2

With some extremely rare exceptions, KNN vs KR endgame should be a draw. However, I am wondering what if this endgame happens in a game with no increments. I think the following will happen at chess.com:

  1. If KNN runs out of time, KR wins.
  2. If KR runs out of time, it is a draw.

Isn't this unfair since both sides do not have reasonable winning chances. It seems to me that in this endgame, time out rule favors KR, though KNN is 1 point up in material (if each knight is counted as 3 points and each rook counted as 5 points).

4

I think the following will happen at chess.com:

  1. If KNN runs out of time, KR wins.
  2. If KR runs out of time, it is a draw.

If chess.com abides by the FIDE Laws of Chess then this is false.

If the side with K+R runs out of time then they lose. K+R vs K+N will also lose. This is the relevant article:

6.9 Except where one of Articles 5.1.1, 5.1.2, 5.2.1, 5.2.2, 5.2.3 applies, if a player does not complete the prescribed number of moves in the allotted time, the game is lost by that player. However, the game is drawn if the position is such that the opponent cannot checkmate the player’s king by any possible series of legal moves.

Here is the relevant helpmate position for K+N vs K+R. No second knight is needed.

[FEn "kr6/2N5/K7/8/8/8/8/8 w - - 0 1"]

Isn't this unfair?

If what you say is correct regarding chess.com's interpretation of the rules then, of course, it is grossly unfair. They would be much fairer if they abided by the FIDE Laws of Chess.

1
  • I completely agree with Brian’s analysis. Building on it KRvKB would be a draw if White times out, as there is no mate for Black. At some level the chess.com app can only approximate the dead position rule and the “half-dead” timeout rule here, using some notion of insufficient material, but maybe they are not even doing that correctly – Laska Feb 25 at 3:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.