Out of curiosity, how many en passants can theoretically be done by both sides, and what is the fastest way to achieve that result?
Every pawn has to take a pawn in an en passant capture, and each pawn taking another pawn will leave the 5th rank. Hence, the total possible number of en passant captures in a game is simply the number of pawns halved, i.e. 8. It can be achieved in the following sequence of moves:
[FEN "rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1"] 1. h4 a5 2. h5 g5 3. hxg6/ep a4 4. b4 axb3/ep 5. a4 h5 6. a5 b5 7. axb6/ep h4 8. g4 hxg3/ep 9. c4 f5 10. c5 d5 11. cxd6/ep f4 12. e4 fxe3/ep 13. f4 c5 14. f5 e5 15. fxe6/ep c4 16. d4 cxd3/ep
This is the minimum number of moves required for 8 en passant captures, and this is due to:
- All moves being pawn moves
- Every pawn advancing 2 squares at its first move