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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.

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This feels like a generalization of the coupon collector problem to me, although this observation may not be helpful. –  Michael Lugo Oct 21 '10 at 22:33
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It is a interesting question, but maybe one reason you haven't seen many responses is that it is a little unclear what you are looking for. I mean, you already have estimates, solutions in special cases, and the exact answer but you are hoping for "a more precise result"? Are you looking for bounds to show when the probability is practically zero? Or are you maybe interested in large y values so the probability is guaranteed to exceed 1/2, say? –  Byron Schmuland Oct 23 '10 at 16:55
    
@Byron I'm sorry to have given the impression that I know more than I do. In fact, I know almost nothing. The special cases I've solved are trivial, e.g. I know the answer for $c=2$ and $n<4$, but not even for $c=2$ with $n$ arbitrary. My estimates are weak or badly justified or both, and, as I said, my expression for the exact answer is useless in practice. Bottom line: I would have given a better picture had I omitted the last paragraph in the original question, and I apologize for the confusion. I will write up my partial results in an answer. –  Larry Denenberg Oct 24 '10 at 7:05
    
Why can't you just find all the possible states after six years, and count the ones where the students all meet each other? –  Justin L. Oct 28 '10 at 0:29
    
@Justin: There are probably way too many possible states, with $\sim 10^{33}$ ways to partition the students into three classes. –  Jens Oct 28 '10 at 10:43

2 Answers 2

As promised in my response to Byron, here is my very partial progress to date on this problem. Though long-winded, it doesn't amount to much. I'm still looking for an answer to the general case, or to other special cases (like $n=2$ or $c=2$) or even just better estimates. I'm also interested to know about any applicable combinatorial tools even without a full answer.

First, a caution: The graphs I'm about to use are not related to the graphs in the general formulation of the original question. Ignore those graphs and just think of students and classrooms!

Let $G$ be a graph with a vertex for each student, $cn$ vertices in all. Say that a given year's assignment of students to classrooms respects $G$ if every edge in $G$ joins two students that are assigned to different classrooms. That is, $G$ encodes constraints on assignments; each edge in $G$ represents two students that must be kept apart. Let $R(G)$ be the probability that a random assignment of students to classrooms respects $G$. Note that $R(G)$ does not depend on $y$ since the definition of "respects" refers to a single year's assignment.

Now let $e(G)$ be the number of edges in $G$. Then the probability that after $y$ years every pair of students has at some point shared a classroom is exactly

$$\sum (-1)^{e(G)} (R(G))^y$$

where the sum is taken over all graphs $G$.

This result follows from a straightforward application of Inclusion/Exclusion. It's very pretty (at least I think so) but useless if you want a real answer. The problem isn't finding $R(G)$; this is tedious but can be automated. The real difficulty is that you have to handle $2^{cn\choose 2}$ graphs, which is far too many. The graphs come in isomorphic bunches (e.g. there are $cn\choose 2$ one-edged graphs that all have the probability of being respected) and you don't have to handle most graphs with lots of edges since $R(G) = 0$ for any graph $G$ that is not $c$-partite. But this doesn't help enough.

We get some use out of this wretched formula by generalizing the problem. Suppose we ask, given the same conditions, for the probability that each student in a given group of $p$ students at some point sees each of the other students in that group. The formula above gives the exact answer if we sum over all graphs with $p$ vertices, each labelled with a student in the group. For example, with $p = 2$ the formula gives the probability that two given students at some point see each other:

$$1 - \left(\frac{n(c-1)}{cn-1}\right)^y$$

(Of course we can get this result by a simpler route.) For $p=3$ the formula gives

$$1 - \left(\frac{N-n+1}{N(N-1)}\right)^y(3(N-1)^y - 3(N-n)^y + (N-2n+1)^y)$$

where $N = cn-1$. I've fought through the formula for $p$ up to 6, but it becomes impossible quickly.

Now some approximations. Define $U_m$ to be the probability that a given student $S$ is at some point in a classroom with each student in a given set of $m$ students not containing $S$. Another straightforward application of inclusion/exclusion gives

$$U_m = \sum_{j=0}^m(-1)^j{m\choose j}{N-j\choose n-1}^y \bigg/ {N\choose n-1}^y$$

where $N = cn-1$ as before. $U_N$ is then the probability that a single given student will see all the others. This is a very weak upper bound on the probability that all the students see all the others, though in trivial cases one student seeing all the others does imply that they all do! $U_m$ is rather easy to calculate; with the original parameters, for example, $U_{74}$ turns out to be a bit less than 1/7792. Note, BTW, that $U_1$ correctly gives the same result as the $p=2$ answer above.

If "student $A$ sees all the others" were independent of "student $B$ sees all the others", then we'd have the answer we seek: $(U_N)^{cn}$. But of course these events are unlikely to be independent. If many students have seen all the others, then surely the probability increases that all have seen the others. So $(U_N)^{cn}$ may be a lower bound on the true answer, if a weak one. Edit: Actually, I don't think it can be a lower bound, since I'm pretty sure one could find a case where one student can see all the others, but it's impossible for all students to do so.

Here's another way of estimating the answer: A group of $p$ students all see each other if and only if both (a) a particular one of these students sees all the other $p-1$ students, and (b) the other $p-1$ students all see each other. The probability of (a) is exactly $U_{p-1}$. Let $C_{p-1}$ denote the probability of (b). If events (a) and (b) were independent, then the exact answer to the original question would be $U_{p-1}C_{p-1}$. But by identical reasoning, $C_{p-1} = U_{p-2}C_{p-2}$. Continuing in this way, the exact answer to the original question would be

$$U_NU_{N-1}U_{N-2}\ldots U_2U_1$$

As above, the events in question are not likely to be independent. But most of the time they're close, right? With the original parameters, the fact that a group of five students all see each other has very little impact on whether a sixth student sees each of the five, though it should increase that probability slightly. This would make the expression above a lower bound on the actual answer, though I don't know how good it is.

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taken from FAQ: "If you do not award your bounty within 7 days, the highest voted answer created after the bounty started with at least 2 upvotes will be awarded half the bounty amount." Will Larry receive half is own bounty since he answered to himself? –  Djaian Nov 3 '10 at 8:01
    
@Djaian: From here: "you do not receive the reputation that was allocated to the bounty. These answers display +0, 'this answer has been awarded bounty worth 0 reputation'." –  J. M. Nov 3 '10 at 8:25
    
@J.M. : thanks ! –  Djaian Nov 3 '10 at 8:39

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.

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Thanks very much for this approach. Can you say a few more words about why this is a lower bound? I might have thought that the very special way in which edges must be chosen makes it harder to color everything. That is, there might be unrestricted choices of edges where everything gets colored, but you can't do it when you have to choose according to classrooms, so your estimate is positive but the actual answer is zero. But maybe this problematic situation never arises, or arises only for relatively small $y$. –  Larry Denenberg Nov 6 '10 at 13:19
    
It was just a guess, but if you want to color all the edges you have a better chance if you don't color any given one too many times, as that uses up more of your tries. In your grouping, no edge gets colored more than 6 times. If we just randomly color edges, some might get colored more than that. –  Ross Millikan Nov 6 '10 at 14:26

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