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Physics, Astronomy & Cosmology / Re: Bell's Inequality error...
« on: 22/11/2016 07:59:19 »
The error lies in the nature of treating the cases where all doors are equal as simply 1/3 of all options and for the other 1/3 where they are the same as equivalent in weight. The probabilities should break down as:
1/3 (all 3 doors equal color and both select the same door) <--> 100% same
1/3 (of non-similar selected doors for 2/3 same color) <--> 50% same
1/3 (of non-similar selected doors for 2/3 different color) <--> 50% different
The bolded are what make up the cases as 2/3 when that first 1/3 must be split or the other two divided.
For any one door selected by both Scully, say, Mulder can pick the same door and automatically is assured of a match. But this 'appears' as it should be treated as ONE probability:
If Scully picks door T and it is blue, and Mulder picks door T, it can ONLY BE 1 x blue.
But if Mulder picked any of the other doors instead, he has 1/2 x blue or 1/2 x red.
So this would be more appropriately be:
1/3 x 1 = probability for identical door selection when all 3 the same = 1/3
1/3 x 1/2 = probability for non-similar door selection for the same = 1/6
1/3 x 1/2 = probability for non-similar door selection for differences = 1/6
Thus, when corrected, should 50% in experiment demonstrate they are the same, you have to treat this as constituting the 1/3 + 1/6 weighted parts (= 1/2) in actuality!
This definitively disproves Bell's theorem to be useful to prove anything!!
1/3 (all 3 doors equal color and both select the same door) <--> 100% same
1/3 (of non-similar selected doors for 2/3 same color) <--> 50% same
1/3 (of non-similar selected doors for 2/3 different color) <--> 50% different
The bolded are what make up the cases as 2/3 when that first 1/3 must be split or the other two divided.
For any one door selected by both Scully, say, Mulder can pick the same door and automatically is assured of a match. But this 'appears' as it should be treated as ONE probability:
If Scully picks door T and it is blue, and Mulder picks door T, it can ONLY BE 1 x blue.
But if Mulder picked any of the other doors instead, he has 1/2 x blue or 1/2 x red.
So this would be more appropriately be:
1/3 x 1 = probability for identical door selection when all 3 the same = 1/3
1/3 x 1/2 = probability for non-similar door selection for the same = 1/6
1/3 x 1/2 = probability for non-similar door selection for differences = 1/6
Thus, when corrected, should 50% in experiment demonstrate they are the same, you have to treat this as constituting the 1/3 + 1/6 weighted parts (= 1/2) in actuality!
This definitively disproves Bell's theorem to be useful to prove anything!!
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