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 Function by Tristan)

It looks like Jer has got it right and you have got it wrong, even though you both get 20 as the final number of people.  Jer's idea of scaling down the possible number of birthdays leads to the replacement of 365 with r (and 364 with r-1, then) and for r=4 and n=3 your formula then gives (1-9/16)(3/4)=21/64. With these values, the 64 possibilities can be readily written down and one finds there are 24 winners, not 21.  Jer's formula does give 24/64.  In your derivation, the events whose probabilities you multiply together to rule out winners earlier in line are not independent, I guess.

Posted by Richard on 2004-04-03 21:31:07