Probabilities of duplicate mail detection by comparing notes among servers
If a given server receives $m \leq M$ copies, it does not receive $M-m$ copies. Also, there are many ways to pick $m$ messages out of $M$; you have to consider all of them. Furthermore, important assumptions are that 


"randomly" means "uniformly at random", that is the probability that any given copy goes to any given server is $\frac{1}{N}$, and that
each copy is routed independently of its siblings, that is we have independent random events for the individual copies.


This is all you need to piece together the probability $\operatorname{Pr}(S_i = m)$ that server $i$ receives $m$ messages.


   If you put the above together, you get

 $\qquad \displaystyle \operatorname{Pr}(S_i = m) = \binom{M}{m} \left(\frac{1}{N}\right)^m \left(1 - \frac{1}{N}\right)^{M-m}$

 for $S_i$ being the number of copies the $i$-th server receives, $1\leq i \leq N$. You should recognize this kind of probability weight.

 If you have doubts that this is correct, run some simulations (not a proof!).


Now it is easy to compute $\operatorname{Pr}(S_i \geq 2) = 1 - \operatorname{Pr}(S_i = 1) - \operatorname{Pr}(S_i = 0)$.


   When using the derived weight in other calcuations, keep in mind that the $S_i$ are not independent because $\sum_{i=1}^N S_i = M$. The underlying probability distribution is the well-studied multinomial distribution.