Joy Christian's colourful exploding balls experiment

[Heinera]

And let me add that this is a verbatim implementation of the analysis step Joy Christian has prescribed in his papers. In no way can Richard Gill be accused of "rigging" the computations. If Joy had just played around with some numerical experiments some seven years ago, we wouldn't be discussing this now.

[Ben6993]

To Minkwe: I will try to follow your last post in detail given more time, and will get back again later. But in general terms I do not follow it at the moment. I may not understand CHSH fully but as it seems to me right now, one could divide the CHSH statistic by 4 and use it as an estimate of the average correlation over the four sets of data. It would have a maximum of +1 but there is no need to look beyond 0.7071. A correlation of 0.61 would be good enough for Joy. I expect the outcome to focus on either 0.5 or 0.7071, and any deviation would be experimental error.

In the many simulations completed, the abs value of the correlation for angle(a-b) = 45 or 135 degrees is very near to 0.7071. An experimental outcome like that would give Joy success. This does not correspond exactly to CSHS because the difference of 45 degrees in (at least some of) the simulations can arise in many different ways, eg (a,b) = (65, 20 degrees) or (165,120). I do not yet understand why limiting (a,b) to a few particular values undermines confidence in the calculation of the correlation of 0.7071. The CHSH statistic seems to me to be very related to the average correlation.

To Heinera: I will also look at your last post in detail given more time, and get back to you. At the moment your position seems to me to be that one cannot achieve a correlation of 0.7071 based on data reported in a table. If one could do that it would already have been done in the Randi challenge, so no surprises there. The many recent simulations all seem to have generated correlations of about 0.7 which are good enough to win the Nobel prize, except for the snag of the 'loophole' interpretations. But none of the simulations AFAIK made self-standing tables of the data which could then be used to generate the correlations in a Randi-like challenge.

So it seems to me that you may want to calculate the correlation coefficients in some kind of graph, as in the simulations, rather than make a self-standing table of raw data. That may be OK as long as none of the data are rejected unnecessarily and raising suspicions
of loopholes. The CHSH cutoff of 2.4 corresponds to an average correlation coefficient of 0.6 and no matter how you calculate the coefficient in an alternative method, I cannot see it being different to the CHSH estimate of it. It should not matter at all to Richard how you calculate the correlation coefficient so long as there are no loopholes.

[Heinera]

To Heinera: I will also look at your last post in detail given more time, and get back to you. At the moment your position seems to me to be that one cannot achieve a correlation of 0.7071 based on data reported in a table. If one could do that it would already have been done in the Randi challenge, so no surprises there. The many recent simulations all seem to have generated correlations of about 0.7 which are good enough to win the Nobel prize, except for the snag of the 'loophole' interpretations. But none of the simulations AFAIK made self-standing tables of the data which could then be used to generate the correlations in a Randi-like challenge.

Exactly. If someone shows up with vectors in a table, the game is over before it began. It is then impossible to beat CHSH or meet the Randi challenge.

[minkwe]

To Minkwe: I will try to follow your last post in detail given more time, and will get back again later. But in general terms I do not follow it at the moment. I may not understand CHSH fully but as it seems to me right now, one could divide the CHSH statistic by 4 and use it as an estimate of the average correlation over the four sets of data. It would have a maximum of +1 but there is no need to look beyond 0.7071. A correlation of 0.61 would be good enough for Joy. I expect the outcome to focus on either 0.5 or 0.7071, and any deviation would be experimental error.

In the many simulations completed, the abs value of the correlation for angle(a-b) = 45 or 135 degrees is very near to 0.7071. An experimental outcome like that would give Joy success. This does not correspond exactly to CSHS because the difference of 45 degrees in (at least some of) the simulations can arise in many different ways, eg (a,b) = (65, 20 degrees) or (165,120). I do not yet understand why limiting (a,b) to a few particular values undermines confidence in the calculation of the correlation of 0.7071. The CHSH statistic seems to me to be very related to the average correlation.

That is what I've tried to explain using the coin analogy. Once you understand the point, the rest will be so obvious you'll slap yourself in the face for having missed it. These questions may help you understand:


Case 1 (single coin):
A = Outcome from tossing just 1 coin (heads = +1, tails=-1)
B = Other outcome we could have gotten but did not (heads = +1, tails=-1)

A + B = 0, why?
<A> + <B> = 0, why?
E(A) + E(B) = 0, why?

