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Khoth comments on A solvable Newcomb-like problem - part 3 of 3 - Less Wrong
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Is the ultimate point of this set of posts this:
If you assume Omega has a probability δ of making an incorrect prediction, then your optimal strategy tends to one-boxing as δ tends to 0, therefore you should one-box in the original problem.
It seems like a lot of words to not quite come out and say that, though, so I'm probably missing something.
One has to be careful how one does these things.
For example, if you are considering what zero divided by zero is, it matters whether you look at x/0 as x tends to 0, 0/y as y tends to zero, t/t as t tends to zero, or various permutations with approaching it from the negative direction. If you do a 3D plot of z = x / y, you see the whole picture.
The chance of Omega making an incorrect prediction isn't the only variable in the problem. It is just the one that's most often concentrated upon.
You would say these words if you would want to check whether you have found a "stable" equilibrium.
If you consider that "0 is not a probability", there is a point in checking for that.