[ExI] Demonstration of Bell's Inequality

[Adrian Tymes]

On Wed, Nov 23, 2016 at 10:54 AM, John Clark <johnkclark at gmail.com> wrote:
> That's the wrong analogy, a better one would need 3 complementary properties
> not just 1
> , so in addition to red/blue lets have heavy/light and
> radioactive/nonradioactive.

You refer to the 3 properties outlined in
http://www.mtnmath.com/whatth/node60.html ?

Treating them as being able to produce 8 (as in 2^3) combinations only
works if the 3 properties are independent, but the 3 properties they
list - one particle detectable at one angle and the other at another,
only differing on specific angle (and reusing specific angles between
the properties) - are clearly dependent on one another.  It'd be like
counting red/purple, purple/blue, and red/blue as the 3 properties
(remembering that purple = red + blue).

In truth, they reduce to 1 property (color for the balls, spin for the
particles), which we know something about (the two particles' spins,
whatever they are, are 90 degrees apart).

> So if you X ray your package and find that it is red you'd expect that on average there would be 2 chances in 8 (1 in 4) that the other package contains a heavy ball; it could be blue heavy and radioactive or
> blue heavy and non-radioactive.

Also, this math doesn't work out.  If you know one ball is red and
therefore the other is blue, there are 4 combinations the other ball
could be, 2 of which include "heavy".  So the chance is 2 in 4, not 2
in 8.