> On 21 Feb 2020, at 04:40, Bruce Kellett <[email protected]> wrote: > > From: Brent Meeker <[email protected] <mailto:[email protected]>> >> On 2/20/2020 5:52 PM, Bruce Kellett wrote: >>> On Fri, Feb 21, 2020 at 12:08 PM 'Brent Meeker' via Everything List >>> <[email protected] >>> <mailto:[email protected]>> wrote: >>> On 2/20/2020 4:26 PM, Bruce Kellett wrote: >>>>> >>>>> 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. >>> >>> >>> I was a little too quick with saying this was the number of bitstrings with >>> equal zeros and ones. What I was actually calculating was the proportion of >>> such strings in the 2^N strings from N trials. Sure, the number of strings >>> with equal zeros and ones increases as 2^N, but so does the total number of >>> strings, so the proportion goes as 1/sqrt(N) as advertised. I think you >>> have you factors of sqrt(pi*M) upside down. >> >> Right. Stirling approximation M! ~ sqrt(2pi*M) (M/e)^M >> Number with M out of 2M = (2M)!/(M!)^2 = >> [sqrt(2pi*2M)(2M/e)^2M]/[sqrt(2pi*M)(M/e)^M]^2 >> = [1/sqrt(p*M)] 2^2M >> (M/e)^2M / (M/e)^2M = 2^2M/sqrt(pi*M) >> >> >>> >>> See below, I say "the proportion of trials with equal numbers of zeros and >>> ones decreases as 1/sqrt(N) as N becomes large." Equal numbers of zeros and >>> ones do not dominate as N increases, so it is not the case that the >>> majority of observers see a probability of 0.5. >> >> Of course that's true. But the more relevant value is the fraction of >> sequences with the proportion of 1s within some narrow range of 0.5. For >> large N, the distribution is Gaussian with std deviation ~sqrt(N) so almost >> equal numbers of 1s and 0s do predominate. > > I was aware of that, but they only dominate in a narrow range when p = 0.5. > My thinking was that since the confidence interval around the estimated > probability shrinks as 1/sqrt(N) for large N, outside a small range of small > deviations from equal numbers of zeros and ones, the confidence interval on > the probability estimates would no longer capture p = 0.5. Also, looking at > numbers of zeros within +- a small number of N/2 would give results for the > asymptotic proportion similar to those for N/2 zeros. Since my calculation > systematically ignores factors of the order of one, I doubt that including > such bit strings with close to equal numbers of zeros and ones would make any > significant difference to the conclusion that such strings do not dominate in > the limit. In other words, I think my conclusion that the majority of the 2^N > observers would not estimate probabilities close to 0.5 is secure. (Ignoring > factors of order one in the calculation!) >
But that argument would work for coin tossing too. That eliminate basically all probabilistic inference, it seems to me. A dwarf and a giant would not accept the Gaussian distribution of height. Bruno > > 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 [email protected] > <mailto:[email protected]>. > To view this discussion on the web visit > https://groups.google.com/d/msgid/everything-list/e12b8637-4dc4-1135-7a89-fec767ccabb5%40optusnet.com.au > > <https://groups.google.com/d/msgid/everything-list/e12b8637-4dc4-1135-7a89-fec767ccabb5%40optusnet.com.au?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 on the web visit https://groups.google.com/d/msgid/everything-list/E4964743-2CD6-493F-A5B6-5AE7EB49BD3D%40ulb.ac.be.
