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

I made the probability of getting the free ticket a function of the number in line.  Let n = the # person in line.  I consider the probability of having a birthday on 2-29 negligible (1:1469, about).

The probability of any person having the same birthday as someone before them is 1-(364/365)^(n-1).  Correct me if I'm wrong there, please!

The probability of being the first to have a dupe birthday, is the same, except multiplying by the probability that no one before already got the free ticket.  This is [1-(364/365)^(n-1)]*[(364/365)^(n-2)*(364/365)^(n-3)....*(364/365)^(1)].  This is the same as raising (364/365) to the triangle number of n-2.  All of this comes out to this:

f(n)=[1-(364/365)^(n-1)]*(364/365)^[(n-2)(n-1)/2]

Again, please correct me if I'm wrong.

I just took my TI-83 and graphed this.  I got the following values.

n  f(n)
1  0
2  .00273973
3  .00545695
4  .00812949
5  .01073577
6  .01325509
7  .01566795
8  .01795627
9  .02010361
10 .02209537
11 .02391896
12 .02556391
13 .02702197
14 .02828714
15 .02935569
16 .03022617
17 .03089926
18 .03137777
19 .03166646
20 .0317719
21 .03170232

After that, it goes downward. So my answer is that the most likely person to get a free ticket is the 20th.

However, if you're birthday is on the leapday, I guess you can't call the probability of anyone else having the same birthday negligible anymore.

Posted by Tristan on 2004-03-30 19:04:08