Case 2 (two coins, each the same type as in case 1):
A = Outcome from tossing the first coin (heads = +1, tails=-1)
B = Outcome from tossing the *second* coin (heads = +1, tails=-1)

A + B <= 2, why?
<A> + <B> <= 2, why?
E(A) + E(B) <= 2, why?

Experiment I: 2 special coins like in case 2:
<A> + <B> = 0.5, appears to violate <A> + <B> = 0, does it mean coins are not local realistic? Experimental error? Spooky business?

Experiment II: Just one of the special coins above, like in case 1:
<A> + <B> = 0, does not violate <A> + <B> = 0, how can the same coins violate <A> + <B> = 0, in one experiment but not another? Experimental error? Spooky business?

[gill1109]

Ben suggests we average + E(0, 135), - E(0, 45), - E(90, 45), and - E(90, 135) and compare it to 0.6 (bigger? or smaller?). I think this is a brilliant idea. For many reasons we should downplay CHSH. We must avoid all careless use of the words "inequality", "bound". We must distinguish theory, in which there can be bounds and inequalities, and experiment, where there can be none.

Instead look at the quantity CHSH/4 thinking of it as an estimate, not a bound. It's an average correlation.

According to QM,

E(0, 45) = E(90, 45) = E(90, 135) = - 0.7
E(0, 135) = + 0.7

According to the formerly best commonly known LHV model (ie, Joy's model excepted)

E(0, 45) = E(90, 45) = E(90, 135) = - 0.5
E(0, 135) = + 0.5

In a CHSH experiment (no inequality! no bound!) we focus on just those four correlations. They are of course typically measured with experimental error and we carefully explain how big that error might be when we write up our experiment for publication in "Nature". If N is about 10 000 for each of the four samples then the experimental error will be of the order of size of 0.01. Excellent! But if it is only about 100 for each of the four samples then the experimental error will be of the order of size of 0.1. Weak!

The reason we look at four correlations and not just one or two is because it is important that both Alice and Bob can and do vary their settings.

The reason we look at four correlations and not a whole "correlation surface" E(alpha, beta) is economics. The most efficient (economical) experiment just looks at four points on this surface. For instance, this paper http://arxiv.org/abs/quant-ph/0307125 shows that if Joy is right, and if he wants to prove to the world that he is right, but he only has a limited amount of cash to pay to the experimenter, then the experiment which we are going to do (CHSH) is the best experiment of all. Better than GHZ, better than Hardy, better than original Bell (aka QRC: quantum Randi challenge).

A first draft ("alpha testing", proof of concept...) of code (implementing Joy's suggestions from his paper) to analyse the data from Joy's experiment has been posted at
http://rpubs.com/gill1109/Bet
It's the proposed actual analysis of the data from the bet, agreed by Joy. It reads directions from two files. At the end it does a little simulation to exhibit and draw the triangle wave. That is not part of the bet, but just a little test/illustration of the code. (I don't know how to make a cosine wave within this framework: Joy asked me for the triangle, because he knows I know how to do that).

Since this code was written by R Gill, in a language called R which lots of people don't know, Joy Christian can be very suspicious. We need that other people look at it and that there are indendent translations by people he trusts into other languages so yet more people can check and double check that the algorithm is OK and all implementations are equivalent. Fantastic that there is now an Excel version too!

When we do the real experiment, Joy has agreed that we will end up with two computer files looking like the two files which are analysed by this code. Two (small) test files are on internet:
http://www.math.leidenuniv.nl/~gill/AliceDirections.txt
http://www.math.leidenuniv.nl/~gill/BobDirections.txt

Code: Select all

My proposed code to analyse the experimental data is

## I use mathematician's notation: theta is the azimuthal angle, phi is the polar
## (aka zenith) angle; both measured in radians.
## Reference: https://en.wikipedia.org/wiki/Spherical_coordinates
## Since my measurement directions all lie in equatorial plane I just extract "theta"


AliceDirections <-
    read.table("http://www.math.leidenuniv.nl/~gill/AliceDirections.txt")
   
names(AliceDirections) <- c("theta", "phi")

head(AliceDirections) ## N pairs theta, phi (N rows, 2 columns)

NAlice <- nrow(AliceDirections)

NAlice

AliceTheta <- AliceDirections$theta # Alice's azimuthal angles

head(AliceTheta)

BobDirections <-
    read.table("http://www.math.leidenuniv.nl/~gill/BobDirections.txt")
   
names(BobDirections) <- c("theta", "phi")

head(BobDirections)

NBob <- nrow(BobDirections)

NBob

if (NAlice != NBob) print("Error: particle numbers don't match") else
    print("Go ahead!")

