Assuming the 128-bit IV is indeed randomly chosen, after encrypting a trillion blocks, you would have roughly a 1.5E-15 (on the order of a quadrillionth) chance of hitting a collision.
Unless I'm mistaken, even if you hit that one-in-a-quadrillion lottery, the result is that the two blocks encrypted with the same IV are more crackable (because you can XOR them together), not that the key itself is easier to obtain, right? (My understanding is that AES is resistant to known-plaintext attacks.)
Assuming the 128-bit IV is indeed randomly chosen, after encrypting a trillion blocks, you would have roughly a 1.5E-15 (on the order of a quadrillionth) chance of hitting a collision.
Unless I'm mistaken, even if you hit that one-in-a-quadrillion lottery, the result is that the two blocks encrypted with the same IV are more crackable (because you can XOR them together), not that the key itself is easier to obtain, right? (My understanding is that AES is resistant to known-plaintext attacks.)