I'm not talking about Marseillais chess (which is similar, but rules are added to reduce the advantage of the white), which is sometimes also called "balanced double-move chess".

In a Double-Move chess both players move two times instead of once. There is no check, capture the king to win.

Is the advantage of white sufficient? I.e. is it feasible to prove that white can always force a win?

Was a question like this examined in the past? I'm looking for references on any related work.

For example, if we were talking about Triple-Move chess (both players move three times instead of once, there is no check, capture the king to win), then white always wins on the "second move" by sacrificing the queen. That is, white makes their "first move" by: pushing e pawn, queen to h5, queen to f7, and black is lost!

To see why black is lost, observe the f1 bishop. It can capture the king on the next move either by b5, d7, e8 or by c4, f7, e8 (the white queen must be captured). If black captures the queen with anything but the king (i.e. spends two moves on the knight), then black does not have enough moves to defend both lines of attack from the bishop. Otherwise, if black captures with the king, then there are again two lines of attack from the f1 bishop and black again does not have enough moves to defend both. Black king gets captured on the "second move" in every scenario.

If we reduce the variant to Double-Move chess, is the advantage for white still significant enough such that we can possibly find the (least?) number of moves in which the white can force a win (capture black king), or is this variant as hard as solving the ordinary chess?

Edit: I'm aware that "If there was a perfect play from both sides, but each player was allowed to do 2 moves every time, would white win?" was asked before, but there it is not clear if the OP refers to the (balanced) Marseillais chess, or not. - Also, the answer there isn't useful as it simply states "White would probably win. However, this is not certain..." without further clarifications. (I.e. they appear to claim that it is as hard as ordinary chess, but this is not obvious.)

  • A first try is opening e3, Qf3 to start the threat on f7. Qh5 fails to Nf6, Nxh5. Tries with the KB can be met by advancing a pawn and capturing, but if black blocks the threat from the queen Bc4 Bb3 may be promising. I used e3 instead of e4 to allow the Q diagonal lines. – Ross Millikan Sep 3 '20 at 15:09
  • After White moves 1.e4 2.Qh5 and 3.Qxf7, Black can defend with 1...Kxf7 2...g5! and 3...Kg7 with a probably winning position – David Sep 4 '20 at 8:27
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    @David Bishop on c1 is attacking g7 and e7, each with two different paths, so the King still gets captured because you don't have time to defend both paths. – Vepir Sep 4 '20 at 12:22
  • @Vepir that's a nice little speech, but you haven't told me what the winning sequence is – David Sep 4 '20 at 18:27
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    @David Follow the second path that isnt' defended: 1. push b pawn, 2. Bishop to b2, 3. Bishop to g7 captures the King. – Vepir Sep 4 '20 at 18:29

We will prove that there exists a strategy for the white player that guarantees them a draw in the case of Double-Move chess.

Assume that such a strategy does not exist. This means that whatever move is made by the whites, there exists a strategy for the blacks leading to a win for them, i.e., the black player has a winning strategy. However, if the whites move a knight forth and back, the position on the board will be the same for the black player, as it was the initial position for the white player! Therefore, after such a move, the black player is not able to guarantee a draw and thus the white player must have a winning strategy. This is a contradiction since we assumed that the whites do not even have a strategy guaranteeing them a draw.

  • The flaw in your argument is that there are two possibilities if white doesn't have a win and so your "proof" fails. One possibility is that black has a forced win but the other possibility is that there is no forced win and best play by both sides is a draw. You have not excluded that possibility. – Brian Towers Mar 25 at 0:27
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    @BrianTowers If there is no forced win and optimal play is a draw, then we already have the result we want (a guaranteed draw). – Edward Mar 25 at 2:11
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    There is still the possibility that white has a forced win but not black. This proof makes the hidden assumption that if white has a forced win, they also have a forced draw. This might be true but we haven't shown it, so our actual result here is that white can never lose with optimal play, or that white has at least a forced draw (if not a win). – Edward Mar 25 at 2:26

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