BobTheta <- BobDirections$theta  # Bob's azimuthal angles

head(BobTheta)

## First pair of measurement directions

Alpha <- 0 * pi / 180
Beta <- 45 * pi / 180
A <- sign(cos(AliceTheta - Alpha))
B <- - sign(cos(BobTheta - Beta))
E11 <- mean(A * B)

## Second pair of measurement directions

Alpha <- 0 * pi / 180
Beta <- 135 * pi / 180
A <- sign(cos(AliceTheta - Alpha))
B <- - sign(cos(BobTheta - Beta))
E12 <- mean(A * B)

## Third pair of measurement directions

Alpha <- 90 * pi / 180
Beta <- 45 * pi / 180
A <- sign(cos(AliceTheta - Alpha))
B <- - sign(cos(BobTheta - Beta))
E21 <- mean(A * B)

## Fourth pair of measurement directions

Alpha <- 90 * pi / 180
Beta <- 135 * pi / 180
A <- sign(cos(AliceTheta - Alpha))
B <- - sign(cos(BobTheta - Beta))
E22 <- mean(A * B)

CHSH <- E12 - E11 - E21 - E22

CHSH

if (CHSH > 2.4) print("Congratulations, Joy") else
    print("Congratulations, Richard")

## Another experiment

AliceTheta <- runif(1000, 0, 360) * pi / 180
BobTheta <- - AliceTheta

Delta <- seq(from = 0, to = 360, by = 10) * pi / 180
Correlation <- numeric(length(Delta))
A <- sign(cos(AliceTheta))
i <- 0
for (delta in Delta) {
    i <- i+1
    B <- - sign(cos(BobTheta - delta))
    Correlation[i] <- mean(A * B)
}
plot(Correlation)

Michel, I am looking forward so much to seeing your Python version.

[gill1109]

PS by the way as well as forgetting about bounds and inequalities (except for Michel's universal bound 4) we could also forget about correlations.

(1 + E(a, b))/2 is a number between 0 and 1, it's the probability of equal outcomes on both sides of the experiment.

(1 - E(a, b))/2 is a number between 0 and 1, it's the probability of unequal outcomes.

CHSH/4 was an average correlation (with some switching of signs).

(4 + CHSH)/8 is a number between 0 and 1, it's an average probability (theory), an average relative frequency (experiment).

Notice how Michel's bound "4" comes in here. Probabilities (theoretical and empirical) lie between 0 and 1.

A CHSH experiment measures an average probability, which according to local realism can at most equal 0.75 +/- experimental error, while according to quantum theory it can be as high as 0.85 +/- experimental error. Of course in an experiment one can see anything from 0 to 1. It is important to know the error bars.

[gill1109]

Nice spreadsheet, Heinera! Brilliant.

Question to Joy: do you accept that we settle our bet by reading your two experimental data files into Excel and doing the calculations by Excel? Or is it fine to read the two experimental data files into R and do the calculations in R. Or do you prefer Java, Python, Mathematica, ....

Michel has offered to do the Python version (and even some others: Ruby for instance) if you would like him to. I think it is very important that we put several equivalent computer programs on the table so that the adjudicators can use whatever they are comfortable with, and that everyone agrees that they all do the right thing, and all do the same thing.

This last point is the most important. There must be no accusations of rigging, of cheating. Everyone who is interested in the experiment has to know in advance what exactly is going to happen. We all go into this, eyes wide open.

Technical question: suppose we have 100 000 exploding balls. Can Excel cope with such large files?

This whole forum is here, thanks to Fred, because of a discussion which started elsewhere between Fred and me. I argued that Joy's experiment was doomed to failure. Fred argued that I had rigged Joy's experiment, or that the whole concept of CHSH was rigged. I argued that Joy had fallen into a trap which he had rigged for himself. Now we are getting close to resolving this whole issue! It's very exciting.

The bet between Joy and me is about four correlations E(0, 45), E(0, 135), E(90, 45), E(90, 135) coming out of Joy's experiment. We both expect three to be negative, one (the second) to be positive. Are they all, up to statistical error, +/- 0.7 (Joy) or +/- 0.5 (Richard)?

