The bet on Christian's experiment

[Joy Christian]

But Joy, if you believe you are home and dry, then you put your 10 000 Euro on the table too, and we take the data file to the adjudicators?

If 10 000 is too much risk for you, we can revert to the original 5 000

No, No, No. I am not falling for this. Your 10,000 Euros offer has nothing to do with the adjudicators or the actual experiment. It is a one-sided offer. You made an offer of 10,000 Euros to anyone who can produce the desired N vectors before June 11 and you are stuck with it. You are not getting out of it by bobbing and weaving.

[gill1109]

I am not bobbing and weaving. I am trying to find out how serious you are. Do you think you have won already, yes or no?

If you still think yes, then we need an independent person to check the four correlations, agree? Because I think "no".

Of course I am interested in getting some money out of this very one-sided bet, after all, but you are within your rights to stick with agreements made so far.

[Joy Christian]

I am not bobbing and weaving. I am trying to find out how serious you are. Do you think you have won already, yes or no?

If you still think yes, then we need an independent person to check the four correlations, agree? Because I think "no".

Of course I am interested in getting some money out of this very one-sided bet, after all, but you are within your rights to stick with agreements made so far.

I will stick with the one-sided offer you have made.

Now that I am beginning to understand how you are testing the correlations, I will learn to test them myself. Then we shall see whether I have won already or not.

[gill1109]

Now that I am beginning to understand how you are testing the correlations, I will learn to test them myself. Then we shall see whether I have won already or not.

I am testing the correlations according to your written instructions, precisely as incorporated in our bet. I still do recommend you ask Fred to calculate them in Mathematica and Michel in Python and Chantal in Java, just in case. I mean ... I might be a lousy R programmer ...

Even your simulation program which created those numbers has parts in it written by me. Have everything checked and double checked by people you trust.

PS also with this one-sided bet: I insist that it is adjudicated. You submit a file, I calculate a correlation. Independent adjudicators tells us who has won. This is not bobbing and weaving. This is a practical necessity. If you deliver by June 11, and claim victory, I'll ask our other board of adjudicators to verify that for us. Is that clear?

[Joy Christian]

PS also with this one-sided bet: I insist that it is adjudicated. You submit a file, I calculate a correlation. Independent adjudicators tell us who has won. This is not bobbing and weaving. This is a practical necessity. If you deliver by June 11, and claim victory, I'll ask our other board of adjudicators to verify that for us. Is that clear?

Clear enough; and also fair enough. But for the record let me note here that this is a new condition, which was not stated in the original offer of 10,000 Euros.

[gill1109]

PS also with this one-sided bet: I insist that it is adjudicated. You submit a file, I calculate a correlation. Independent adjudicators tell us who has won. This is not bobbing and weaving. This is a practical necessity. If you deliver by June 11, and claim victory, I'll ask our other board of adjudicators to verify that for us. Is that clear?

Clear enough; and also fair enough. But for the record let me note here that this is a new condition, which was not stated in the original offer of 10,000 Euros.

True, it was not originally stated. Nor was its negation originally stated. But we have chosen one of those two, and agreed.

[Joy Christian]

PS also with this one-sided bet: I insist that it is adjudicated. You submit a file, I calculate a correlation. Independent adjudicators tell us who has won. This is not bobbing and weaving. This is a practical necessity. If you deliver by June 11, and claim victory, I'll ask our other board of adjudicators to verify that for us. Is that clear?

Clear enough; and also fair enough. But for the record let me note here that this is a new condition, which was not stated in the original offer of 10,000 Euros.

True, it was not originally stated. Nor was its negation originally stated. But we have chosen one of those two, and agreed.

Yes, but note that I will also submit my simulation and/or calculations to the adjudicators, with hopes that they will look at all pieces of evidence without prejudice.

