In any given week, there's a 0.0000036% chance 2 students will need an appendectomy. Assuming a 3 year program, with an academic year consisting of, let's say 32 weeks per year (96 total), that makes for a 0.00035% chance that 2 out of 6 MFA students would need an appendectomy in the same week, or about 1 in 300,000.
I don't think this is correct. Essentially you're looking for the probability that (X = 2), where X is a binomial distribution B(6, p), and p is the probability that a single given student needs an appendectomy in a single given week. (Strictly speaking, I think you want the probability that (X >= 2), but the difference is utterly negligible.)
Accepting the same starting data, we have p = 1/(7 * 4103) = 0.000035. This won't be quite right, as I strongly doubt that the risk of appendicitis is constant throughout one's life. But it's our best estimate right now. Let q = (1 - p) = almost 1. By the binomial distribution,
Pr(X = 2) = (6 2) * p^2 * q^4
That (6 2) is supposed to be combinatorial notation - that is, the number of ways you can select two students from a pool of 6. It equals 15.
Anyway, you run through that formula, and you find that the probability of two afflicted students in any one given week is about 1.82 * 10^-8, or 1 in 55 million. (I'll call it P.) Which is to say, not very likely. Assuming that there are 96 weeks where this could happen, and that each week is more or less independent of the others (not quite true, if in one week two students need an appendectomy, there are only at most four students left at risk in any future weeks; but close enough for us), then the odds that you'll be in here one week asking "What are the odds?" are:
1 - (1 - P)^96 = 1 in 573,000.
So I think the Facebook friend overstated the odds by close to double. I suspect this is because he neglected to notice that his method of picking students (first one at random, then the second one at random) double-counts every possible pair. You could wind up with, say, students B and E by picking B and then E, or by picking E and then B. The friend's methodology adds the chance of B-then-E, and then also adds the chance of E-then-B; but really, this is the same event (B-and-E together), and should only be counted once.