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Common Birth Dates II (Posted on 2021-11-17) Difficulty: 3 of 5
Common Birth Dates asked for the expected number of pairs of common birth dates for a cohort of 1000 people whose ages were clustered around a certain value.

Now consider a similar question to the traditional birthday problem: If you attend a concert or other event where people tend to be of a similar age, how large would such an event need to be in terms of number of attendees, so that you'd have at least a 50% probability that there would be at least two people born on the exact same day (year, month and day)?

Again to make things specific: assume the standard deviation about whatever the mean age to be is 12 years and it follows a normal distribution.

Again feel free to vary the assumptions about the distribution.

See The Solution Submitted by Charlie    
Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
analytic attempt | Comment 2 of 6 |
Take P as the Normal pdf, mean age in days, mu = 60 * 356 and sigma = 12 * 365 +3,
and 120 years being max, m=43825 days, then the likelihood of any pair of audience members _not_ sharing  birthdays is:

ppair = sum{i=0 to m} P(i) (1-P(i))

There are n (n-1)/2 ~ (n^2)/2 possible pairs. Setting probability 0.5 to them all not sharing a birthday is:

ppair^( (n^2) /2) = 0.5

0.5 n^2  ln (ppair) = ln 0.5

n = sqrt (2 ln (0.5) /ln (ppair)   )  = 146

This disagrees with  the simulation, which gave n = 157

The n-1 ~ n approximation does not explain the discrepancy...

Program follows. What's my error? 
program ana
implicit none
real*8 num,p,g,ppair,term
integer i,m
m=43825
ppair=0
   do i=0,m
   p=g(i*1d0)
   ppair=ppair+p*(1d0-p)
   enddo
num=sqrt(2d0*log(0.5d0)/log(ppair))
print 1,num
1 format('House = ',f5.1)
end
 
function g(x)
implicit none
real*8 g,mu,sigma,pi,x
pi=4d0*atan(1d0)
mu=60d0*365d0
sigma=12d0*365d0+3d0
g = ( 1d0/ (sigma * sqrt(2d0*pi)) ) *   
1   exp ( -(x-mu)**2d0 / (2d0*sigma**2) )
return
end 

lord@rabbit 12478 % ana 
House = 146.1


Edited on November 18, 2021, 5:48 pm
  Posted by Steven Lord on 2021-11-18 07:29:10

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