Answers.Body: <p>I have never heard your point before, and it might be original.</p>
  <p>There is a famous theorem by Ernst Zermelo (more famous for his set theory work), see <a href="https://en.m.wikipedia.org/wiki/Zermelo%27s_theorem_(game_theory)" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Zermelo%27s_theorem_(game_theory)</a> which when applied to chess means that either White can win, Black can win, or both players can force <em>at least</em> a draw. To regular chess players, and even apparently to Wikipedians, the words “at least” might not seem necessary but to problemists they are essential, as the questioner’s elegant counter-examples show.</p>
  <p>If Black is forced to mate, then he/she has gained at least a draw, so Zermelo’s theorem is not violated.</p>
  
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Answers.Body: <p>If I understood correctly, the position that you are after is a <strong>dead position</strong> because White and Black can't force a checkmate (including self-inflicted) nor a stalemate. Thus, if no draw is agreed and FIDE rules (dead-position, three-fold and fifty-move) don't apply (edited thanks to comment below), both sides would keep playing for ever and without any possible outcome.</p>
  <p>The position below is the minimal example of a dead position:</p>
  <pre><code>[fen &quot;k7/8/8/p2p2p1/P2P2P1/8/8/K7 w - - 0 1&quot;]
  </code></pre>
  <h4>Example</h4>
  <p>White to play and force a dead position:</p>
  <pre><code>[fen &quot;1k4n1/1p2p3/1P1pPp1p/pP1P1PpP/P5P1/8/2Q5/K6B w - - 0 1&quot;]
  </code></pre>
  <p>Where the solution is 1. Qc8+</p>
  <p>IMO, the example I show is overly easy and has no &quot;wow&quot; factor. So, I am curious to see a good composition that arises from your observation.</p>
  
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Answers.Text: If no mate is possible, even with total co-operation, then the position is a draw by dead position, under FIDE rules, and the game ends immediately
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Answers.Text: Isn't the position in the OP "Position where neither player can force a win and neither player can force a draw" a **dead position** by definition? If not, how is it different?
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Answers.Text: No, that was not my intent. In a dead position, both players can force a draw; in fact, they have no choice but to force a draw. Please study carefully the examples I provided. The point is that if White has a forced selfmate then Black cannot force a draw, because White can force Black to win the game, and a win is not a draw.
Answers.UserDisplayName: Timothy Chow
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Answers.Text: Thanks for the clarification. I guess I jumped to conclusions too quickly from the title and I will need to re-read the text at least 10 more times and focus on the diagrams. Leaving the answer here in case someone else misinterprets the OP like me.
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Answers.Body: <blockquote>
  <p>One sometimes sees the claim that in every chess position, either White has a forced win, or Black has a forced win, or both players can force a draw.</p>
  </blockquote>
  <p>One doesn't. The claim is patently nonsense. Here is a position, which I am sure we are all familiar with, which is neither a forced win nor a forced draw for anyone. All 3 results are possible and everything is to play for:</p>
  <pre><code>[fen &quot;&quot;]
  </code></pre>
  <blockquote>
  <p>While this claim is &quot;morally&quot; correct</p>
  </blockquote>
  <p>No, it's not. It's not correct in <em>any</em> sense.</p>
  <blockquote>
  <p>My question is, has this observation been made before (most likely, by chess problem composers)?</p>
  </blockquote>
  <p>Only if they've had far too much to drink.</p>
  <p>The argument starting &quot;but if I had a 32-piece tablebase then ...&quot; is also nonsense. Since there are more possible chess positions than there are atoms in the universe a 32-piece tablebase is a physical impossibility. If you could store one position per atom the whole universe would still be too small to hold your tablebase.</p>
  
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Answers.Text: I have never seen the claim "in every chess position, either White has a forced win, or Black has a forced win, or both players can force a draw.", and although somewhat opinion based, the claim seems to be completely bogus. There are many positions where one color is playing for two results, while the other is playing for just one.
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Answers.UserDisplayName: Brian Towers
Body: <p>One sometimes sees the claim that in every chess position, either White has a forced win, or Black has a forced win, or both players can force a draw.</p>
  <p>While this claim is &quot;morally&quot; correct, it is technically not quite correct.  To be pedantically correct, in the third case, we should say that both players can force <b>at least</b> a draw.  The point is that there can exist positions in which neither White nor Black can force a win, but White can force Black to win, and Black can force White to win.  In such a scenario, neither player can &quot;force a draw&quot; in the sense of guaranteeing that the game will not terminate in a win for either player.</p>
  <p>My question is, <strong>has this observation been made before (most likely, by chess problem composers)?</strong> I would be surprised if the observation is original, but I have not seen anyone state it explicitly before.</p>
  <p>Below is a position (with White to move) that I came up with to illustrate the point.  The position is theoretically drawn, and this can be seen without tablebases; the only defense to Black's threat of 1...d1Q+ is 1.f4, after which Black can avoid defeat only with 1...d1Q+ 2.Qxd1 Qxf4+ or 1...Qxf4+ 2.Qxf4 d1Q+ with a draw in either case.  But White can force Black to win with 1.Qg2+ Qxg2#, and if White tries to &quot;force a draw&quot; with 1.f4 then Black can force White to win with 1...Qg2+ 2.Qxg2#.</p>
  <pre><code>[FEN &quot;8/8/8/8/6Q1/5P2/3p3q/5K1k w - - 0 1&quot;]
  </code></pre>
  <br>
  I contacted Noam Elkies and he said he had not seen this stipulation before, but he noted that the  (known) five-man minimum for a selfmate (see below) also achieves "my" task.  White can selfmate with 1.Qg7+ Qxg7#.  Either of the alternatives 1.Kf6 or 1.Kh6 leads to a tablebase draw, but also allows Black to selfmate with 1...Qg7+.
  <pre><code>[FEN &quot;6qk/4Q3/6K1/8/6b1/8/8/8 w - - 0 1&quot;]
  </code></pre>
  
