Answers.Body: <p>There is likely not a way to force a loss. Selfmates are relatively rare, and one can probably not be forced from the opening position.</p>
  
  <p>The best plan, paradoxically, is likely to try to capture as many of the opponent's pieces (but not pawns) as possible, while keeping your own pieces to be better able to force your opponent. The opponent's pawns are kept because they have very little choice as to where they can move, and can also be easily blocked to avoid the opponent moving them before you want them to. For example, in the following position, the f-pawn is blocked and the king cannot move, so Black has no choice but to advance the g-pawn and will have to checkmate White in two moves:</p>
  
  <pre><code>[FEN "3Q4/8/8/4N3/6p1/7k/5p2/5N1K b - - 0 1"]
  </code></pre>
  
  <blockquote>
    <p>If not, can one side force a draw</p>
  </blockquote>
  
  <p>Well, if checkmate indeed cannot be forced, then almost by definition a draw could be forced. The game can't go on forever, after all. If nothing else, the players will eventually run out of pawns to move or pieces to capture, and the 50 move rule will come into play.</p>
  
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Answers.Text: The question of whether one would be able to force stalemate in the absence of the 50-move and repetition rules would still be relevant.  Obviously even without the 50-move rule, any game which doesn't result in a checkmate or stalemate would necessarily result in an eventual draw by repetition, since there are only a finite number of possible positions.
Answers.UserDisplayName: supercat
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Answers.Text: Try it against [WorstFish](https://lichess.org/@/WorstFish) (yes, Stockfish playing to lose), surprisingly hard!
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Answers.Body: <p>As far as I know, there is no chess variant that has been investigated that is exactly as you describe. However, <a href="https://en.wikipedia.org/wiki/Losing_chess" rel="noreferrer">antichess</a> is a quite similar concept. The objective is to lose all your pieces/be stalemated, and it is required for you to take a piece if you can. It was weakly solved to be a win, starting with 1. e3. It is still interesting, however, as many of the lines are extremely long (over 100 moves). </p>
  
  <p>In order to answer your exact question, it would be fairly difficult, but might be slightly easier to solve than regular chess. A powerful computer would likely be needed in order to figure it out.</p>
  
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Answers.Text: The antichess variant is different because it requires capturing when possible, but without this constraint, I guess it becomes a question of whether one side can get a [selfmate](https://en.wikipedia.org/wiki/Selfmate) position while the other side tries to move a piece back and forth until it's a draw or to get a stalemate position.
Answers.UserDisplayName: Victor
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Answers.Text: Note that [*weakly solved* is specific game theory terminology](https://en.wikipedia.org/wiki/Solved_game), which means that the solution gives a full strategy starting from the beginning of the game against all possible opponent choices (but not from positions which you never would reach if you followed that strategy, that's the "weakness"). Not quite as weak as someone would think if they don't know the term!
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Answers.Text: http://magma.maths.usyd.edu.au/~watkins/LOSING_CHESS/
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Answers.Body: <p>Finding the optimal strategy requires the investigation of pretty much all (pretty much, in the grand scale of things) the positions that are possible to get to from the start.</p>
  
  <p>While developing a computer program that plays extremely well in chess (or other similar games) is a solvable task—actually, it is a solved task somewhere in the beginning of the XXI century—finding and proving an optimal strategy <strong>is a totally different problem</strong>. Chess is hard to fully analyze because of the large number of possible positions. </p>
  
  <p>To go more formal, consider <a href="https://en.wikipedia.org/wiki/Shannon_number" rel="noreferrer">Shannon number</a> <code>10^120</code> which is a lower-bound on the game-tree complexity of chess. It is true that the sensible games would go much lower to around <code>10^40</code>, but it will not matter too much.</p>
  
  <p>For the game variant you describe, one has to find the <em>Shannon-like</em> number to say about the feasibility of finding the optimal strategy. However, simply by looking at the rules, I doubt it would result in something that is feasible.</p>
  
  <p>Thus, I don't expect an optimal playing strategy for this variant to be found any time soon. While it might be slightly easier than for regular chess, the complexity will still be enormous.</p>
  
  <p>NB: <a href="https://en.wikipedia.org/wiki/Losing_chess" rel="noreferrer">Antichess example</a> is different because of the introduction of the compulsory capture, which severely reduces the number of possible variations.</p>
  
