billschnieder said:
No, I provided the full universe for my hidden "variable" z, which were clearly defined.
But the way
you defined z was not sufficient to give any reason for Bell or any of his advocates to
expect that the equation P(AB|abs, bbs, z) = P(A|abs, z) P(B|bbs, z) would be valid, thus you are just engaging in a silly strawman. Bell only says that in a local realist universe it is
possible to find a set of variables to define z such that this equation will hold, not that it should hold for any arbitrary definition of z. And given your argument that the response to pushing the switches depended only on a simple internal mechanism in each ball which either activated the same-color light or the opposite-color light, it is indeed possible to define such a z that guarantees the equation will hold, as I showed.
billshnieder said:
The term "variable" is actually not appropriate since it carries a connotation of something with multiple values. Maybe that is what is confusing you. That is why t'Hooft prefers to use the terms "beables" as different from "changeables".
t'Hooft did not mean to imply anything about whether the values can "change", either in the sense of changing with time for a single particle, or in terms of being different for different particles in different trials.
billschnieder said:
No it must not! "z" was clearly defined and my full universe includes all possibilities given that "z" is true.
There's nothing "wrong" with your definition of z in itself, but as I said, if you think Bell would have said P(AB|abs, bbs, z) = P(A|abs, z) P(B|bbs, z) for
your definition of z, then this is either a strawman or a complete misunderstanding of the physical meaning behind equation (2) in his paper.
billschnieder said:
P(B|bbs, aObN) is a nonsensical expression with no meaning. It is similar to saying. What is the probability of Bob getting a blue light if Bob gets a blue light.
No, it's "what is the probability of Bob getting a blue light given that he pushed the blue switch, and given that his ball contains a mechanism that causes the ball to light up blue if the blue switch is pressed and red if the red switch is pressed, while Alice's ball contains a mechanism that causes the ball to light up red if the blue switch is pressed and blue if the red switch is pressed".
Do you think any case where P(Y|X)=1, i.e. X implies Y with probability 1, is "a nonsensical expression with no meaning"? If you don't have a problem with conditional probabilities where the condition implies the result with probability 1, what is your special objection to P(B|bbs, aObN)?
billschnieder said:
No surprise that you will never get any result other than 1 or 0 (certainties) because your full universe will consist of exactly one case and no more, in which case it makes no sense to compare such a result with QM where values other than 1 and 0 are obtained.
You really don't understand local hidden variables theories then! In a case where the experimenters always obtain identical results with probability 1 whenever they choose identical detector settings, a local hidden variables theory says hidden variables associated with each particle on a given trial must completely determine their responses to each detector setting, and they are always created with identical predetermined responses. However, these identical predetermined responses may be different on different trials, which is what accounts for the fact that any given detector setting can yield different results on different trials, and thus the experimenter (who doesn't know the hidden variables on each trial) has a probability estimate for the result on a given detector setting which is not 0 or 1.
Note that all this is directly analogous to the scratch lotto card analogy I gave you on
post #18 of the other thread, which I repeatedly asked you to address but you kept ignoring. Can you tell me whether you agree that both the following would be true?
1. On each trial, if you know the full set of "hidden fruit" behind each box on the cards given to Alice and Bob, then the probability Alice gets a cherry if she scratches a given box is also known with probability 1.
2. If the source generating the cards picks the "hidden fruits" randomly on each trial, then if you
don't know the hidden fruits, the probability of cherry vs. lemon when a given box is scratched would not be 0 or 1.
billschnieder said:
I'm afraid it is you who is badly misunderstanding probability theory. The conditioning variable must be specific, not just a vague concept of "everything in past light cone".
Nothing vague about it, in a local realist universe. Do you agree that in a local realist universe, the laws of physics associate some well-defined set of basic physical facts with each point in spacetime? Do you agree that for some event A, the set of points in spacetime which are in the past light cone of A at some specific time t is also completely well-defined and unambiguous?
billschnieder said:
Besides Bell was not dealing with certainties but with probabilities. Your approach MUST NECESSARILY result in a certainties (0, or 1) rather than probabilities.
0 and 1 are allowable probabilities, surely you don't disagree? Would you like some examples appearing in textbooks where the probability of something is 0 or 1?
billschnieder said:
Again, "z" does not represent all different possible hidden variable states. You are confusing functional notation with Probability notation. When I write P(A|ab) in a generic equation, a and b are not variables but place-holders for "beables" not variables.
