on 07/10/10 at 18:03:05:
Was a conclusion ever reached in the Kaufman/Pinski line: 1.e4 e5 2.Nf3 Nc6 3.Nc3 Nf6 4.Nxe5 Nxe5 5.d4 Nc6 6.d5 Bb4 7.dxc6 Nxe4 8.Qd4 Qe7.
Pinski gives 1.e4 e5 2.Nf3 Nc6 3.Nc3 Nf6 4.Nxe5 Nxe5 5.d4 Nc6 6.d5 Bb4 7.dxc6 Nxe4 8.Qd4 Qe7 9.Qxg7
(?!) Nxc3+ 10.Be3 Nd5+ 11.c3 Rf8 12.cxb4 Nxe3 13.fxe3 Qxb4+
as "very close to winning," and I don't have Kaufman on hand to see if he adds anything (IIRC, he succinctly repeats Pinski).
More analysis is given in this very extensive PDF file
on the Halloween Gambit by Paul Keiser.
In short, I think most lines are roughly equal or thereabouts after 9.Be3! Nxc3
[9...0-0 10.Bd3 Nxc3 11.bxc3 Bd6 12.cxd7 (12.cxb7 Bxb7 13.0-0)
Bxd7 13.Qe4 Qxe4 14.Bxe4
9...f5 10.Bd3 Bc5 11.Qc4 Bxe3 12.Bxe4]10.a3 Nd5
[10...Bd6 11.cxd7+ Bxd7 12.Qxc3]11.axb4 Nxe3
[11...Qxb4+ 12.Qxb4 Nxb4 13.cxb7 Bxb7 14.0-0-0]12.cxd7+
(or 12.fxe3) Bxd7 13.Qxe3 Qxe3 14.fxe3
At any rate, this antidote isn't very troublesome, and there are better ways to give the piece back if one doesn't want to try to refute the gambit outright.