that it's not possible, but i'll give it a little more thought these days.
If you haven't already. Please check out the above article from De Raedt.

 Quote by billschnieder Interesting you throw that accusation given that you do exactly the same thing when analyzing the EPR experiment.
Ha! I knew it would come to that! I didn't even mention EPR, did I? It was obvious from the very first post that you don't have a genuine question to ask, that this is yet another silly attempt at sneak attack on Bell's inequalities.

 Quote by billschnieder Now even a cave man can understand that is the expectation value of paired product of outcomes a*b for a pair of dice thrown a very large number of times, just like in the OP example.
No, this is not always the case, and it is specifically not true in your example. Having a large sample is not enough, sample selection procedure has to be unbiased. I can give you plenty of examples where biased sampling procedure will give you all sorts of different average values from the same random sequence. But I'm afraid this is too much for a caveman.

Bill, I do not get your line of reasoning at all. You present your own 'proof' and then you present a counterexample which invalidates your proof. Then you somehow drag EPR into discussion, and then you blame Bell for your own stuff-ups, while at the same time refusing to pay attention to his "vital assumption". Please stop that.

 Quote by Delta Kilo ... sample selection procedure has to be unbiased.
Aren't the sample selection procedures in bill's coin experiment and in Bell tests both biased, ie., both outcome dependent?

 Quote by billschnieder How come?
The inapplicability of the inequality to the coin experiment seems obvious. I remember you phrasing this in a succinct way in one of your posts (in another thread), but I forgot exactly how it goes.

 Quote by billschnieder How come the inequalities which were supposed to be valid for the "coin-toss" experiment as demonstrated earlier in the OP, get violated by the experiment?
The inequality is based on three simultaneously existing values. The experiment can only generate two values at a time.

 Quote by billschnieder ... we assume the inability to measure all three simultaneously is inconsequential.
I take this to be your vital assumption. That is, this is the assumption that is contradicted via the violation of the inequality by the coin-toss test.

I also take it that this is what you consider to be the effective cause of BI violation in Bell tests. Which would mean that what Bell stated as being the vital assumption was not the vital assumption, and the locality (or independence) condition encoded in Bell's formulation is precluded from being the effective cause of BI violation.

Bell inequalities can be formulated without encoding a locality or independence assumption. So, for now I suppose I agree with your analysis and assessment -- until somebody explains it better.

 Quote by Delta Kilo Bill, I do not get your line of reasoning at all. You present your own 'proof' and then you present a counterexample which invalidates your proof. Then you somehow drag EPR into discussion, and then you blame Bell for your own stuff-ups, while at the same time refusing to pay attention to his "vital assumption". Please stop that.
Doesn’t Bell do the exactly the same. That is, Bell presented his own proof and then he used quantum predictions as a counterexample to disprove it. Then concluded the locality assumption was impossible or false.

Could you convincingly explain what assumption Bill made that is impossible or false?

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 Quote by rlduncan Could you convincingly explain what assumption Bill made that is impossible or false?
This was already done, and virtually anyone should be able to see that Bill's example is not apropos. The below reasoning would be cited in one form or another in any response to Bill's "question". I am just sorry that Bill's deception has caused you confusion.

 Quote by DrChinese 4. The root rule is not the CHSH inequality. The root requirements are: i) the possibility of any permutation must be within the range of 0 to 1 (realism); ii) the outcome cannot change based on what the observers choose to look at (locality or separability or whatever you want to call it). In your example, i) is satisfied but ii) is not. Note that QM predicts values outside the range of i). You can see that negative probabilities are such a prediction at: http://drchinese.com/David/Bell_Theo...babilities.htm To convince yourself that your example, if properly respecting ii), would not violate any type of Bell inequality, just write down a set of data points for your a/b/c (we have had this discussion previously of course). To make things work out for you, I will make it simple: I will always select a for one of the two, and will randomly pick between b and c. That way, you can respect ii) (since the outcome b does not change). Below are the only 2 permutations: a b c + - - - + + This has =-1, =-1, =+1, and the inequality is not violated (as it was when requirement i was violated). QED.

