## Birthday Problem

Anything goes, but keep it seemly...

### Birthday Problem

Replying OT to this post ...
coloin wrote:see http://mathforum.org/dr.math/faq/faq.birthdayprob.html
There is a 50% chance that 2 out of 23 people will share the same birthday - therefore number of days in the year = 365 !

That's not quite correct, assuming that you meant to say "at least 2 out of 23 people", as per the standard bday problem (i.e. that you're simply inverting the standard bday problem, with the usual idealisations). Inverting the well-known solution for P(n,d), given the numbers you state -- i.e. solving for d in P(23,d)=0.5 -- does not lead to 365 as the number of days in the year:

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`#people  #days/year  n        d           P(n,d)-------  ----------  ----------23       365         0.50729...23       366         0.50632...23       367         0.50535...23       368         0.50438...23       369         0.50342...23       370         0.50246...23       371         0.50150...23       372         0.50055...23       373         0.49960...where P(n,d) = 1 - d!/((d-n)!*d^n)`

... "therefore there are about 373 days in the year."
r.e.s.

Posts: 337
Joined: 31 August 2005

Thankyou
I was quoting the birthday problem inexactly........

But to extrapolate .......

Can you complete it for these hypothetical figures ?

Lets say the year was very long and it turned out that the average figure was at least 2 in 10000 had the same birthday ?
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`#people      #days/year       P(n,d)n               d-----        ---------        --------10000           ?               0.50`

How many days in the year then ?

C
coloin

Posts: 1725
Joined: 05 May 2005

Using Stirling's series, it looks like the integer d that comes closest to solving P(10000,d) = .5 is 72,130,872 (days).
r.e.s.

Posts: 337
Joined: 31 August 2005