Sure. Consider a sequence of n=4 Bernoulli trials. Let h be the number of
heads. Then we can make a table of the number of all possible sequences bc
with exactly h heads and with the corresponding observed proportion h/n
h bc h/n
0 1 0.0
1 4 0.25
2 6 0.5
3 4 0.75
4 1 1.0
So each possible sequence will correspond to one of Everett's worlds. For
example hhht and hthh belong to the fourth line h=3. There are sixteen
possible sequences, so there will be sixteen worlds and a fraction
6/16=0.3125 will exhibit a prob(h)~0.5.
But suppose it was an unfair coin, loaded so that the probability of tails
was 0.9. The possible sequences are the same, but now we can apply the
Born rule and calculate probabilities for the various sequences, as follows:
h bc h/n prob
0 1 0.0 0.656
1 4 0.25 0.292
2 6 0.5 0.049
3 4 0.75 0.003
4 1 1.0 0.000
So most of the observers will get empirical answers that differ
drastically from the Born rule values. The six worlds that observe 0.5
will be off by a factor of 1.8. And notice the error only becomes greater
as longer test sequences are used. The number of sequences peak more
sharply around 0.5 while the the Born values peak more sharply around 0.9.
Brent
*By the above paragraph, it seems you've already falsified the MWI, except
that you could claim that's what no-collapse yields in this-world. I don't
see any reason for claiming each sequence is observed in different worlds.
AG*
There's no unique sequence "in this world" because there's no unique "this
world" in MWI.
Brent
*IMO this is ridiculous. How can you disprove the MWI when you accept its
foolish claim of many worlds? All that's required is to show that the
no-collapse hypothesis gives wrong results compared to Born's rule in the
only world you know for sure, THIS-WORLD. AG*
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