JesseM said:
That's because your version of H is too vague and doesn't actually specify whether the red card was the one that was picked to send to Alice and the white card was the one that was picked to send to Bob, or vice versa. If you completely specified the hidden properties of the envelope that was sent to Bob--namely "the person who picked the cards from the box put the red card in Bob's envelope, and the envelope continued to have that hidden card on its journey to Bob"--then in that case it would be true that P(B|H)=P(B|AH).
billschnieder said:
I am not sure I agree with this. In probability theory, when we write P(A|H), we are assuming that we know H but not A. If we knew A already (a certainty), there is no point calculating a probability is there?
I think I explained clearly in the text above that I was calculating the probability of B, not A, given either just H or given both H and A. If you want to calculate the probability of A rather than B, then you can easily modify the paragraph above:
That's because your version of H is too vague and doesn't actually specify whether the red card was the one that was picked to send to Alice and the white card was the one that was picked to send to Bob, or vice versa. If you completely specified the hidden properties of the envelope that was sent to Alice--namely "the person who picked the cards from the box put the red card in Alice's envelope, and the envelope continued to have that hidden card on its journey to Alice"--then in that case it would be true that P(A|H)=P(A|BH).
billschnieder said:
In this case when Bell obtains cos(theta), his equation will only be valid for two angles when cos(theta)= 0 or 1! How then can this equation apply to other angles? Therefore I don't think that is the reasoning here.
Huh? The argument is about what probabilities would be calculated by an ideal observer
if they had access to the hidden variables H (which are assumed to have well-defined values at all times in a local realist theory), not just what probabilities are calculated by normal observers who don't know the values of the hidden variables. How could it be otherwise, when H explicitly appears in the conditional probability equations?
billschnieder said:
Furthermore, the variables are hidden from the perspective of the wise men, it is not God trying to calculate the probabilities but the wise men, because they do not have all the information. We are only looking from God's perspective to verify that the equation the wise men choose to use corresponds to the factual situation and in the example I gave it does not appear to.
No, you're completely confused, the argument is about taking a God's-eye-view and saying that no matter how we imagine God would see the hidden variables, in a local realist theory God would necessarily end up making predictions about the statistics of different correlations that are different from what we humans actually observer in QM.
billschnieder said:
But in my example, with the information known by God, this is the case, every box only contains two cards, one red and one white and in each iteration of the experiment one of the cards is sent to Bob and the other to Alice. The equation P(AB|H) = P(A|H)P(A|BH) always works but P(AB|H) = P(A|H)P(A|H) works only in very limited case in which H is no longer hidden and calculating probabilities is pointless.
It's not "pointless" if you can use this hypothetical God's-eye-perspective (where nothing is hidden) to show that if the hidden variables are such that Alice and Bob always get the same result when they perform the same measurement, that
must imply certain things about the statistics they see when they perform different measurements--and that these statistical predictions are falsified in real quantum mechanics! This is a reductio ad absurdum argument showing that the original assumption that QM can be explained using a local realist theory must have been false.
Perhaps you could take a look at the scratch lotto analogy I came up with a while ago and see if it makes sense to you (note that it's explicitly based on considering how the 'hidden fruits' might be distributed if they were known by a hypothetical observer for whom they aren't 'hidden'):
Suppose we have a machine that generates pairs of scratch lotto cards, each of which has three boxes that, when scratched, can reveal either a cherry or a lemon. We give one card to Alice and one to Bob, and each scratches only one of the three boxes. When we repeat this many times, we find that whenever they both pick the same box to scratch, they always get the same result--if Bob scratches box A and finds a cherry, and Alice scratches box A on her card, she's guaranteed to find a cherry too.
Classically, we might explain this by supposing that there is definitely either a cherry or a lemon in each box, even though we don't reveal it until we scratch it, and that the machine prints pairs of cards in such a way that the "hidden" fruit in a given box of one card always matches the hidden fruit in the same box of the other card. If we represent cherries as + and lemons as -, so that a B+ card would represent one where box B's hidden fruit is a cherry, then the classical assumption is that each card's +'s and -'s are the same as the other--if the first card was created with hidden fruits A+,B+,C-, then the other card must also have been created with the hidden fruits A+,B+,C-.
