I recently came up with the following chess puzzle (which has almost nothing to do with one's Chess skills):

Puzzle: Consider a variant of chess where black has to start with $1...e5$ regardless of white's first move. All other rules remain unchanged. Prove that in this variant, white can at least force a draw.

I have told this it to several people, and some of them seem to think that my solution is wrong.

Solution: consider the position where white has a pawn on $e4$ and black has a pawn on $e5$, with all other pieces at their starting positions. Let's refer to this position as $P$, the player with the next move as player $1$, and the other player as player $2$. Then according to Zermelo's Theorem, starting from position $P$, either player $1$ can force a win, or player $2$ can force a win, or they can both force a draw. We consider them one by one:

  • If they can both force a draw starting with position $P$, all white has to do is to reach this position by playing $1.e4$, as black will have to go $1...e5$ which creates position $P$.

  • If player $1$ can force a win, white has to reach position $P$ and be the player with the next move. In this case, white will still play $1.e4$, which is followed by $1...e5$. At this point, position $P$ has been reached and white is the player with the next move.

  • If player $2$ can force a win. White will play $1.e3$ and after black plays $1...e5$, white will play $2.e4$. Now we're in position $P$, and it's black to move. This means that black is player $1$, which makes white player $2$.

They argue that position $P$ is not the same position when player $1$ is white and when it is black. When player $1$ is black, their king is on their right side, whereas when player $1$ is white, their king is on their left (same goes for the queen).

My answer is that starting with any symmetric position (like $P$, or the starting position in standard chess or this position) if white had a winning (drawing) strategy $\mathcal{A}$ starting with position $P$, then if black were to start instead, it would have a winning (drawing) strategy $\mathcal{A}'$ where for every move (black's or white's) in $\mathcal{A}$, row $i$ is replaced with $9-i$, where $1 \leq i \leq 8$ (e.g. $Nc6$ would become $Nc3$). Thus position $P$ is the same regardless of who starts.

Is my proof correct?

  • $\begingroup$ Strictly speaking, you should notice that there are no possible en passant captures (since that's the one potential difference between your 1. e4 and 2. e4 positions). $\endgroup$ – Micah Feb 10 at 6:08

Yes, this is a fine strategy stealing argument. The mirror image of the board does not matter as castling is defined in a way to respect the mirror image and all other moves are symmetric.

  • $\begingroup$ Thanks for the name! Apparently, I still cannot upvote :) $\endgroup$ – PkT Feb 10 at 6:28

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