Highest Possible Crit Range?

[ocdscale]

If you only gain one attack per crit, that's analogous to flipping a coin until you hit tails. Did you mean getting a free attack, in addition to another roll?

I understand terminally's analogy, so I hope he doesn't mind if I explain it.
I still disagree with his proof however, but I have no means to verify it. My main disagreement being the proof I can see is better than the proof I cannot.

Anyway, the analogy
Imagine the number you stand on is always equal to the number of attacks you posses.
You start on the number corresponding to your full attack string.
Roll one attack. If it crits, move one step in the positive direction (you gain two attacks to roll, but you used up one). If it misses or otherwise fails to crit, move one step in the negative direction (you lose one attack).

It's a random walk. The proofs I've seen suggest a biased random walk does not have a 100% probability of crossing each value (in this case 0), analogizing from a biased gambler's ruin proof.

To respond to Kalirren:
Edit: Cleaned up my post,
First: The Ap^2 term is clever, but I'm not sure the Bp term is right. Shouldn't it just be B (failing to confirm is just as bad as missing outright)?

Second: I've convinced myself that if anything, the p=1 must be the extraneous root.
Simple proof: Set A=1, B and C to 0.
This simulates the build that never misses, always crits, and always generates double attacks. I think everyone in the thread would agree this hypothetical build is capable of infinite hits.
p = Ap^2 + B + C
Still has p=1 as a root, which of course makes no sense.

The reason p=1 exists as a root is because the roots of the equation aren't the "answer" to what p is in the system modelled, they only represent values of p that are consistent with the model. In other words, the 'right' value of p must be consistent with the model, but not all values that are consistent are 'right'.

The reason p=1 is consistent with a model that never misses is because the model is calculating p in a circular fashion. It's a very clever method of course (I wish I had thought of it first), but the circularity is the flaw.
I'm fairly certain the non p=1 root does represent the "answer" simply because the quadratic can only have two roots, and if one is logically unsound, it other must be right.

Edit 2:
Sigh. Threads like this make me wish I continued to pursue mathematics instead of going into law. Math is fun.