Jackson Whipps Showalter vs. Albert Beauregard Hodges Match
? 1894 - 4/10/1894
1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7
Showalter 1 ½ 1 ½ 0 ½ 0 0 1 1 1 0 ½ 0 1 +7-6=4
Hodges 0 ½ 0 ½ 1 ½ 1 1 0 0 0 1 ½ 1 0 +6-7=4
Showalter finished third at New York, 1893, behind 20 year old
Harry Nelson Pillsbury, and Albert Beauregard Hodges. Pillsbury
had no interest in challenging him to a match, but Hodges did.
The match was to be won by the first player to win 7 games.
Stakes were $100 per side. Showalter was White in the odd
numbered games. Play took place at the Manhattan Chess Club.
Games 15 and 16 were both decisive, one being won by Showalter,
the other by Hodges. I'm just not sure which was which.
When the score was tied at 6-6, both players agreed to play a
second match later, regardless of who won the 7th and deciding game.
| #
|
Opening
|
Ended
|
| 1
|
Ruy Lopez
|
?
|
| 2
|
Queen Pawn Opening
|
?
|
| 3
|
French
|
2/17/1894
|
| 4
|
Queen Pawn Opening
|
?
|
| 5
|
Sicilian Defense
|
2/19/1894
|
| 6
|
Queen Pawn Opening
|
2/21/1894
|
| 7
|
Ruy Lopez
|
3/4/1894
|
| 8
|
Stonewall
|
3/7/1894
|
| 9
|
Sicilian Defense
|
3/13/1894
|
| 10
|
Queen's Gambit Declined
|
3/19/1894
|
| 11
|
?
|
?
|
| 12
|
Queen's Gambit Declined
|
3/23/1894
|
| 13
|
?
|
?
|
| 14
|
Colle Opening
|
4/1/1894
|
| 15
|
?
|
?
|
| 16
|
?
|
?
|
| 17
|
Ruy Lopez
|
4/10/1894
|
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