[gill1109]

Richard said "PS also with this one-sided bet: I insist that it is adjudicated. You submit a file, I calculate a correlation. Independent adjudicators tell us who has won. This is not bobbing and weaving. This is a practical necessity. If you deliver by June 11, and claim victory, I'll ask our other board of adjudicators to verify that for us. Is that clear?"

Clear enough; and also fair enough. But for the record let me note here that this is a new condition, which was not stated in the original offer of 10,000 Euros.

True, it was not originally stated. Nor was its negation originally stated. But we have chosen one of those two, and agreed.

Yes, but note that I will also submit my simulation and/or calculations to the adjudicators, with hopes that they will look at all pieces of evidence without prejudice.

Sure. But the 10 000 Euro one-sided bet is about processing one (or maybe two) data-files according to an agreed protocol. So actually, how you created those numbers seems to me to be fairly irrelevant. Because this is what we agreed:

File of numbers -> four specific correlations calculated separately according to agreed formulas -> Joy wins or I win

So the only thing that we could disagree on would be the calculation (since not everyone knows the same computer languages etc etc and who knows maybe there is an ambiguity in our agreement, what that calculation should be) and that would be the only thing they would have to adjudicate on. But of course, everything will be in the public eye, so everyone who is interested can check, too. But, in case you and I still disagree, we need a board of adjudicators in order to utter the last word, even if it is a formality, because in the eyes of the world it is already clear who has won.

[minkwe]

Four separate calculations, according to the agreed formulas, implemented in R by Zen:

Code: Select all

N <- 10^7
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("JoyVector.txt", nlines = N))
alpha <- 0 * pi / 180
beta <- 45 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))

Code: Select all

N <- 10^7
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("JoyVector.txt", nlines = N))
alpha <- 0 * pi / 180
beta <- 135 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))

Code: Select all

N <- 10^7
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("JoyVector.txt", nlines = N))
alpha <- 90 * pi / 180
beta <- 45 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))

Code: Select all

N <- 10^7
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("JoyVector.txt", nlines = N))
alpha <- 90 * pi / 180
beta <- 135 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))

Results:

Code: Select all

[1] -0.7501332

[1] 0.2500942

[1] -0.7498742

[1] -0.2498984

I'm clearly not blind but I do not see sampling without replacement in the above.

[gill1109]

I'm clearly not blind but I do not see sampling without replacement in the above.

Correct. Joy Christian agreed to do the data-analysis according to his own instructions in his own experimental paper. You can read right at the beginning of this thread exactly what he and I agreed, both of us with our eyes wide open. We are not blind either.

That's why the sum is exactly 2

If instead each correlation would have been calculated on the basis of a disjoint random sample of about one quarter of the 10 million runs, then, with very large probability, the sum would have been very close to 2. In fact, 50% of the time it would have been a tiny bit more, 50% of the time it would have been a tiny bit less.

*If* Christian had asked for four correlations based on random disjoint subsamples then (a) I would have insisted that I was the one who did the sampling (it would have been honest random sampling, completely independent of the numbers in the computer files), and (b) I would have insisted that all the observed correlations were within 0.1 of their targets, not within 0.2 of their targets.

So I think, Michel, you agree that there is no way Joy can win the new bet?

Note: I am talking here about the recently agreed one-sided bet (I'll call it "the new bet"): 10 000 Euro from me for a computer file which wins the original bet about the experiment (see beginning of this thread, call it "the old bet"), it can be put together however you like. eg. by a simulation of Joy's model. To be delivered before 11 June. Thereafter only 5000 Euro. The offer is open to everyone. First come first served.

Do you also agree that if we switch to independent random subsampling for the four correlations, and if I insist on a greater accuracy (0.1 instead of 0.2), and if N remains as large as it now is, then I will win the modified new bet with very large probability?

Basically this is the silly computer experiment which I wanted you to do, some time back. It's good that we are now doing it "live", so to speak.

[Joy Christian]

Correct. Joy Christian agreed to do the data-analysis according to his own instructions in his own experimental paper.

