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Out of curiosity, how many en passants can theoretically be done by both sides and the fastest way to achieve that result?

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8

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
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  • 1
    You reach the correct conclusion but your reasoning is unsound. A pawn captured ep never got further than its 4th rank. But a pawn capturing ep thereby reaches its 6th rank, and so can't capture more than one pawn ep and can't get captured ep. So no pawn can both capture ep and get captured ep. So each ep capture involves two pawns (one white and one black) which are not involved in any other ep capture. So there are no more than 16/2=8 ep captures.
    – Rosie F
    Dec 13 '18 at 10:05
  • 1
    The reasoning is correct, but it does not go with the example. My reasoning implies that only one side will take the other side's pawns en passant.
    – Wais Kamal
    Dec 13 '18 at 17:29

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