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Recently a game was played where two GMs swapped kingside and queenside knights and then agreed a draw (and got their just dessert): How often can the starting position be repeated in a game without automatic draw? Since even four castling rights can be forfeited, the answer surely is "75 moves".

EDITED OUT: The not 100% FIDE-compliant, but mathematical more interesting question is that you assume automatic draw on repetition but NOT on 75 moves without pawn move or capture, and no player claims. This will be a very large number, in the order of (32 over 4).

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  • For the second question (repetition not number of moves) I suggest you look at chess.stackexchange.com/questions/43548/… which poses the same question in in the context of the Codex where it is actually the case that 3-fold repetition does end the game, while 50-moves do not. This question really has nothing to do with FIDE
    – Laska
    Feb 23 at 7:29
  • In that case my "intended" question is a factual duplicate, and I accept the answer below. I hesitatingly minimally edit my question. Feb 23 at 8:25
  • Cool - please do have a shot at my own mathematical question some time
    – Laska
    Feb 23 at 8:42

1 Answer 1

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You appear to ask several different questions, so I will answer your headline question:

How often can the initial position be repeated under FIDE tournament rules?

You give a very big clue to the correct answer with this:

two GMs swapped kingside and queenside knights and then agreed a draw (and got their just dessert)

To elaborate, the arbiter intervened and forfeited both players for bringing the game of chess into disrepute. This is now a thing with arbiters at all levels of FIDE rated tournaments. In last year's Major&Minor tournament in the Isle of Man (from which you would be excluded because of your high rating), run alongside the Grand Swiss, two players played:

[fen ""]

1. e3 e6 2. Ke2 Ke7 3. Qe1 Qe8 4. Kd1 Kd8

at which point the arbiter came along and tried to replace the kings and queens on their correct squares on the assumption that the board had just been set up wrong. The players corrected him and said they were about to agree a draw. The arbiter stopped the clocks and told them he would have to consult with the chief arbiter of the Grand Swiss. After doing so he meted out the same punishment - both players forfeited for bringing the game into disrepute.

Hence under FIDE tournament rules with an arbiter present the game would last until the arbiter decided the players were bringing the game into disrepute, which would be a lot fewer than 75 moves and the result would not be a draw.

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  • Brian, could you edit in the link to the recent incident? My googlefu was weak, I couldn't find it. (Also, your answer cries for the next question: when will an arbiter decide that not chess but shenanigan is played? 1.Nf3 Nc6 - unusual, 2.Ng5 Nb4 - aw cmon, 3.Ne4 Nd5 - time for a double forfeit? §whatnot tells arbiters to act with utmost tact, but I would be tempted to pull the plug already at move pair 2.) Feb 19 at 7:30
  • @HaukeReddmann If you are talking about the IoM incident then you won't find any but the most indirect evidence on Google. Had I been the arbiter (and not subject to external pressure) the incident would have appeared in chess-results as 0F-0F. However it was recorded as both players not paired and not playing. The only reason I know about it is because I now live and play my club chess on the IoM.
    – Brian Towers
    Feb 19 at 11:35

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