[Joy Christian]

I am not sure what all the fuss is about.

Why are people getting excited about producing little computer programs which do nothing more than implement four elementary mathematical equations (the four equations below) which I can write in literally 20 seconds?

**********************************************************************************

For the record, let me repeat that equation (16) of my attached
experimental paper describes exactly how the expectation values
E(a, b), E(a', b), E(a, b'), and E(a', b') are to be computed in my
proposed experiment. Four separate sums are to be calculated as
follows

E(a, b) = 1/N Sum_j A_j B_j ,

E(a, b') = 1/N Sum_j A_j B'_j ,

E(a', b) = 1/N Sum_j A'_j B_j ,

and

E(a', b') = 1/N Sum_j A'_j B'_j .

It is a matter of indifference whether N here is chosen to be the same
or different for each of the four alternatives.

The experimental procedure described in my paper is unambiguous.

**********************************************************************************

Further details of my proposed experiment can be found on my blog.

[Heinera]

I am not sure what all the fuss is about.

Why are people getting excited about producing little computer programs which do nothing more than implement four elementary mathematical equations (the four equations below) which I can write in literally 20 seconds?

Because in order to raise money for the experiment by crowd funding, donors would ask for proof of concept. "I hear some people are saying that it is mathematically impossible to generate quantum correlations from a list of pairs of vectors in the way you specify. If that is true, the whole experiment will be pointless, of course. So can you show us a hypothetical list that generates the results you hope to see?"

Now you have the tools to go hunting for that list.

[Joy Christian]

I am not sure what all the fuss is about.

Why are people getting excited about producing little computer programs which do nothing more than implement four elementary mathematical equations (the four equations below) which I can write in literally 20 seconds?

Because in order to raise money for the experiment by crowd funding, donors would ask for proof of concept. "I hear some people are saying that it is mathematically impossible to generate quantum correlations from a list of pairs of vector in the way you specify. If that is true, the whole experiment will be pointless, of course. So can you show us a hypothetical list that generates the results you hope to see?"

Now you have the tools to go hunting for that list.

No need to go hunting. You can find several such lists on my blog.

[gill1109]

No need to go hunting. You can find several such lists on my blog.

I do not see the two files there, which we need to input to my program.

Please give us the links to the two data files.

If you have them, and if they work, then as far as I am concerned the experiment is superfluous and you have already won your bet from me, and a multitude of public apologies, just like that.

Regarding the whole discussion about inequalities and bounds ...

I made a little R script to draw the correlation surfaces E(a, b) and just four points on them (a) under QM and/or Joy's model, (b) under the traditional "best" LHV model.

The point being: a CHSH-syle experiment tests four points on the surface. Let's forget the word "bound" and the word "inequality". They lead to endless misunderstanding.

But please do let's realize that an experiment is always subject to experimental error, statistical error. We don't determine those four points exactly, but only approximately. If N = 100 (per point) the error will be about 0.1. Not good. If N = 10 000 (per point) the error will be about 0.01. Plenty enough.

http://rpubs.com/gill1109/Wireframe

[Joy Christian]

http://rpubs.com/gill1109/Wireframe

Nice graphics! I haven't checked the four "red" points, but I am sure someone will.

Can you superimpose the two 3D plots and colour the four points differently so that we can immediately see the difference in the respective predictions?

[gill1109]

http://rpubs.com/gill1109/Wireframe

Nice graphics! I haven't checked the four "red" points, but I am sure someone will.

Can you superimpose the two 3D plots and colour the four points differently so that we can immediately see the difference in the respective predictions?

Thanks for the compliment! I am going to try to do this, but it is difficult, and I am using a part of R which I'm not familiar with. Seems to me that this ought to be more easy. Maybe someone can do a better job with Mathematica or GnuPlot or whatever?

But my next try will be to put both sets of points in the same plots, with different colours.

UPDATE: I now have both sets of points on both surfaces. You can see the experimental challenge. These are the points where the difference is largest!

I think that drawing two surfaces which are so close together in one plot is not going to work. Or require major tricks with semi-transparency and so on. (My grandson, helping me at the bottle bank, recently surprised me by describing the colour of plain glass bottles as "invisible" - as opposed to green and brown bottles).