"beables" is a physical term, nothing to do with a particular type of notation in probability theory. If you disagree, find me an example of a probability text
not relating to discussions of fundamental physics which uses the term "beables".
It is true that there is sometimes a distinction between uppercase and lowercase in probability theory notation, so X could represent a random variable and x would then represent some particular possible value of that random variable (so X might have possible values x
1, x
2, etc.). See
here. Of course if we want to use this convention, it'd be better not to have capital A and B represent particular measurement outcomes as we have done so far!
billschnieder said:
I don't need to multiply by P(z) because my full universe is already defined by z, ie within my full universe, P(z) is already 1. I don't need to add anything from different hidden variables because I am dealing with a specific hidden variable. I don't need to consider all possible hidden variables because being omniscient about the workings of my machine, I know for sure that it only operates as I described. Yet, according to Bell's choice of equations, my machine is non-local.
No, this is the strawman and/or misunderstanding. Bell's equation is not meant to apply to any arbitrary definition of the extra variable (z in your equation), it's based on the idea that in a local realist universe it's
possible to define the variable in such a way as to make the equation hold. And this was indeed true in your example when I defined z to give the information about which mechanism was in Alice's ball and which was in Bob's.
billschnieder said:
P(A|abs, z) means the Probability that A is observed, given that abs is True and z is true. So when you talk about summing over all values of z,what does the term P(A|abs, z) mean to you.
It means the probability for any
single value z of the random variable Z. When I talked about summing over all the different values of z, that was for the purposes of eliminating it from the equation to get P(A|abs). Suppose for example the random variable Z has only two possible values z
1 or z
2, so on a large set of N trials, we'd expect the number of trials with z
1 to be N*P(z
1), and the number of trials with z
2 to be N*P(z
2), with P(z
1) + P(z
2) = 1. Then if we want to know P(A|abs), do you disagree that the following equation would hold? P(A|abs) = P(A|abs, z
1)*P(z
1) + P(A|abs, z
2)*P(z
2)
billschnieder said:
If you are thinking of adding up the results from say z1 with those of z2, ..., zn then clearly the results from z1 are due to a completely different context from z2 etc Therefore you can not add them up legitimately.What is the rule of probability theory that permits you to do that addition?
I would have to look a bit for a formal statement of the rule, but first tell me, do you disagree that if each trial in the sample space has some specific value of the random variable Z drawn from the set {z
1, z
2, ... z
n}, then it must be true that P(A) = (sum over all values of i from 1 to n) P(A|z
i)*P(z
i) ? If you don't disagree with that, then do you disagree that for some other fact b which is true on some trials but false on others, it would be true that P(A|b) = (sum over all values of i from 1 to n) P(A|b, z
i)*P(z
i) ?
billschnieder said:
The value of the hidden variable z in my example is the description of the mechanism of the machine I provided.
But your z did not explicitly detail which mechanism was in the ball given to Alice and which was in the ball given to Bob. You defined z like this:
Here is the so-called 'full universe' of possibilities defined by z: where abs means Alice presses blue switch and brs means Bob presses red switch. The color after the hyphen, is what Alice and Bob see as a result of their pressing their switches.
Alice, Bob
1: abs-blue, brs-blue
2: abs-blue, bbs-red
3: abs-red, bbs-blue
4: abs-red, brs-red
5: ars-blue, bbs-blue
6: ars-blue, brs-red
7: ars-red, brs-blue
8: ars-red, bbs-red
If each of these supposed to be a distinct z, so z
1 would represent possibility 1, z
2 would represent possibility 2, then of course this would allow us to infer which mechanism was in Alice's ball on each trial. But from your later statements it seems you didn't mean z to have distinct values, but just to represent the knowledge that
one of these possibilities will hold on each trial, without knowing which one will hold on any given trial.
billschnieder said:
The concept of multiple values for the hidden variable is a misunderstanding carried over from functional notation and probably over-reading into the term "variable".
This comment reveals a complete lack of understanding of
Bell's paper. His λ was of course supposed to represent a variable that could take different values on different trials; if that wasn't true, equation (2) in which he
integrates over all possible values of λ would make no sense at all!