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Answers.UserDisplayName: Anton Menshov
Answers.Body: <p>What a great question.</p>
  <p>One important mathematical model for chess is Combinatorial Game Theory. Under the vanilla version of this beautiful theory, developed by Conway, Berlekamp &amp; Guy, one loses when one has no legal moves. One modification of the theory is so-called &quot;misère&quot; (a term taken from card-games) where one <em>wins</em> when one has no legal moves.</p>
  <p>A game and its misère form thus share the same game tree. They just have opposite values assigned to the leaf nodes. One can derive values for all other nodes in the game trees by working back from the leaf nodes. One amazing feature is that the value of non-leaf positions between misere and non-misère versions of the same game can be very similar.</p>
  <p>For example, the winning strategies for the well-known game of nim are essentially the same. It's only with the winning player's final choice that they decide to take one more or one less match from the last non-singleton pile.</p>
  <p>Another way to say this: players vie for strategic dominance. Once that is achieved, then one could force the opponent to do a number of different tasks: be checkmated, to checkmate, to send king on journey to occupy every square of the board, etc. The difference between these tasks would be secondary - once strategic dominance is achieved. In practice, it's not as clear cut: part of achieving strategic dominance in chess involves threats of mating. An attack may sacrifice strategic dominance for a winning tempo advantage in order to checkmate quickly. But nevertheless, regular chess and misère have much in common.</p>
  <p>Now chess problemists have long had a misère genre they term &quot;selfmates&quot;. (The problem database PDB contains over 58,000 of these today! Type g='s#' in the window <a href="https://pdb.dieschwalbe.de" rel="nofollow noreferrer">https://pdb.dieschwalbe.de</a>) These are about the dynamics of forcing a nearly-dominated opponent to execute the victory task, in an interesting way. They tend to involve long series of checks (so-called entailing moves in combinatorial game theory). For example <a href="https://pdb.dieschwalbe.de/P1014331" rel="nofollow noreferrer">https://pdb.dieschwalbe.de/P1014331</a> takes 10 moves. All this means that the initial game array is unpromising as a start for a selfmate. If I had to bet (and of course we can never know), I would hazard that misère chess is a positional draw from the starting position. I think this is true even if it turns out that orthodox chess is a win for White, because in misère, tempo does not have the same value.</p>
  
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Answers.Text: tl;dr version of the same: Forcing selfmate needs *far* more material advantage than forcing mate. (This would be a nice question for the problemists: What is the minimum material advantage to force selfmate against bK+bPb7 - to make the mate most easy - I bet you need at least wBb1 to block and Q+R to force Blacks hand. Of course, a twomover is less demanding, R+B suffice. Did I say tl;dr? :-) (EDIT: Actually, with heavy material, Q+promotable P can force selfmate against Q in some cases, see PDB.)
Answers.UserDisplayName: Hauke Reddmann
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Answers.Text: @Hauke - I was trying to get a handle on what might happen in the opening position. The culture in stackexchange is indeed to have more comprehensive and self-contained answers than in a chat forum
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Answers.Body: <p>Forcing self-mate can be done from certain positions, the one forcing their opponent to checkmate them is usually far ahead in material. When I was a teenager (a long time ago) I actually studied the art of self-mate.</p>
  
  <p>There is a game in Chess Companion where one player gave the other queen odds and the one with the extra queen had to force self-mate (and achieved it!). From that game I learned the technique.</p>
  
  <p>Obviously it is not forced from the opening position.</p>
  
  <p>To achieve it you have to be far ahead in material to restrict your opponent's legal moves. Their final move will be to capture one of yours that is giving check. (I can't unfortunately remember how it's done).</p>
  
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Answers.Text: But is there any proof that White can't force a selfmate from the opening position and there is a first move that prevents Black from forcing a selfmate? A proof for that would imply that neither side can force a loss.
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Answers.Text: It's unlikely white can force a self-mate but if so white can also force a win. Because to force a self-mate your own side has to be totally in control of the position.
  
  Whether or not it is forced given queen odds (we can assume a win is forced given queen odds) is something I do not know, it would need a serious solver.
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Answers.Body: <p><strong>Almost surely not</strong></p>
  <p>See Wikipedia on <a href="https://en.wikipedia.org/wiki/Handicap_(chess)" rel="nofollow noreferrer">chess handicaps</a>, in particular <a href="https://en.wikipedia.org/wiki/Handicap_(chess)#Odds_of_queen" rel="nofollow noreferrer">this game</a>.</p>
  <blockquote>
  <p>In response for receiving the queen, Black undertakes to force White to checkmate Black.</p>
  <p>Paris vs. Marseilles, correspondence 1878 (remove White's queen)</p>
  </blockquote>
  <p>Black successfully loses, but only by starting with an extra queen. This means Black can achieve a &quot;winning position&quot;, pin White's remaining pieces, and set up a self-mate. Without the extra queen it's extremely unlikely Black can lose against an opponent who's trying to not win - that's why it's a handicap.</p>
  