Indeed. This is what the instructions in my experimental paper say: viewtopic.php?f=6&t=31&p=1651&hilit=replacement#p1651

Note that Richard Gill and I seem to read these instructions rather differently.

[gill1109]

Correct. Joy Christian agreed to do the data-analysis according to his own instructions in his own experimental paper.

Indeed. This is what the instructions in my experimental paper say: viewtopic.php?f=6&t=31&p=1651&hilit=replacement#p1651
Note that Richard Gill and I seem to read these instructions rather differently.

I think we agreed on the crucial elements of these instructions, though we obviously disagree on many other things. I personally do not see any ambiguity in them, either.

Anyway, this is what we agreed on (I mean, this is from the text of our bet):

Let's call the directions of angular momentum in Alice's file u_k, k=1,...,N, and in Bob's file v_k, k = 1, ..., N

If I pick measurement directions a and b, thinking now of directions as unit vectors in R^3, then according to Christian's experimental paper the outcomes left and right are

A_k = sign(a . u_k) and B_k = sign(b . v_k),

and the estimated (observed, sample, experimental ...) correlation is

E(a, b) = 1/N sum_k A_k B_k = ( N(++) + N(--) - N(+-) - N(-+) ) / ( N(++) + N(--) + N(+-) + N(-+) )

in the obvious notation.

There are just two computer files. Four pairs of directions a and b. Four correlations, each to be computed separately.

However Joy now only supplies one file, but we understand that is because the directions of angular momentum come in opposite pairs.

The following R code, written by Zen and checked by me, to the best of my capabilities, calculates just one of those correlations exactly as prescribed. It thereby assumes that Christian's single file contains the x and y coordinates of Alice's directions of angular momentum. Two numbers per row, 10^7 rows. Please just 10^5 next time!

Code: Select all

N <- 10^7
e <- matrix(nrow = 2, ncol = N, byrow = FALSE, data = scan("JoyVector.txt", nlines = N))
alpha <- 0 * pi / 180
beta <- 45 * pi / 180
a <- c(cos(alpha), sin(alpha))
b <- c(cos(beta), sin(beta))
mean(sign(colSums(e * a)) * -sign(colSums(e * b)))


If Christian wants to calculate each correlation with different code he had better inform us so that various people can independently check that my code, his code, and our agreed formula, all give the same outcome on the same data file and the same pair of directions a and b.

[minkwe]

I'm clearly not blind but I do not see sampling without replacement in the above.

Correct.

Why claim that you are calculating correlations separately and sampling without replacement when you are clearly not?

[gill1109]

I'm clearly not blind but I do not see sampling without replacement in the above.

Correct.

Why claim that you are calculating correlations separately and sampling without replacement when you are clearly not?

I do not claim that I am sampling without replacement.

I am doing exactly what Joy prescribes in his experimental paper, and what he has agreed with me that we will do, when we evaluate the outcome of his experiment. Read the first posting in this thread, carefully, please. Maybe you should read the relevant section of Joy's experimental paper, carefully, too. Not according to what you think Christian ought to have written, but according to what he actually did write.

I calculate the correlations "separately" in the sense that Christian intended. Not in one Nx4 spreadsheet. But based on the same set of directions of angular momentum of pairs of particles. Which as Joy's paper makes abundantly clear is completely OK by him. Great. So I took what he offered. I am not a fool. When I make bets like this I make them because I know I will win. Comprende?

Michel, if you think that this is a crazy bet which I am certain to win, please try to explain that to Christian. I am not responsible for what he wrote in his papers.

But if you think I would suddenly lose, if each correlation were computed, instead, on a disjoint random subsample (as long as you grant me also the freedom to reduce Joy's allowed margin of error to +/- 0.1), then I suggest you do the little R experiment I suggested some time ago, in order to learn a little bit about statistics, statistical error, standard errors, the one over square root of N law, and all that.

