TITLE: Can we determine the game-theoretically best first move by White in chess without solving chess?
QUESTION [5 upvotes]: In turn-based board games with high branching factor (such as chess) are there any arguments that could ascertain the ideal first move but not solve the entire game?
I am asking because solving chess entirely seems infeasible but conceivably one could hope to determine the ideal first moves.

REPLY [8 votes]: Setting aside chess in particular, we can certainly come up with games for which we can determine the game-theoretically best move without solving the game in general.  For example, it is easy to prove, via a strategy-stealing argument, that a game such as Hex is a first-player win.  So you can invent a game called "meta-Hex" in which the first player's first move is to choose whether to be the first player or the second player of a game of Hex, and then to play a game of Hex accordingly.  Obviously, the game-theoretically optimal best first move in meta-Hex is to choose to be the first player in a game of Hex.  We have figured this out without figuring out how to play Hex (or meta-Hex) optimally in general.
For chess in particular, any kind of rigorous proof about the best first move appears to be hopelessly out of reach.  As Carlo Beenakker noted in a comment, the conventional wisdom is that chess is a draw; if true, this might seem to be a favorable situation for mathematical analysis since it would mean that any reasonable first move (and perhaps any first move at all) would probably be game-theoretically optimal.  But in practical terms, we cannot expect a mathematical proof.  (It is also morally certain that if White gives Black queen odds, then Black has a win, but a mathematical proof of even this seemingly obvious fact seems out of reach.)  At the same time, of course, it is even more hopeless to prove that we cannot find such a proof.
It's perhaps worth mentioning Fraenkel and Lichtenstein's paper Computing a perfect strategy for $n\times n$ chess requires time exponential in $n$.  This result provides strong circumstantial evidence that there is not going to be a short proof that $8\times 8$ chess is a draw, but of course their result does not actually say anything about $8\times 8$ chess.