On 2/20/2020 4:26 PM, Bruce Kellett wrote:
On Thu, Feb 20, 2020 at 10:32 PM Bruno Marchal <[email protected]
<mailto:[email protected]>> wrote:
On 20 Feb 2020, at 06:12, Bruce Kellett <[email protected]
<mailto:[email protected]>> wrote:
On Mon, Feb 17, 2020 at 4:13 PM 'Brent Meeker' via Everything
List <[email protected]
<mailto:[email protected]>> wrote:
On 2/16/2020 2:17 PM, Bruce Kellett wrote:
No,that argument is mistaken, as Kent's general argument in
terms of the binomial expansion shows. All 2^N persons will
use the frequency operator to conclude that their
probabilities are the correct ones. Some will be seriously
wrong,
But almost all will intersubjectively agree that p is near
0.5. Science theories are based on intersubjective
agreement...not personal exepriences.
This is a point that has come up several times in this discussion
of Adrian Kent's argument that all observers in an Everett world
will think that their observed probabilities are the correct
ones, and that other observers will agree. Brent has argued that
all will agree that p is near 0.5.
This argument has worried me, so I thought that some serious
calculations were in order. If you have an even number of trials,
N = 2M, the the number of binary strings that have equal numbers
of ones and zeros (both equal to M) is given, from the binomial
distribution, as N!/M!*M!. Using the Stirling expansion for the
factorial, as N gets large,
N! ~ sqrt(N) N^N.
So (2M)!/M!^2 ~ 1/sqrt(N).
I think you misplaced a 2. I get sqrt(pi*M) 2^2M. Anyway the number
obviously goes to infinity. And you can see from the Gaussian approx
that the distribution of 1s becomes more concentrated near N/2 as N->oo.
Brent
In other words, the proportion of trials with equal numbers of
zeros and ones decreases as 1/sqrt(N) as N becomes large. This
means that it is simply not true that the majority of observers
will find a probability at or near p = 0.5. In fact, the
proportion who find any specific probability decreases as N
increases. This makes sense, since the number of binary strings
increases exponentially, as 2^N, but the number of instances with
any particular proportion of zeros and ones increases only
linearly with N.
That makes not much sense. To say that the probability of having a
girl or a boy is roughy 1/2 (slightly different in China) does not
mean that the probability of having exactly 5 girls and 5 boys is
one, in a row of ten kids, as you argue correctly above. In fact
that is (binomial 5 10) times (1/2)^10 = 0,246…, and get lower and
lower when repeating the trials.
You just give the good argument for inferring P=1/2 in the
(iterated) self-duplication. It seems to me.
You have missed the point in a spectacular fashion. I calculated the
number of times in the 2^N binary strings that there are equal numbers
of zeros and ones -- not the probability that you will get equal
numbers. If you multiply that probability by the number of strings,
you get the number I calculated, which is N!/M!*M! (for N=2M).
It is the fact that this number goes to zero as N goes to infinity
that shows that equal probabilities do not dominate the complete set
of binary strings. This complete the proof that assuming p = 1/2 in
the iterated WM-duplication case is completely unjustified by the data.
Bruce
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