[minkwe]

I am doing exactly what Joy prescribes in his experimental paper, and what he has agreed with me that we will do, when we evaluate the outcome of his experiment.
...
I calculate the correlations "separately" in the sense that Christian intended.

Not according to Joy, you are not.

Not in one Nx4 spreadsheet. But based on the same set of directions of angular momentum of pairs of particles.

This is a joke right? You must think I'm stupid or something. I'm beginning to doubt whether you understand what Joy means by calculating correlations separately.

Michel, if you think that this is a crazy bet which I am certain to win, please try to explain that to Christian.

I'll advise you not to presume what I think. I'm tired of silly games, Joy says calculate them separately, and sample without replacement, just do that already. As a statistician you should know what he means. If you can't even do something as simple as that then what's the point. Haven't you been distracted from the main issue too long already by peripheral nonsense and bobbing and weaving about methodology? Isn't it tiring to still be arguing about method of calculation when the real interest is in whether the correlations can be reproduced or not? Unless the real issue is method of calculation rather than whether Joy's model can reproduce QM, in which case you can spare me from the fool's errand.

But if you think I would suddenly lose

Again I'll advise you not to presume what I think. Just calculate the correlations separately and sample without replacement. Enough is enough! What are you afraid of? That is the starting point anything else is just garbage. Until then see ya.

[Joy Christian]

OK:

I have revised the simulation, http://rpubs.com/jjc/16415, with the sample size 10^5.

I have unfixed the "b" directions and removed the "good" line. Now it cannot possibly be interpreted as having anything to do with the detection loophole.

2 x N vectors, e_k and -e_k are generated in the simulation. They produce a very good approximation to the -cosine curve, with neither "a" nor "b" fixed.

I claim that I have won the 10,000 Euros offered by Richard Gill, at least morally. Of course, he can still demand the two data files, which I am trying to produce.

Please do check out the last plot of the simulation to recognize that none of the four correlations, namely

E(0, 45) = - 0.7071...,

E(0, 135) = + 0.7071...,

E(90, 45) = - 0.7071...,

and

E(90, 135) = - 0.7071....,

deviate by more than 0.2 from the predicted values. That is all that matters to recognize the fact that I have already won the 10,000 Euros from Richard Gill.

[gill1109]

Michel, please keep on topic. No silly games.

What is the topic? It's the agreement, at the beginning of this thread, which Joy made with me. Read it carefully. If you have something new to say about it, go ahead. If you think that Christian is seriously disadvantaging himself, then you can tell him. Actually you already told him that but he took no notice.

But on a side issue: you think it makes any huge difference whether a correlation is computed on N pairs of numbers, or only on a random sample of about one quarter of them? (N is rather large, the numbers are all +/- 1). Why is that? We were earlier talking about a bet where Christian believes that some correlation is equal to 0.7071... while I think it is less than 0.5. Earlier, we took 0.6, the half way mark, as criterion for who wins, who loses.

Now that Joy is not interested at all in disjoint subsamples, so there is less randomness, less noise in the whole experiment, I have magnaminiously given Christian even more lee-way: now the threshhold (for who wins, who loses) is 0.5071... I could hardly be more generous than that!

Christian has generously agreed to dispense with disjoint random subsamples. (They were not mentioned in his experimental paper, anyway). Christian does not retract a word of his experimental paper, and we have gone right back to the roots. That's what we are betting on. You had better read his paper if you want to know what he might mean by "separately". I don't care much what he might have meant. I only care about what he actually wrote. Especially since he still stands by every word. I have generously expanded my earlier proposed criterion of 0.7071... +/- 0.1 to 0.7071... +/- 0.2. I have even given Christian the opportunity to win the bet three times over:

(a) if he can create the data files by simulation before June 11, he gets 10 000 Euro
(b) if the data files generated by his experiment with the exploding balls later also win the original bet, he gets 5 000 Euro.

