On Thu, Feb 20, 2020 at 10:32 PM Bruno Marchal <marc...@ulb.ac.be> wrote:
> On 20 Feb 2020, at 06:12, Bruce Kellett <bhkellet...@gmail.com> wrote: > > On Mon, Feb 17, 2020 at 4:13 PM 'Brent Meeker' via Everything List < > everything-list@googlegroups.com> 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). > > 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 -- 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 everything-list+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLRk_NJqse18RMdhdpRb%2B0nZWWuYDejfoHrwOJkgbaq24g%40mail.gmail.com.