By the way my real motivation for being here is not winning bets or pursuing some kind of personal feude or anything like that, but learning new tools for science outreach. For that purpose the very best people to talk to are clever people who strongly disagree with you but who don't quite know everything you know. You don't quite know everything they know, either. The biggest problem is communication. This can only be a win-win-win situation: learn how to communicate, share knowledge, gain knowledge.

[minkwe]

Regarding the whole discussion about inequalities and bounds ...

I made a little R script to draw the correlation surfaces E(a, b) and just four points on them (a) under QM and/or Joy's model, (b) under the traditional "best" LHV model.

Richard, make an effort to understand my little coin example. Once you do, you'll immediately withdraw your paper.

The point being: a CHSH-syle experiment tests four points on the surface. Let's forget the word "bound" and the word "inequality". They lead to endless misunderstanding.

You keep talking of CHSH. CHSH is fatally flawed as I've already demonstrated. Why do you keep bringing up CHSH.

But please do let's realize that an experiment is always subject to experimental error, statistical error. We don't determine those four points exactly, but only approximately. If N = 100 (per point) the error will be about 0.1. Not good. If N = 10 000 (per point) the error will be about 0.01. Plenty enough.

Again make an effort to understand the coin example, and you'll see that experimental/statistical error are red-herrings. Bell's theorem is just one big silly mistake.

[minkwe]

Ben suggests we average + E(0, 135), - E(0, 45), - E(90, 45), and - E(90, 135) and compare it to 0.6 (bigger? or smaller?). I think this is a brilliant idea.

Really?! Because it allows you to hide the fact that you are using the same debunked CHSH logic? Why is it not enough to show that for any angle pair you pick E(a,b) agrees with QM. Why-o-why are you tethered to that CHSH expression?

[Ben6993]

Richard, I have looked at your R program and you are using the same formula for the sawtooth curve as me i.e. -1+ 2* angle /180.

I assumed, and noted so in an earlier post, that the maximum difference was 0.707-0.5 = 0.207 at angle = 45 degrees.

But now my calculations show that the maximum is for angle = 40 degrees.

Angle xxxxx -Cosine(angle) xxxxx -1+ 2* angle /180 xxxxx Difference
40 xxxxx -0.766 xxxxx -0.556 xxxxx -0.210
45 xxxxx -0.707 xxxxx -0.5 xxxxx -0.207

I had assumed that 45 deg was chosen because it gave the maximum difference, but maybe not? I may have made a silly calculation slip up, but I have checked it several times and cannot find an error.

[Heinera]

Ben suggests we average + E(0, 135), - E(0, 45), - E(90, 45), and - E(90, 135) and compare it to 0.6 (bigger? or smaller?). I think this is a brilliant idea.

Really?! Because it allows you to hide the fact that you are using the same debunked CHSH logic? Why is it not enough to show that for any angle pair you pick E(a,b) agrees with QM. Why-o-why are you tethered to that CHSH expression?

And when will you understand that if for any angle pair you pick, E(a,b) agrees with QM, will imply that E(0, 135), E(0, 45), E(90, 45), and E(90, 135) must also agree with QM, since 0, 45, 90, and 135 is part of "any"?

[minkwe]

Ben suggests we average + E(0, 135), - E(0, 45), - E(90, 45), and - E(90, 135) and compare it to 0.6 (bigger? or smaller?). I think this is a brilliant idea.

Really?! Because it allows you to hide the fact that you are using the same debunked CHSH logic? Why is it not enough to show that for any angle pair you pick E(a,b) agrees with QM. Why-o-why are you tethered to that CHSH expression?

And when will you understand that if for any angle pair you pick, E(a,b) agrees with QM, will imply that E(0, 135), E(0, 45), E(90, 45), and E(90, 135) must also agree with QM, since 0, 45, 90, and 135 is part of "any"?

Heinera,
You too should make an effort to understand the coin analogy. Your question shows that you have not understood it. If any angle pair you pick agrees with QM then of course (0, 135), (0, 45), (90, 45) and (90, 135), being angle pairs also agree with QM. Nobody denies this trivial obvious fact. Angle pairs are angle pairs. That you will think I don't understand that angle pairs are angle pairs is just silly.