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Body: <p>Chess is unsolved, so there is no known optimal strategy for one side to force a win or a draw.</p>
  
  <p>However, is there an optimal strategy to lose? Consider a variant where checkmate results in a loss, but all other rules are the same as regular chess. Is there a known optimal strategy to win or force a draw?</p>
  
  <p>For example, can one side force the opponent to checkmate (assuming the opponent tries to avoid checkmate)? If not, can one side force a draw by rules like stalemate, threefold repetition, or the fifty-move rule? Or is this as hard to solve as regular chess?</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2020-05-28T11:48:00.773Z
Comments.Text: If both players avoid winning, it's going to be a long game no matter what. If just 1 side is trying to lose, it's fairly simple to do so. Fool's mate, Scholar's mate, etc.
Comments.UserDisplayName: Mast
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Comments.Text: nitpick: I think you mean "Can any side force a loss...?". Because the answer to "Can either side..." is an obvious no, from game theory. At most only one side can.
Comments.UserDisplayName: Jeffrey
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Comments.CreationDate: 2020-05-29T15:00:15.720Z
Comments.Text: you could lose by simply running out of time
Comments.UserDisplayName: BenTaylor
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Comments.Text: @Jeffrey Not necessarily, because the two sides are not quite equal.  I could imagine a possibility where white might be able to force it but black not, or vice versa.
Comments.UserDisplayName: WBT
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Comments.Text: @WBT Maybe it's my english, but I read "Can either side" as "Does there exist strategies for both black and white that allow each to force a loss, if they want".
Comments.UserDisplayName: Jeffrey
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Comments.Text: @BenTaylor, if it's timed, it's trivial for White to force a loss by running out of time, but if it's untimed, then it's harder to solve
Comments.UserDisplayName: Victor
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Comments.Text: @Jeffrey Rewording to "can one side" should make it easier to intrepret, as "can either side" can seemingly be interpreted as "can both sides"
Comments.UserDisplayName: Victor
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Comments.CreationDate: 2020-05-30T19:04:02.567Z
Comments.Text: @Victor - yes, I agree that being able to force a loss is conditional on it being timed. But, strictly, chess should always be timed - competition chess always is.
  If it were not timed, then I could force a draw against the best player in the world by simply sitting there not moving.
Comments.UserDisplayName: BenTaylor
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Comments.Text: @BenTaylor Unlimited time doesn't allow forcing a draw. Before the game ends, the game result is indeterminate, so "not moving" doesn't result in a draw. To force a draw, the current player has to make a move (because the game has to end to have a result), but if the opponent has a winning strategy regardless of that move, there is no way to force a draw.
Comments.UserDisplayName: Victor
Tags: <chess-variants><selfmate>
Title: Can one side force a loss in regular chess?
ThreadText: # Can one side force a loss in regular chess?
  
  (Asked by: username_0 on 2020-05-28)
  
  Chess is unsolved, so there is no known optimal strategy for one side to force a win or a draw.
  
  However, is there an optimal strategy to lose? Consider a variant where checkmate results in a loss, but all other rules are the same as regular chess. Is there a known optimal strategy to win or force a draw?
  
  For example, can one side force the opponent to checkmate (assuming the opponent tries to avoid checkmate)? If not, can one side force a draw by rules like stalemate, threefold repetition, or the fifty-move rule? Or is this as hard to solve as regular chess?
  
  **Tags:** chess-variants, selfmate
  
  > **Comments:**
  > **username_1**: If both players avoid winning, it's going to be a long game no matter what. If just 1 side is trying to lose, it's fairly simple to do so. Fool's mate, Scholar's mate, etc.  
  > **username_2**: nitpick: I think you mean "Can any side force a loss...?". Because the answer to "Can either side..." is an obvious no, from game theory. At most only one side can.  
  > **username_3**: you could lose by simply running out of time  
  > **username_4**: @username_2 Not necessarily, because the two sides are not quite equal.  I could imagine a possibility where white might be able to force it but black not, or vice versa.  
  > **username_2**: @username_4 Maybe it's my english, but I read "Can either side" as "Does there exist strategies for both black and white that allow each to force a loss, if they want".  
  > **username_0**: @username_3, if it's timed, it's trivial for White to force a loss by running out of time, but if it's untimed, then it's harder to solve  
  > **username_0**: @username_2 Rewording to "can one side" should make it easier to intrepret, as "can either side" can seemingly be interpreted as "can both sides"  
  > **username_3**: @username_0 - yes, I agree that being able to force a loss is conditional on it being timed. But, strictly, chess should always be timed - competition chess always is. If it were not timed, then I could force a draw against the best player in the world by simply sitting there not moving.  
  > **username_0**: @username_3 Unlimited time doesn't allow forcing a draw. Before the game ends, the game result is indeterminate, so "not moving" doesn't result in a draw. To force a draw, the current player has to make a move (because the game has to end to have a result), but if the opponent has a winning strategy regardless of that move, there is no way to force a draw.  
  
