TITLE: What is the probability that every pair of students studies together at some point?
QUESTION [15 upvotes]: A cohort in a school consists of 75 students who study for 6 years.  Each year, the students are randomly distributed into 3 classrooms of 25 students each.  What is the probability that, after 6 years, each student has at some point been in a classroom with every other student?
More generally:  Starting with an edgeless (undirected) graph on cn vertices, a round consists of first randomly partitioning the vertices into c disjoint sets of n vertices each, then adding an edge between every pair of not-yet-joined vertices that lie in the same set.  What is the probability that, after y rounds, the result is a complete graph on cn vertices?
I have estimates and solutions to special cases, and it's straightforward to find
the probability that a single given student sees all the others, but I don't know how to tackle the question in general.  (I do have a very pretty but completely useless expression for the exact answer, which I can supply if there's interest.)  In the case c=3, n=25, y=6 it's clear that the answer is "so close to zero that nobody can tell the difference" but I was hoping for a more precise result.  Any guidance appreciated.

REPLY [2 votes]: The only help I can give is to suggest a lower bound.  You have cn students and the study together graph could then have cn(cn-1)/2 edges.  Each session you pick $\frac{cn(n-1)}{2}$ edges to color in and you ask whether after y sessions all the edges are colored.  If you ignore the class grouping, you can do the same problem randomly choosing $\frac{ycn(n-1)}{2}$ edges independently with replacement to color.  I think your case will color the whole graph with higher probability, as no edge can claim more than $y$ of the colorings, but it should be close.  Now this is a nice Poisson distribution.  Each edge is colored with probability $1-\exp(-\lambda)$, where $\lambda$ is the average number of colorings each edge receives, here $\frac{ycn(n-1)}{cn(cn-1)}$, or just about $\frac{y}{c}$.  The chance that all edges are colored is then $(1-\exp(-\lambda))^{\frac{cn(cn-1)}{2}}$, which as you say is very small.