Have mathematicians created any theorem that determines the outcome of a endgame? If so, how could it be used to win a specific endgame?


3 Answers 3


Unfortunately the (original) question was a bit ambiguous and arguably still is. I try to sort the different interpretations out:

  • Chess fulfils all requirements of "standard" game theory (zero-sum, two players, full information, no chance). This is your main theorem. Thus Chess could be solved principially - "just" create a 32 piece tablebase :-) In praxis, this of course is hopeless, cf. Allure's points. At the moment (2021), Chess is completely solved for 7 piece endgames, 8 is the next goal and in the works.
  • OK, assume you meant "solve some specific given position" instead. In this case, an unconditional "yes". For only one example, look up Noam Elkies' work on simple pawn endings (although identical results can be obtained with classical chess analysis - arguably it looks more elegant in Conway notation instead - at least to a mathematician).
  • "Just" create a 32 piece tablebase? Is it possible to create a complete, finite move tree for chess? There could be infinite looping between various positions if neither player claimed a stalemate by the fifty move rule, couldn't there? Commented Dec 17, 2021 at 11:38
  • If there is a finite tree of all the legal moves then yes, there is a theorem about how to win any chess game: "just" build the tree and prune back from all the leaves that don't result in a win. Commented Dec 17, 2021 at 11:41
  • 3
    @ChristopherHamkins Yes. First, 75 move rule is mandatory so the game is not infinite. Besides, even if you ditch that rule so you can force one of long checkmate sequences, number of possible board states is finite - so you require a loop for an endless game. But any repeat of a position trivially means draw as no move in a loop gets you any closer to the win or lose, so you can simply prune those loops to "draw", keeping tree size finite. Commented Dec 17, 2021 at 12:41
  • @ChristopherHamkins Creating the tree is problematic, there are supposedly 2*10^46 possible positions. So, about 1k-10k atoms of Earth per position. That said, despite this difficulty of listing moves, a proof of draw or any side winning is not impossible despite that - heck, it might be even possible to find a compact solution that proves "X moves to checkmate" for every possible white and black move (assuming perfect play from that point onward). Commented Dec 17, 2021 at 12:57
  • 1
    @ZizyArcher We're talking about math, not physics. Mathematically, it doesn't matter if there's more possible positions than atoms on the earth. It's about provability, not practicality. Commented Dec 17, 2021 at 17:02

Such a theorem doesn't exist.

The closest is the strategy-stealing argument that can be used to show that for many games, the second player does not have a winning strategy. The gist of the argument is that if the second player did have a winning strategy, then the first player can play a random move and then adopt that strategy. If the random move is never a disadvantage (e.g. in Hex) then the first player would have a winning strategy, thereby proving by contradiction that the second player cannot have a winning strategy and the first player always wins (or the game ends in a draw).

This argument does not work for chess because there are positions where you don't want to make a move since any move is fatally weakening (zugzwang).

  • It's widely believed that, if chess were solved to the extent that TicTacToe is, then a player playing white perfectly against a perfect black player, white would either be able to force a win or at least a draw, but no such thing has been proven yet. I do wonder if it's been proven that there is some sort of perfect <forced win / draw> for at least one of the sides, though.
    – TKoL
    Commented Dec 16, 2021 at 16:50
  • Edit to my previous answer: it looks like the common opinion is that both players playing perfectly would be a draw. A perfect white player cannot force a win.
    – TKoL
    Commented Dec 16, 2021 at 16:53

Your question is a bit ambiguous, so I will attempt to answer it in the manner I think you will find most useful.

  1. Well the whole of chess / 9LX isn't solved, but there are endgame sort of 'theorems' in the sense that there are positions that are provably wins/draws and so if you know the theorems and their proofs, then theoretically you know how to win/draw those positions.

See the 1st 40 seconds of this video:


  1. Loophole: these are mathematical theorems in the sense that chess / 9LX is a mathematical game

  2. Karsten Müller talks about something like this at timestamp 5:20 in an interview with chessbase india: Endgames are like pure maths in the sense that they are something to prove. In contrast, openings are like applied maths or statistics like 'which opening has a high/er draw rate?' or 'which opening leads to early queen trades (or reaches endgame early)?'


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.