Wayne Conover vs Alex Yermolinsky
Philadelphia, 1993 · Result 0–1 · Indian Defense: Normal Variation (A50).
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Game details
- White
- Wayne Conover (2310)
- Black
- Alex Yermolinsky (2615)
- Result
- 0–1
- Event
- Philadelphia
- Year
- 1993
- Opening
- Indian Defense: Normal Variation (A50)
About this chess game
This chess game between Wayne Conover (2310) and Alex Yermolinsky (2615) was played at Philadelphia in 1993 and finished 0–1. The opening was the Indian Defense: Normal Variation (A50). You can replay the full game move by move on the interactive board above, or open it on the CipherChess analysis board to study every move with the Stockfish engine.
Looking for more Wayne Conover games or Alex Yermolinsky games? This Wayne Conover vs Alex Yermolinsky encounter is one of millions of chess games indexed in the CipherChess mega database. Browse both players' full records, the openings they play most, and head-to-head results, then load any game onto the board to prepare your own lines against the Indian Defense: Normal Variation.
Frequently asked questions
Who won Wayne Conover vs Alex Yermolinsky?
Wayne Conover vs Alex Yermolinsky (1993) finished 0–1, a win for Alex Yermolinsky.
What opening was played in Wayne Conover vs Alex Yermolinsky?
The game opened with the Indian Defense: Normal Variation (ECO A50).
Can I replay this chess game move by move?
Yes. Use the interactive board on this page to step through every move of Wayne Conover vs Alex Yermolinsky, or open it on the CipherChess analysis board to review it with the Stockfish engine.