The problem is that if this were true, it would force you to the conclusion that on those trials where Alice and Bob picked different boxes to scratch, they should find the same fruit on at least 1/3 of the trials. For example, if we imagine Bob and Alice's cards each have the hidden fruits A+,B-,C+, then we can look at each possible way that Alice and Bob can randomly choose different boxes to scratch, and what the results would be:
Bob picks A, Alice picks B:
opposite results (Bob gets a cherry, Alice gets a lemon)
Bob picks A, Alice picks C:
same results (Bob gets a cherry, Alice gets a cherry)
Bob picks B, Alice picks A:
opposite results (Bob gets a lemon, Alice gets a cherry)
Bob picks B, Alice picks C:
opposite results (Bob gets a lemon, Alice gets a cherry)
Bob picks C, Alice picks A:
same results (Bob gets a cherry, Alice gets a cherry)
Bob picks C, Alice picks picks B:
opposite results (Bob gets a cherry, Alice gets a lemon)
In this case, you can see that in 1/3 of trials where they pick different boxes, they should get the same results. You'd get the same answer if you assumed any other preexisting state where there are two fruits of one type and one of the other, like A+,B+,C- or A+,B-,C-. On the other hand, if you assume a state where each card has the same fruit behind all three boxes, so either they're both getting A+,B+,C+ or they're both getting A-,B-,C-, then of course even if Alice and Bob pick different boxes to scratch they're guaranteed to get the same fruits with probability 1. So if you imagine that when multiple pairs of cards are generated by the machine, some fraction of pairs are created in inhomogoneous preexisting states like A+,B-,C- while other pairs are created in homogoneous preexisting states like A+,B+,C+, then the probability of getting the same fruits when you scratch different boxes should be somewhere between 1/3 and 1. 1/3 is the lower bound, though--even if 100% of all the pairs were created in inhomogoneous preexisting states, it wouldn't make sense for you to get the same answers in less than 1/3 of trials where you scratch different boxes, provided you assume that each card has such a preexisting state with "hidden fruits" in each box.
But now suppose Alice and Bob look at all the trials where they picked different boxes, and found that they only got the same fruits 1/4 of the time! That would be the violation of Bell's inequality, and something equivalent actually can happen when you measure the spin of entangled photons along one of three different possible axes. So in this example, it seems we can't resolve the mystery by just assuming the machine creates two cards with definite "hidden fruits" behind each box, such that the two cards always have the same fruits in a given box.
And you can modify this example to show some different Bell inequalities, see post #8 of
this thread for one example.
billschnieder said:
But that is not quite true. The justification for reducing P(A|BH) to P(A|H) is not based on whether B gives [i[additional[/i] information but on whether B gives any information. (see Conditional Independence in Statistical Theory, J.R Statist. Soc B, 1979 41, No. 1, pp. 1-31)
I don't have access to that reference (could you quote it?) but I'm confident it doesn't say what you think it does. In a situation where the probability of A is completely determined by H and the probability of B is also completely determined by H, then it would naturally be true that P(A|BH) would be equal to P(A|H), even if the P(A) was not equal to P(A|B) (i.e. if you don't know H, knowing B does give some information about the probability of A). Do you claim the reference somehow contradicts this?
For example, suppose we have two identical-looking flashlights X and Y that have been altered with internal mechanisms that make it a probabilistic matter whether they will turn on when the switch is pressed. The mechanism in flashlight X makes it so that there is a 70% chance it'll turn on when the switch is pressed; the mechanism in flashlight Y makes it so there's a 40% chance when the switch is pressed. The mechanism's random decisions aren't affected by anything outside the flashlight, so whether or not flashlight X turns on doesn't change the probability that flashlight Y turns on.
Now suppose we do an experiment where Alice is sent one flashlight and Bob is sent the other, by a sender who has a 50% chance of sending X to Alice and Y to Bob, and a 50% chance of sending Y to Alice and X to Bob. Let H1 and H2 represent these two possible sets of "hidden" facts (hidden to Alice and Bob since the flashlights look identical from the outside): H1 represents the event "X to Alice, Y to Bob" and H2 represents the event "Y to Alice, X to Bob". Let A represent the event Alice's flashlight turns on when she presses the switch, B represents the event that Bob's flashlight turns on when she presses the switch.
