Probability of 3 streams of data on same Compute Node Assume there are 10 compute nodes and 6 sources of data which randomly assign a destination within 1:10 compute nodes.  What is the probability that three sources (or more) go to the same compute node at the same time?
I tried using the 'birthday' paradox Poisson concept and came up with the following:
For [n=1] you have (10/10) or 100% probability that a single stream will be sent to a compute node 
For [n=2] you have (10/10)*(1/10) or 10% probability that a two streams will be sent to the same compute node
For [n=3] you have (10/10)(1/10)(2/10) or 2% probability that three streams will be sent to the same compute node
Is this logic correct with a 2% probability? 
 A: The following solution uses an exponential generating function (EGF).
It may be simpler to solve the complementary problem.  What is the probability that all nodes have no more than 2 of the 6 sources assigned to them?  The sources may be assigned to the nodes in $10^6$ possible ways, all of which we assume are equally likely.  We would like to count the number of ways in which each node has 0, 1, or 2 sources assigned.  More generally, we might count the number of ways that $r$ sources can be assigned.  The EGF for the number of ways sources can be assigned to a single node is $1+x+(1/2!)x^2$, so the EGF for a sequence of $10$ such assignments is 
$$f(x) = \left( 1 + x + \frac{1}{2!} x^2 \right)^{10}$$
The EGF gives the number of ways any number of sources can be assigned to the nodes, but all we really need is the number of ways 6 sources can be assigned.  Expanding $f(x)$, we find the coefficient of $x^6$ is $1,170$, so the number of ways 6 sources can be assigned to the nodes subject to the given constraints is $6! \times 1170 = 842,400$, and the associated probability is $$\frac{842,400}{10^6} = 0.8424$$
(I used a computer algebra system for the expansion, but I think that with a little work one could compute the coefficient with pencil and paper.)
So the answer to the original problem, the probability that at least one node has three or more sources assigned to it, is
$$1 - 0.8424 = \boxed{0.1576}$$