What you still do not understand is that results of *all the 4 angle pairs from a single set of particles* is different from results from 4 angle pairs from 4 different sets of particles. EVEN IF those 4 sets of particles are from the same population as the single set. That you still do not understand this point is shown clearly in this question you asked in the other thread:

So if the original correlations (all computed on the whole set) didn't violate the CHSH inequality (CHSH<2), and the correlations computed on four disjoint random subset would not change much, we can now conclude that the four latter correlations would still not significantly violate the CHSH inequality, since term by term, they are approximately equal to the original correlations?

viewtopic.php?f=6&t=44

[gill1109]

Ben suggests we average + E(0, 135), - E(0, 45), - E(90, 45), and - E(90, 135) and compare it to 0.6 (bigger? or smaller?). I think this is a brilliant idea.

Really?! Because it allows you to hide the fact that you are using the same debunked CHSH logic? Why is it not enough to show that for any angle pair you pick E(a,b) agrees with QM. Why-o-why are you tethered to that CHSH expression?

What do you mean by "any", Michel? Do you mean all, or some, or a few? Is one enough? Must it be a million? If so how must the adjudicating committee decide the winner? We have to give them a simple unambiguous recipe. What would yours be?

It is actually enough to show that for four angle pairs, experimentally observed/calculated E(a,b) agrees with QM theoretical E(a,b), provided the four pairs are chosen sensibly. Moreover, if Joy wants to maximize his chance of winning of proving that his correlations do not fit to my theory, he should go for the smallest possible number of pairs, and if he takes the smallest number, four, then he should go for the optimal four pairs: (0, 45), (0, 135), (90, 45), and (90, 135).

I want to do the most effective experiment. It is better to look closely at a small number of angle pairs where the difference is large, than to look at all of the angle pairs at the same time. This has been known for a long long long long time...

Why the hell should we waste time measuring the correlation for pairs of angles where there is almost no difference between Joy's theory and my theory?

It is a question of economics. If we can only afford to observe 10 000 pairs of particles, and if we can only measure each particle once, what is the smartest way to distribute our resources? Answer: using the *four* pairs (0, 45), (0, 135), (90, 45), and (90, 135)

I have no interest whatever anymore in discussing CHSH inequality or Bell theorem here. I am interested in discussing an experiment, and I want the experiment to be as powerful as possible, whatever the truth might be. It must have the biggest possible chance of proving Joy right, if he is right. It must have the biggest possible chance of proving me right, if I am right. The best experiment of all just measures four correlations, at setting pairs (0, 45), (0, 135), (90, 45), and (90, 135). (or equivalent... there are other optimal choices. They are essentially the same. You know the various symmetries in this experiment, which everyone agrees on).

And I am still waiting for Michel's solution to my exercise, and after that, his Python translation of the program which evaluates the winner of the bet.

[Heinera]

Really?! Because it allows you to hide the fact that you are using the same debunked CHSH logic? Why is it not enough to show that for any angle pair you pick E(a,b) agrees with QM. Why-o-why are you tethered to that CHSH expression?

And when will you understand that if for any angle pair you pick, E(a,b) agrees with QM, will imply that E(0, 135), E(0, 45), E(90, 45), and E(90, 135) must also agree with QM, since 0, 45, 90, and 135 is part of "any"?

Heinera,
You too should make an effort to understand the coin analogy. Your question shows that you have not understood it. If any angle pair you pick agrees with QM then of course (0, 135), (0, 45), (90, 45) and (90, 135), being angle pairs also agree with QM. Nobody denies this trivial obvious fact. Angle pairs are angle pairs. That you will think I don't understand that angle pairs are angle pairs is just silly.

So you say that E(0, 135), E(0, 45), E(90, 45), and E(90, 135) will have the quantum correlations in Joy's experiment. Richard (and I as well) says E(0, 135) will be 0.5, the three others -0.5. Where is the "debunked CHSH logic" in that? The criterion that the average should be compared to 0.6 is just a way to still settle the bet if the correlations should turn out to be somewhere in between 0.5 and 0.7071.

What you still do not understand is that results of *all the 4 angle pairs from a single set of particles* is different from results from 4 angle pairs from 4 different sets of particles. EVEN IF those 4 sets of particles are from the same population as the single set. That you still do not understand this point is shown clearly in this question you asked in the other thread:

So if the original correlations (all computed on the whole set) didn't violate the CHSH inequality (CHSH<2), and the correlations computed on four disjoint random subset would not change much, we can now conclude that the four latter correlations would still not significantly violate the CHSH inequality, since term by term, they are approximately equal to the original correlations?

viewtopic.php?f=6&t=44

I understand very well that the answer to the question I asked you is "yes". I hoped you understood it as well.