Probability any $2$ will be the same at the moment I have a debate going with a friend about how possibility works and it is: 
Say at any one time you can see $30$ snow flakes and there are $49000$ diffident possible snow flakes that can be found what is the possibility that any $2$ will be the same at the moment?
 A: This is similar to the Birthday Problem.  Given $N$ distinct snow flakes, the probability that in a group of $r \leq N$ snowflakes there are two identical is:
$$
  P = 1 - \frac{{}_N P_r}{N^r} = 1 - \frac{N(N-1)(N-2) \cdots (N-r+1)}{N^r}
$$
Thus, when $N = 49 000$ and $r = 30$, we have
$$
  P \approx 0.00884
$$
So, about $0.884\%$ chance, which is fairly close to $1\%$.
A: The assumption of independence is not really plausible, since similar weather conditions presumably tend to produce snowflakes that fall roughly in the same family. But if we do assume independence, the probability the $30$ are not all different is
$$1-\frac{N(N-1)\cdots (N-29)}{N^{30}},$$
where $N=49000$.
A: There are $49000^{30}$ different possible combinations of 30 snowflakes.
There are $49000\cdot (49000-1) \cdots \cdot (49000-29) = \frac{49000!}{(49000-30)!}$ ways to choose 30 unique snowflakes out of 49000.
The probability that all the snowflakes you see are unique is therefore
$$ \frac{\frac{49000!}{(49000-30)!}}{49000^{30}},$$
so the probability that at least two are the same is one minus this number.