The bet under (a) is even one-sided: if he doesn't do that, he still doen't have to pay anything
The bet under (b) is two-sided: if he loses, he does have to pay.

On the other hand, the bet under (a) is an open challenge: anyone can try to win the 10 000 from me, at no cost to themselves (after June 11, only 5 000).

The computer files coming out of Joy's experiment, if they do the job (0.7071... +/-0.2 for all four correlations of interest, but no random subsampling) win both (a) and (b) for Joy, if no-one has earlier claimed (a).

Seems to me that I am incredibly generous.

[gill1109]

OK:

I have revised the simulation, http://rpubs.com/jjc/16415, with the sample size 10^5.

I have unfixed the "b" directions and removed the "good" line. Now it cannot possibly be interpreted as having anything to do with the detection loophole.

2 x N vectors, e_k and -e_k are generated in the simulation. They produce a very good approximation to the -cosine curve, with neither "a" nor "b" fixed.

I claim that I have won the 10,000 Euros offered by Richard Gill, at least morally. Of course, he can still demand the two data files, which I am trying to produce.

Splendid! I hope for your sake that you are right. When you have the two data files ready, I am ready to process them. If they do the job then you indeed already morally won the 10 000. On the other hand, if they don't do the job, then you were wrong about that.

[gill1109]

Python code for computing the correlations (contributed by Zen). Same results as before.

Code: Select all

from numpy import pi, sign, sin, cos, radians, loadtxt
N = 10**7
e = loadtxt("JoyVector.txt", delimiter = " ")
alpha = radians(0)
beta = radians(45)
a = [cos(alpha), sin(alpha)]
b = [cos(beta), sin(beta)]
sum = 0
for i in range(0, N - 1):
    sum = sum + sign(e[i, 0] * a[0] + e[i, 1] * a[1]) * -sign(e[i, 0] * b[0] + e[i, 1] * b[1])
print sum / N
print -cos(alpha - beta)


Results:

Code: Select all

# Angles:           (0, 45)
# File correlation: -0.7501332
# QM  correlation:  -0.707106781186548
 
# Angles:           (0, 135)
# File correlation: 0.2500942
# QM  correlation:  0.707106781186546
 
# Angles:           (90, 45)
# File correlation: -0.7498742
# QM correlation:   -0.707106781186548
 
# Angles:           (90, 135)
# File correlation: -0.2498984
# QM  correlation:  -0.707106781186548


[gill1109]

Perl code for calculating the correlations (contributed by Zen). Same results yet again.

Code: Select all

#!/usr/bin/perl
 
use strict;
use constant PI => 3.14159265358979;
 
my $N = 10**7;
 
my $alpha = 0;
my $beta = 45;
my @a = (cos($alpha * PI / 180), sin($alpha * PI / 180));
my @b = (cos($beta * PI / 180), sin($beta * PI / 180));
 
my $sum = 0;
open(F, "JoyVector.txt") or die $!;
while (<F>) {
    chomp;
    my ($e_0, $e_1) = split(/ /, $_);
    $sum += sign($e_0 * $a[0] + $e_1 * $a[1]) * -sign($e_0 * $b[0] + $e_1 * $b[1]);
}
close(F);
 
print "Angles:           ($alpha, $beta)\n";
print "File correlation: ", $sum / $N, "\n";
print "QM correlation:   ", -cos(($alpha - $beta) * PI / 180), "\n";
 
sub sign { $_[0] >= 0 ? 1 : - 1; }


Results:

Code: Select all

# Angles:           (0, 45)
# File correlation: -0.7501332
# QM  correlation:  -0.707106781186548
 
# Angles:           (0, 135)
# File correlation: 0.2500942
# QM  correlation:  0.707106781186546
 
# Angles:           (90, 45)
# File correlation: -0.7498742
# QM correlation:   -0.707106781186548
 
# Angles:           (90, 135)
# File correlation: -0.2498984
# QM  correlation:  -0.707106781186548