  ---
  
  ## 6 answers
  
  ### Answer by username_5 on 2020-05-28. Score: 14
  
  There is likely not a way to force a loss. Selfmates are relatively rare, and one can probably not be forced from the opening position.
  
  The best plan, paradoxically, is likely to try to capture as many of the opponent's pieces (but not pawns) as possible, while keeping your own pieces to be better able to force your opponent. The opponent's pawns are kept because they have very little choice as to where they can move, and can also be easily blocked to avoid the opponent moving them before you want them to. For example, in the following position, the f-pawn is blocked and the king cannot move, so Black has no choice but to advance the g-pawn and will have to checkmate White in two moves:
  
  ```
  [FEN "3Q4/8/8/4N3/6p1/7k/5p2/5N1K b - - 0 1"]
  
  ```
  
  > If not, can one side force a draw
  
  Well, if checkmate indeed cannot be forced, then almost by definition a draw could be forced. The game can't go on forever, after all. If nothing else, the players will eventually run out of pawns to move or pieces to capture, and the 50 move rule will come into play.
  
  > **Comments:**
  > **username_6**: The question of whether one would be able to force stalemate in the absence of the 50-move and repetition rules would still be relevant.  Obviously even without the 50-move rule, any game which doesn't result in a checkmate or stalemate would necessarily result in an eventual draw by repetition, since there are only a finite number of possible positions.  
  > **username_7**: Try it against [WorstFish](https://lichess.org/@/WorstFish) (yes, Stockfish playing to lose), surprisingly hard!  
  
  ---
  
  ### Answer by username_8 on 2020-05-28. Score: 9
  
  As far as I know, there is no chess variant that has been investigated that is exactly as you describe. However, [antichess](https://en.wikipedia.org/wiki/Losing_chess) is a quite similar concept. The objective is to lose all your pieces/be stalemated, and it is required for you to take a piece if you can. It was weakly solved to be a win, starting with 1. e3. It is still interesting, however, as many of the lines are extremely long (over 100 moves).
  
  In order to answer your exact question, it would be fairly difficult, but might be slightly easier to solve than regular chess. A powerful computer would likely be needed in order to figure it out.
  
  > **Comments:**
  > **username_0**: The antichess variant is different because it requires capturing when possible, but without this constraint, I guess it becomes a question of whether one side can get a [selfmate](https://en.wikipedia.org/wiki/Selfmate) position while the other side tries to move a piece back and forth until it's a draw or to get a stalemate position.  
  > **username_9**: Note that [*weakly solved* is specific game theory terminology](https://en.wikipedia.org/wiki/Solved_game), which means that the solution gives a full strategy starting from the beginning of the game against all possible opponent choices (but not from positions which you never would reach if you followed that strategy, that's the "weakness"). Not quite as weak as someone would think if they don't know the term!  
  > **username_10**: http://magma.maths.usyd.edu.au/~watkins/LOSING_CHESS/  
  
  ---
  
  ### Answer by username_11 on 2020-05-28. Score: 7
  
  Finding the optimal strategy requires the investigation of pretty much all (pretty much, in the grand scale of things) the positions that are possible to get to from the start.
  
  While developing a computer program that plays extremely well in chess (or other similar games) is a solvable task—actually, it is a solved task somewhere in the beginning of the XXI century—finding and proving an optimal strategy **is a totally different problem**. Chess is hard to fully analyze because of the large number of possible positions.
  
  To go more formal, consider [Shannon number](https://en.wikipedia.org/wiki/Shannon_number) `10^120` which is a lower-bound on the game-tree complexity of chess. It is true that the sensible games would go much lower to around `10^40`, but it will not matter too much.
  
  For the game variant you describe, one has to find the *Shannon-like* number to say about the feasibility of finding the optimal strategy. However, simply by looking at the rules, I doubt it would result in something that is feasible.
  
  Thus, I don't expect an optimal playing strategy for this variant to be found any time soon. While it might be slightly easier than for regular chess, the complexity will still be enormous.
  
  NB: [Antichess example](https://en.wikipedia.org/wiki/Losing_chess) is different because of the introduction of the compulsory capture, which severely reduces the number of possible variations.
  