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 Quote by billschnieder Why.
 Quote by billschnieder This has been done by the group of Hans De Raedt. See http://arxiv.org/abs/1112.2629 However, many people still do not understand their result and simply wave it off. The OP was meant as way to re-cast their results in a manner that is simple enough to highlight the just the major issues while eliminating the peripheral stuff which often obstructs understanding of the major issues.
You pretend to ask "why" when it is obvious you are not asking a question at all. You are trying to TELL us something which again, is non-standard physics and represents your personal theories.

So to answer the title question and wrap things up: This example has NO connection to Bell and is not applicable in any way. To show you how inapplicable it is, let's morph it to this example, which is exactly equivalent. There is a bag, and in it are 3 marbles. Some are red and some are green. We reach in and get 2 out, always 1 red and 1 green. By your [sarcastic adjective omitted] reasoning, this too violates the CHSH inequality but is local and realistic. I hope it is clear that this is EXACTLY the ORIGINAL EPR paradox and was considered explicitly by Bell and rejected over 40 years ago.

I'm sorry Bill, but you are approaching childishness here and yet another new low. If it were up to me (and it is not), I would shut this thread down now that it is completely clear what your true purpose is. This is not really the place for debate on your personal beliefs. I will definitely report you if you continue this charade.

 Quote by DrChinese You pretend to ask "why" when it is obvious you are not asking a question at all.
I'm clearly asking a question. The inequality is valid and yet it is violated, there is a reason for it and that is the question. To discuss that reason. And you do understand the question from your responses so far, although you do not like the question because it reveals your misunderstanding. That is hardly grounds to close a thread because DrC does not like it.

[quote]You are trying to TELL us something which again, is non-standard physics and represents your personal theories.[/b]
Everything in the openning post is standard physics. The inequality proven in the OP post is Bell's inequality and the treatment of data from the experiment in the OP is very similar to how data is treated in EPR experiments. It is funny that you are willing to spend time discussing in threads about an unpublished "Herbet's proof" and your own personal proofs from your non-peer-reviewed website, and it is OK for you to refer other readers to your non-peer-reviewed personal theories about a nonsensical idea such as "Negative Probabilities" but as soon as I start discussing valid published peer-reviewed material which you don't like, you start throwing suggestions to moderators to lock the thread.

The views discussed here are published in the following articles which apparently you are unware of:
EPL, 87 (2009) 60007, http://arxiv.org/abs/0907.0767
J. Comp. Theor. Nanosci. 8, 1011 - 1039 (2011), http://arxiv.org/abs/0901.2546
Optics Communications 170 (1999) 55-60 http://arxiv.org/abs/quant-ph/0101094
Optics Communications 170 (1999) 61-66 http://arxiv.org/abs/quant-ph/0101087
 So to answer the title question and wrap things up: This example has NO connection to Bell and is not applicable in any way.
You are wrong, it has a connection. You do not argue that Bell's inequality should not apply to three coins in the manner described in the first part of the OP because you know that it should. In fact, you have often used similar arguments to push your unpublished so-called "DrC Challenge". In any case, feel free to wrap up yourself. We will do just fine without your whining.
 To show you how inapplicable it is, let's morph it to this example, which is exactly equivalent. There is a bag, and in it are 3 marbles. Some are red and some are green. We reach in and get 2 out, always 1 red and 1 green. By your [sarcastic adjective omitted] reasoning, this too violates the CHSH inequality but is local and realistic.
That makes no sense, especially as I've made no such argument.
 I'm sorry Bill, but you are approaching childishness here and yet another new low.
That's a funny accussation coming from someone who throws a tantrum whenever their beliefs are challenged.
 If it were up to me (and it is not), I would shut this thread down now that it is completely clear what your true purpose is. This is not really the place for debate on your personal beliefs. I will definitely report you if you continue this charade.
Be my guest. In fact it would more useful if you please direct all your off-topic complains about the thread to the moderators rather than littering the thread with unfounded accusations. I will definitely report you if you continue to disrupt the discussion with your accussations. It is very clear what your purpose is. You want to shutdown any discussion that goes against your personal beliefs. Why are you so afraid?

 Tags bell's inequality, bell's theorem, epr paradox