ON THIS LINK

A great description of the concept/math behind the Birthday problem. Basically, if you have 23 people in a room, odds are better than 50-50 that 2 folks will share the same birthday. [With a group of 7, odds are better than 50-50 that two folks have a birthday within a week of each other.] Also the basis for the "Birthday attack" hacking vector. Given a hashing (1-way encryption) algorithm of fixed length output, multiple inputs can result in the same hash. It is very difficult to guess an alternate input that will hash into a match of a single given result. But given a large known set of results (e.g., a user account database table) it become more likely that an alternate input might result in a match of at least one existing hash. Thinking on such things keeps the mind sharp :)