Here, P(A) = P(A|H1)*P(H1) + P(A|H2)*P(H2) = (0.7)*(0.5) + (0.4)*(0.5) = 0.55
and P(B) = P(B|H1)*P(H1) + P(B|H2)*P(H2) = (0.4)*(0.5) + (0.7)*(0.5) = 0.55
Since P(A|B) = P(A and B)/P(B), we must have P(A|B) = (0.7)*(0.4)/(0.55) = 0.5090909...
So you see that P(A|B) is slightly lower than P(A), which makes sense since if Bob's flashlight lights up, that makes it more likely Bob got flashlight X which had a higher probability of lighting, and more likely A got flashlight Y with a lower probability of lighting.
But despite the fact that B does give some information about the probability of A, it is still true that P(A|B and H1) = P(A|H1) = 0.7, since H1 tells us that Alice got flashlight X, and that alone completely determines the probability that Alice's flashlight lights up when she presses the switch, the fact that Bob's flashlight lit up won't alter our estimate of the probability that Alice's lights up. Likewise, P(A|B and H2) = P(A|H2) = 0.4.
I'm sure that whatever the reference you gave says, it doesn't imply that this reasoning is incorrect.
billschnieder said:
You re saying pretty much the same thing there, that A gives no additional information to H. But that is not the meaning of conditional independence. Conditional independence means that A gives us no information whatsoever about B.
I gave a pretty detailed argument in posts #61 and 62 on
that thread, starting with the paragraph towards the end of post #61 that says "Let me try a different tack". If you aren't convinced by my comments so far in this post, perhaps you could identify the specific point in my argument on the other thread where you think I say something incorrect? For example, do you disagree with this part?
I'd like to define the term "past light cone cross-section" (PLCCS for short), which stands for the idea of taking a spacelike cross-section through the past light cone of some point in spacetime M where a measurement is made; in SR this spacelike cross-section could just be the intersection of the past light cone with a surface of constant t in some inertial reference frame (which would be a 3D sphere containing all the events at that instant which can have a causal influence on M at a later time). Now, let \lambda stand for the complete set of values of all local physical variables, hidden or non-hidden, which lie within some particular PLCCS of M. Would you agree that in a local realist universe, if we want to know whether the measurement M yielded result A, and B represents some event at a spacelike separation from M, then although knowing B occurred may change our evaluation of the probability A occurred so that P(A|B) is not equal to P(A), if we know the full set of physical facts \lambda about a PLCCS of M, then knowing B can tell us nothing additional about the probability A occurred at M, so that P(A|\lambda) = P(A|\lambda B)?
In case we are dealing with a local realist universe that is not deterministic, I think I should add here that the PLCCS of M is chosen at a time
after the last moment of intersection between the past light cones of M and B, so that no events that happen after the PLCCS can have any causal influence on B. Continuing the quote:
If so, consider two measurements of entangled particles which occur at spacelike-separated points M1 and M2 in spacetime. For each of these points, pick a PLCCS from a time which is prior to the measurements, and which is also prior to the moment that the experimenter chose (randomly) which of the three detector settings under his control to use (as before, this does not imply the experimenter has complete control over all physical variables associated with the detector). Assume also that we have picked the two PLCCS's in such a way that every event in the PLCCS of M1 lies at a spacelike separation from every event in the PLCCS of M2. Use the symbol \lambda_1 to label the complete set of physical variables in the PLCCS of M1, and the symbol \lambda_2 to label the complete set of physical variables in the PLCCS of M2. In this case, if we find that whenever the experimenters chose the same setting they always got the same results at M1 and M2, I'd assert that in a local realist universe this must mean the results each of them got on any such trial were already predetermined by \lambda_1 and \lambda_2; would you agree? The reasoning here is just that if there were any random factors between the PLCCS and the time of the measurement which were capable of affecting the outcome, then it could no longer be true that the two measurements would be guaranteed to give identical results on every trial.