Game 215 in Wilhelm Steinitz’s The Book of the Sixth American Chess Congress (1891), printed pages 276–277. Played in the second round at New York on May 10, 1889. The original notes retain descriptive notation.

1. 2…d6 Black is already at some disadvantage, owing to his KB being confined.
2. 4…Be7 B — K3 is now better. The doubled Pawn that White could create would be compensated by the open KB file, and the probability that White would have to advance his own KBP two squares sooner or later, thus enabling Black to dissolve that doubled Pawn.
3. 9.Qd2 B—K3 instead would have given White a stronger attack. If then 9...B—R5 ch.; 10 K—Q2, whilst now we prefer the adverse game.
4. 11…Nd4 Kt—K4 was better, for White could not well take that Knight, as a hole would be formed at his KB4 which would be strongly commanded by Black’s minor pieces. If then 11 Q—K3, 11 P—QB3; 12 KB—QKt3, 12 Q—B3, with the superior game.
5. 12…Nxf4 Black loses all his advantage now, and he has the inferior position for the ending, his Pawns not being so compact as those of the opponent. Kt—K3 was by far better.
6. 18…a5 B—Kt3, threatening to win a piece by Kt x B, would have gained an important move at least.
7. 27…c5 The hole here formed makes his game soon untenable. B — Q sq., with the view of playing P — KKt 4 in reply to P—B5, was his best defence.
8. 33.Ra2 A quicker way of winning was 33 Kt—K7 ch., 33 K—Kt sq.; 34 Kt—B6, 34 B—Kt sq. (or 34...K—Q2; 35 P—Q5, followed by R—KB3, with a winning game); 35 P—K5, 35 P x P; 36 P x P, 36 K—K2; 37 Kt x B, 37 R x Kt; 38 R x R, 38 K x R; 39 P—K6, 39 P x P; 40 P—B6, and wins.
9. 35…g4 His game becomes quite hopeless after this. K—K3, followed by B x P, gave him a better chance of prolonging the struggle.