At a movie theater, the manager announces that they will give a free ticket to the first person in line whose birthday is the same as someone who has already bought a ticket. You have the option of getting in line at any time. Assuming that you don't know anyone else's birthday, that birthdays are distributed randomly throughout the year, etc., what position in line gives you the greatest chance of being the first duplicate birthday?

from http://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml

(In reply to re(3): Function [with observations] by Jer)

Thank you for your reply.  I have been doing some comparing. Your formula is [(n-1)*r!]/[(r^n)*(r+1-n)!] whereas Tristan's is
[1-u^(n-1)]*u^[(n-2)(n-1)/2] where u=(r-1)/r. Setting r=6 so that u=5/6 gives for n=3 the values 60/216 and 55/216, resp. -- the formulas do not really agree exactly. As to who is right, I am not quite sure, but I have a suspicion that Tristan multiplied probabilities of events that are not really independent when he calculated the probability of no winner ahead of the nth person.

Posted by Richard on 2004-04-03 14:53:00