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Consider a standard game of chess. We make the following modification: on a turn, if a player makes a move which neither captures a piece nor puts their opponent’s king in check, then they may make a second move on the same turn with the same properties. (This does not stack, so at most two moves are made per turn.) Can Black guarantee victory in a finite number of turns?

From OTIS by Evan Chen

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No - if there were a strategy to guarantee victory in a finite number of turns, White could make their first move Nc3 Nb1, and then copy that strategy to guarantee victory against Black - a contradiction.

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  • $\begingroup$ Some additional explanation might be helpful as I'm failing to see how... for instance, what does it mean by >! "copy that strategy"? The strategy of moving both knights on each turn? How does the guarantee victory? $\endgroup$
    – Michael
    Commented Dec 4, 2023 at 8:41
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    $\begingroup$ @Michael. Nc3 Nb1 is a "no-op". It leaves white's pieces exactly where they started ... and now the other player (black) has to move. Which is exactly the position that black was in at the start of the game. Essentially White can choose their first move to be "I want to let you go first"/ "I want to be black instead". $\endgroup$
    – Brondahl
    Commented Dec 4, 2023 at 9:02
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    $\begingroup$ "Copy that strategy" means "apply whatever winning strategy Black possesed, to win from the starting board. $\endgroup$
    – Brondahl
    Commented Dec 4, 2023 at 9:08
  • $\begingroup$ Doesn't the arrangement of the king and queen prevent White from using some strategies Black might have? In a standard game of chess, the king starts to the right of the queen for White, but from Black's perspective with the board flipped, the king starts to the left of the queen. $\endgroup$
    – zaen
    Commented Dec 5, 2023 at 7:24
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    $\begingroup$ @zaen Every strategy can be copied - just reflect the strategy over the line separating the 4th rank from the 5th. You might have been thinking about reflecting across the center of the board, which would not work indeed. $\endgroup$
    – Adayah
    Commented Dec 5, 2023 at 13:58

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