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Is it proved that white can guarantee at least draw in chess?

A while ago I was told that it was proved using strategy-stealing, but I cannot find a reference.

Postscript. Please accept my apology --- most likely there is no such theorem, otherwise the reference would be already founded.

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    $\begingroup$ I am almost sure that it is not proved. There are quite subtle positions with mutual zugzwang, and why not the initial one? $\endgroup$
    – Fedor Petrov
    Commented Jan 7, 2022 at 21:07
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    $\begingroup$ There is a known problem in which the stealing strategy works: chess in which each move consists of two usual moves. Then White may steal the strategy by Kc3+Kb1, although unlikely this helps you to draw against very good chess player $\endgroup$
    – Fedor Petrov
    Commented Jan 7, 2022 at 22:28
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    $\begingroup$ @user, which letter stands for which piece may depend on what language you learned your chess in. $\endgroup$
    – Gerry Myerson
    Commented Jan 8, 2022 at 3:37
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    $\begingroup$ Anton, is not "White can at least draw" just equivalent to "initial position is not a mutual zugzwang"? $\endgroup$
    – Fedor Petrov
    Commented Jan 8, 2022 at 4:05
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    $\begingroup$ I think this is a well-known open problem (apparently attributed to Tutte!). It has been discussed on MO before: mathoverflow.net/questions/107385/a-chess-question-of-w-t-tutte . That question was closed, I think basically because it's a well-known open problem. $\endgroup$
    – HJRW
    Commented Jan 11, 2022 at 17:12

3 Answers 3

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Usually the strategy-stealing argument would go: suppose Black has a winning strategy. Then White can make an arbitrary first move, and then follow Black's winning strategy (with the adjustment that White makes an arbitrary move if the move White is suppose to make has already been played as a result of a previous arbitrary move).

However, this argument only works for games where playing an arbitrary move does not hurt you; this is not the case with chess, where zugzwang exists. So, the strategy-stealing argument does not go through.

It is mentioned in the answers to If there is a winning strategy, is it for White? that a proof of "White has a guaranteed draw or better" does not exist, and if it does, would be very difficult to find.

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  • $\begingroup$ "this argument only works for games where playing an arbitrary move does not hurt you" --- that is not true. $\endgroup$
    – Anton Petrunin
    Commented Jan 7, 2022 at 23:54
  • $\begingroup$ @AntonPetrunin Perhaps there are games where playing an arbitrary move can hurt you, but the strategy stealing can be modified in such a way that this does not end up being an issue; however, I am unaware of any such examples. Do you have one in mind? $\endgroup$
    – ckefa
    Commented Jan 8, 2022 at 0:15
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    $\begingroup$ What about chomp? $\endgroup$
    – Anton Petrunin
    Commented Jan 8, 2022 at 0:29
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    $\begingroup$ @AntonPetrunin Ah, I see. That's a good example of how strategy stealing can be modified to avoid the issue of making arbitrary moves that hurt you. However, note that Chomp is not a counterexample to the claim I made in my answer: the basic "make an arbitrary move and then strategy steal" argument does not work for Chomp, since that includes the case where your arbitrary move is to take the poisoned square. I'm not saying that a modified strategy stealing argument cannot exist for chess, only that the basic one does not work. $\endgroup$
    – ckefa
    Commented Jan 8, 2022 at 1:07
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I think the real essence of the question is not so much in whether zugzwang exists in chess, but whether it exists in the first few (say 5) moves, and whether in that time white can effectively lose a move. I feel like the tree diagram for this shouldn't actually be that deep, if it's done in some clever way. You shouldn't have to do the whole game tree.

I'm thinking something along the lines of (assuming black can win), if the winning response to 1. e4 is e5, then the response to 1. e3 cannot also be e5, because then 2. e4 strategy steals. If you can piece together enough of those, perhaps you can prove that white can always strategy steal.

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    $\begingroup$ I'm not downvoting, but this seems immensely unlikely and I'm a little surprised it's gotten this much positive feedback. $\endgroup$
    – Steven Stadnicki
    Commented Jan 11, 2022 at 1:11
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The type of strategy-stealing in chomp (mentioned by the OP in a comment) is exactly the one which could, in principle, be applied to chess: you don't want to neglect the first move because "it won't hurt", you actually aim at "reabsorbing" it so as to make W play exactly the same winning positions you are assuming B has at his disposal.

But, as it was said in the answer by ckefa, this, even if possible, would be very difficult. However the converse direction seems perhaps more tractable: is it possible to show that, if a winning strategy for B exists, it is not possible to steal it with W in the above-mentioned sense? For instance: if copying W moves as far as possible (e.g. 1. e4 e5 2. Nf3 Nf6 3. Nc3 Nc6 4.Be2 Be7 5. d4 d5 ecc... ) is the beginning of a winning strategy for B, it's plausible that it cannot be stolen by W.

(Of course this is of scarce practical relevance, because it seems so unlikely that B wins...but maybe it could be interesting for more abstract reasons).

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