Yes, I got that, I was just curious about what the specific family was.

The convergence to a Gaussian that you describe is essentially due to the central limit theorem, so you basically seem to be looking for a non-positive analog of a stable distribution. What is that picture of yours a plot of, anyway?

@Barium: Ah, yes, I missed the "What if we make it increasingly likely for us to select the previously flipped bit" part in your question. I think we could deal with that by defining the process over the set of edges $E$ instead of the vertices $V$; in particular, in the specific case in your question, it should be enough to store $(n_0, n_1, \Delta n_0, \Delta n_1)$, where $(\Delta n_0, \Delta n_1) \in \{(0,1),(0,-1),(1,0),(-1,0)\}$ denotes the change in $(n_0, n_1)$ during the previous iteration.

I couldn't possibly know, but my suspicion would be that a copyeditor bowdlerized the article without the author's knowledge or permission, and that the author, upon finding out, complained strongly enough for the magazine to give in and publish the correction.

@Francesco: I believe the only problem with your shorter definition of equality is that it turns the axiom of extensionality, as usually written, into a tautology. It should be sufficient if you also rewrite that axiom as $\forall x \forall y ( \forall z (z \in x \leftrightarrow z \in y) \rightarrow \forall w (x \in w \leftrightarrow y \in w) )$. Otherwise, though, you could end up with sets x and y and a set (or, in MK, a class) w such that x = y (according to your definition of "="), $x \in w$ but $y \not\in w$. That would not be good.

When you say "the n-th suffix of the sequence has n for prefix", do you mean that the first n digits in the sequence evaluate to n in some base (i.e., denoting the sequence by $a_1, a_2, \ldots$ and the base by b, that $n = \sum_{k=1}^n b^{n-k} a_k$)?