  
  ---
  
  ### Answer by username_12 on 2020-12-23. Score: 3
  
  What a great question.
  
  One important mathematical model for chess is Combinatorial Game Theory. Under the vanilla version of this beautiful theory, developed by Conway, Berlekamp & Guy, one loses when one has no legal moves. One modification of the theory is so-called "misère" (a term taken from card-games) where one *wins* when one has no legal moves.
  
  A game and its misère form thus share the same game tree. They just have opposite values assigned to the leaf nodes. One can derive values for all other nodes in the game trees by working back from the leaf nodes. One amazing feature is that the value of non-leaf positions between misere and non-misère versions of the same game can be very similar.
  
  For example, the winning strategies for the well-known game of nim are essentially the same. It's only with the winning player's final choice that they decide to take one more or one less match from the last non-singleton pile.
  
  Another way to say this: players vie for strategic dominance. Once that is achieved, then one could force the opponent to do a number of different tasks: be checkmated, to checkmate, to send king on journey to occupy every square of the board, etc. The difference between these tasks would be secondary - once strategic dominance is achieved. In practice, it's not as clear cut: part of achieving strategic dominance in chess involves threats of mating. An attack may sacrifice strategic dominance for a winning tempo advantage in order to checkmate quickly. But nevertheless, regular chess and misère have much in common.
  
  Now chess problemists have long had a misère genre they term "selfmates". (The problem database PDB contains over 58,000 of these today! Type g='s#' in the window <https://pdb.dieschwalbe.de>) These are about the dynamics of forcing a nearly-dominated opponent to execute the victory task, in an interesting way. They tend to involve long series of checks (so-called entailing moves in combinatorial game theory). For example <https://pdb.dieschwalbe.de/P1014331> takes 10 moves. All this means that the initial game array is unpromising as a start for a selfmate. If I had to bet (and of course we can never know), I would hazard that misère chess is a positional draw from the starting position. I think this is true even if it turns out that orthodox chess is a win for White, because in misère, tempo does not have the same value.
  
  > **Comments:**
  > **username_13**: tl;dr version of the same: Forcing selfmate needs *far* more material advantage than forcing mate. (This would be a nice question for the problemists: What is the minimum material advantage to force selfmate against bK+bPb7 - to make the mate most easy - I bet you need at least wBb1 to block and Q+R to force Blacks hand. Of course, a twomover is less demanding, R+B suffice. Did I say tl;dr? :-) (EDIT: Actually, with heavy material, Q+promotable P can force selfmate against Q in some cases, see PDB.)  
  > **username_12**: @username_13 - I was trying to get a handle on what might happen in the opening position. The culture in stackexchange is indeed to have more comprehensive and self-contained answers than in a chat forum  
  
  ---
  
  ### Answer by username_14 on 2020-05-28. Score: 3
  
  Forcing self-mate can be done from certain positions, the one forcing their opponent to checkmate them is usually far ahead in material. When I was a teenager (a long time ago) I actually studied the art of self-mate.
  
  There is a game in Chess Companion where one player gave the other queen odds and the one with the extra queen had to force self-mate (and achieved it!). From that game I learned the technique.
  
  Obviously it is not forced from the opening position.
  
  To achieve it you have to be far ahead in material to restrict your opponent's legal moves. Their final move will be to capture one of yours that is giving check. (I can't unfortunately remember how it's done).
  
  > **Comments:**
  > **username_0**: But is there any proof that White can't force a selfmate from the opening position and there is a first move that prevents Black from forcing a selfmate? A proof for that would imply that neither side can force a loss.  
  > **username_14**: It's unlikely white can force a self-mate but if so white can also force a win. Because to force a self-mate your own side has to be totally in control of the position.  Whether or not it is forced given queen odds (we can assume a win is forced given queen odds) is something I do not know, it would need a serious solver.  
  
  ---
  
  ### Answer by username_15 on 2023-09-24. Score: 0
  
  **Almost surely not**
  
  See Wikipedia on [chess handicaps](https://en.wikipedia.org/wiki/Handicap_(chess)), in particular [this game](https://en.wikipedia.org/wiki/Handicap_(chess)#Odds_of_queen).
  
  > In response for receiving the queen, Black undertakes to force White to checkmate Black.
  >
  > Paris vs. Marseilles, correspondence 1878 (remove White's queen)
  
  Black successfully loses, but only by starting with an extra queen. This means Black can achieve a "winning position", pin White's remaining pieces, and set up a self-mate. Without the extra queen it's extremely unlikely Black can lose against an opponent who's trying to not win - that's why it's a handicap.
  
  
  ---
  
  