On 8/27/2025 5:07 PM, Bruce Kellett wrote:
On Thu, Aug 28, 2025 at 5:16 AM Brent Meeker <[email protected]>
wrote:
I think some specificity would help this debate. Suppose N=6, so
there are 64 different sequences in 64 different worlds. The
number of observers is irrelevants; we can suppose the results are
recorded mechanically in each world. Further suppose that a=b so
there is no question of whether amplitudes are being respected.
Then in one of the worlds we have 011000. Per the Born rule its
probability is 0.2344. In MWI it is 1/64=0.0156. The difference
arises because the observers applying the Born rule looks at it as
an instance of 2 out of 6 successes.
The trouble with this is that you are treating this as an instance of
Bernoulli trials with probability p= 0.5. When every outcome occurs
with every trial we no longer have a Binomial distribution The
Binomial distribution assumes that you have x successes out of N
trials. In the Everettian case you have one success on every trial.
So your probability above for 2 successes applies to Bernoulli trials
with one as the 'success'. The thing is that the probability of
getting a zero is also 1/2, so we also have four successes out of six
trials in your example. The Binomial probability for this result is
also 0.2344. Actually, if we regard this experiment as a test of the
Born rule, we have four zeros in 6 trials, which gives an estimate of
the probability as 4/6 = 0.667, or as two ones in 6 trials which gives
an estimate of the probability as 2/6 = 0.333, neither estimate is
equivalent to |a|^2 = 0.5. The difference becomes more pronounced as N
increases. The problem with your analysis is that you are assuming a
binomial distribution. and we do not have any such distribution.
That's how an experimenter will compute the probability of 2 successes
in 6 trials given p=0.5. Yes, the probability estimate is 0.33 given 2
in 6. The probability of the result given p is not generally the same
as the estimate of p given the result. The former is binomial
distributed. The estimates of p aren't part of a distribution since
they don't add up to 1.
Brent
Bruce
So why can't the MWI observer do the same calculation? He
certainly can. /He can apply the Born rule./ But when he does so,
it can't be interpreted as a probability of his branch since such
probabilities would add up to much more than 1.0 when summed over
the 64 different worlds. From the standpoint of statistics 011000
is the same as 001010 and their probabilities sum. Their
difference is just incidental, but they are different worlds in
MWI and summing them makes no sense.
Brent
--
You received this message because you are subscribed to the Google
Groups "Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send
an email to [email protected].
To view this discussion visit
https://groups.google.com/d/msgid/everything-list/CAFxXSLRfM7Gh8pVsvwqT6cTiOGFavvRSXncFmhzTiAk7VAtvDA%40mail.gmail.com
<https://groups.google.com/d/msgid/everything-list/CAFxXSLRfM7Gh8pVsvwqT6cTiOGFavvRSXncFmhzTiAk7VAtvDA%40mail.gmail.com?utm_medium=email&utm_source=footer>.
--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
To view this discussion visit
https://groups.google.com/d/msgid/everything-list/7625d067-bf99-4769-a874-2f735c0cf554%40gmail.com.
