On Fri, Feb 21, 2020 at 12:08 PM 'Brent Meeker' via Everything List < [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. 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. Bruce 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. >> >> -- 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/CAFxXSLRyfMDX8G%2BLxoy4mw1KX2kVmGLR1BH0fXjucMZ_VmsdrQ%40mail.gmail.com.
