On 3/22/2022 11:05 PM, Bruce Kellett wrote:

On Wed, Mar 23, 2022 at 4:01 PM Brent Meeker <meekerbr...@gmail.com>wrote:On 3/22/2022 7:55 PM, Bruce Kellett wrote:Actually, that is where I started. I assumed that Alice and Bob were both able to collect results from N trials before they met. Then there are 2^N copies of each experimenter, and a potential (2^N)^(2^N) pairs when they meet. The trouble to be explained is that there are actually only 2^N pairs in a real experiment, each with inequality-violating correlations. What has happened to all the extra pairings that MWI must produce? (Most of which have correlations violating the quantum predictions.)Don't you mean (2^N)x(2^N).Probably. Both Alice and Bob split into 2^N copies for binary results.So each of the 2^N Alice's splits into 2^N further copies, one foreach copy of Bob.

`Why would Alice split again for each copy of Bob. There's no QM`

`involved. Each Alice and each Bob has seen N events and has recorded`

`them as a binary number of N bits in their notebook. They are`

`quasi-classical observers and they're going to meet in the future. If`

`there are not (2^N)x(2^N) meetings there must be some quantum magic that`

`it is eliminating or otherwise weighting them.`

Brent

That seems to give (2^N)*(2^N) pairings. That is still a lot ofredundant copies when only one copy of Bob meets each copy of Alice inthe actual experiment.Bruce. --You received this message because you are subscribed to the GoogleGroups "Everything List" group.To unsubscribe from this group and stop receiving emails from it, sendan email to everything-list+unsubscr...@googlegroups.com.To view this discussion on the web visithttps://groups.google.com/d/msgid/everything-list/CAFxXSLSZkh8_YNySPduxeo9rB_nc9_5XbFM%2B2dtnZiFCFbyVfw%40mail.gmail.com<https://groups.google.com/d/msgid/everything-list/CAFxXSLSZkh8_YNySPduxeo9rB_nc9_5XbFM%2B2dtnZiFCFbyVfw%40mail.gmail.com?utm_medium=email&utm_source=footer>.

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