Yeah well someone reminded me that I never did find out the solution to that birthday's problem:

How many people do you have to stick into a cage before it's guaranteed that two have the same birthday (and that's birthday every year not just the year they were born)?

My reasoning:

Given n people in the cage, there are nC2 different pairs that you can make. Given a pair of people the probability they share a birthday is 1/3652 = 1/133225. So if P is the probability that two people in the cage share a birthday,

P = nC2 / 133225

But P >= 1 so:

133225 >= nC2

266450 >= n(n - 1)

n2 - n - 266450 <= 0

This gives n >= 516.7

So n = 517 people.

I'm not pretending to know the flaw there (i.e. I'm not asking ppl this just for kicks), and I would genuinely like to know what it is...

Oh yeah and if you don't know what you're talking about (like you think a probability can be greater than 1, or don't understand why if A and B aren't independant P(A) * P(B) = P(A & B) is bollocks), please don't bother wasting our (the people who do or want to) time.