user-1: Have mathematicians created any theorem that determines the outcome of a endgame? If so, how could it be used to win a specific endgame?

user15223: 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).

user11144: 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).


-1: <p>Have mathematicians created any theorem that determines the outcome of a endgame? If so, how could it be used to win a specific endgame?</p>
 15223: <p>Such a theorem doesn't exist.</p>
<p>The closest is the <a href="https://en.wikipedia.org/wiki/Strategy-stealing_argument" rel="noreferrer">strategy-stealing argument</a> 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).</p>
<p>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).</p>
 11144: <p>Unfortunately the (original) question was a bit ambiguous and arguably still is. I try to sort the different interpretations out:</p>
<ul>
<li>Chess fulfils all requirements of &quot;standard&quot; game theory (zero-sum, two players, full information, no chance). <a href="https://en.wikipedia.org/wiki/Minimax_theorem" rel="noreferrer">This</a> is your main theorem. Thus Chess <em>could</em> be solved principially - &quot;just&quot; 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.</li>
<li>OK, assume you meant &quot;solve some <em>specific</em> given position&quot; instead. In this case, an unconditional &quot;yes&quot;. For only one example, look up Noam Elkies' work on <a href="https://arxiv.org/abs/math/9905198" rel="noreferrer">simple pawn endings</a> (although identical results can be obtained with classical chess analysis - arguably it looks more elegant in <a href="https://en.wikipedia.org/wiki/Surreal_number" rel="noreferrer">Conway notation</a> instead - at least to a mathematician).</li>
</ul>
