Algorithm to generate deviate for a Poisson distribution According to a previous post, the following algorithm generates deviates of a Poisson distribution:
static Random random = new Random();
public static final int generatePoissonDeviate(double a) {
    double limit = Math.exp(-a);
    double product = random.nextDouble();
    int n;
    for( n = 0; product >= limit; n++ ) product *= random.nextDouble();
    return n;
}

What is "a" in this algorithm? It does not appear to be a mean.
When I run this algorithm it does not generate any zeroes.
I want to generate deviates of a Poisson distribution with a mean value N that includes zeroes. For example, assume number of children per family has a Poisson distribution and the mean number of children per family is 2.5. How can I generate deviates for individual families? Obviously the deviates must include the possibility of a family having zero children.

Here is a link to a really efficient implementation of a routine taking a mean, using a flat random generator (on (0,1) and delivering a Poisson deviate:
http://www.fnal.gov/docs/working-groups/fpcltf/Pkg/CLHEP/Random/RandPoisson.cc
I wrote this some years ago.  It is in C++, but you should be able to discern the algorithm from this code even if you don't want to use C++.  Lhe lnGamma routine is based on Numerical Recipes. 