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Comments.CreationDate: 2022-11-03T22:03:56.637Z
Comments.Text: @Evargalo: Not Sam Loyd, but I know the problem too. (google google) Mortal, kneel before my google-fu! :-) [White to force the end of the game in two moves](https://chess.stackexchange.com/questions/13332/white-to-force-the-end-of-the-game-in-two-moves)
Comments.UserDisplayName: Hauke Reddmann
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Comments.Text: I have never seen the claim "in every chess position, either White has a forced win, or Black has a forced win, or both players can force a draw.", could you please provide a reference?
Comments.UserDisplayName: Pablo S. Ocal
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Comments.CreationDate: 2023-07-20T03:00:55.103Z
Comments.Text: Are the players allowed to let the clock run out in a position where they are forced to mate their opponent to claim a draw?
Comments.UserDisplayName: Lucenaposition
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Comments.Text: @Lucenaposition Very clever! But even if we allow that sort of trick, I cannot *force* my opponent to lose on time, and so I cannot *force* a draw.
Comments.UserDisplayName: Timothy Chow
Tags: <problems><selfmate>
Title: Position where neither player can force a win and neither player can force a draw
ThreadText: # Position where neither player can force a win and neither player can force a draw
  
  (Asked by: username_0 on 2022-09-18)
  
  One sometimes sees the claim that in every chess position, either White has a forced win, or Black has a forced win, or both players can force a draw.
  While this claim is "morally" correct, it is technically not quite correct.  To be pedantically correct, in the third case, we should say that both players can force at least a draw.  The point is that there can exist positions in which neither White nor Black can force a win, but White can force Black to win, and Black can force White to win.  In such a scenario, neither player can "force a draw" in the sense of guaranteeing that the game will not terminate in a win for either player.
  My question is, has this observation been made before (most likely, by chess problem composers)? I would be surprised if the observation is original, but I have not seen anyone state it explicitly before.
  Below is a position (with White to move) that I came up with to illustrate the point.  The position is theoretically drawn, and this can be seen without tablebases; the only defense to Black's threat of 1...d1Q+ is 1.f4, after which Black can avoid defeat only with 1...d1Q+ 2.Qxd1 Qxf4+ or 1...Qxf4+ 2.Qxf4 d1Q+ with a draw in either case.  But White can force Black to win with 1.Qg2+ Qxg2#, and if White tries to "force a draw" with 1.f4 then Black can force White to win with 1...Qg2+ 2.Qxg2#.
  [FEN "8/8/8/8/6Q1/5P2/3p3q/5K1k w - - 0 1"]
  
  
  I contacted Noam Elkies and he said he had not seen this stipulation before, but he noted that the  (known) five-man minimum for a selfmate (see below) also achieves "my" task.  White can selfmate with 1.Qg7+ Qxg7#.  Either of the alternatives 1.Kf6 or 1.Kh6 leads to a tablebase draw, but also allows Black to selfmate with 1...Qg7+.
  [FEN "6qk/4Q3/6K1/8/6b1/8/8/8 w - - 0 1"]
  
  **Tags:** problems, selfmate
  
  > **Comments:**
  > **username_1**: @Evargalo: Not Sam Loyd, but I know the problem too. (google google) Mortal, kneel before my google-fu! :-) [White to force the end of the game in two moves](https://chess.stackexchange.com/questions/13332/white-to-force-the-end-of-the-game-in-two-moves)  
  > **username_2**: I have never seen the claim "in every chess position, either White has a forced win, or Black has a forced win, or both players can force a draw.", could you please provide a reference?  
  > **username_3**: Are the players allowed to let the clock run out in a position where they are forced to mate their opponent to claim a draw?  
  > **username_0**: @username_3 Very clever! But even if we allow that sort of trick, I cannot *force* my opponent to lose on time, and so I cannot *force* a draw.  
  
  ---
  
  ## 3 answers
  
  ### Answer by username_4 on 2022-11-19. Score: 1
  
  I have never heard your point before, and it might be original.
  
  There is a famous theorem by Ernst Zermelo (more famous for his set theory work), see <https://en.m.wikipedia.org/wiki/Zermelo%27s_theorem_(game_theory)> which when applied to chess means that either White can win, Black can win, or both players can force *at least* a draw. To regular chess players, and even apparently to Wikipedians, the words “at least” might not seem necessary but to problemists they are essential, as the questioner’s elegant counter-examples show.
  
  If Black is forced to mate, then he/she has gained at least a draw, so Zermelo’s theorem is not violated.
  
  
  ---
  
  ### Answer by username_5 on 2022-11-19. Score: -1
  
  If I understood correctly, the position that you are after is a dead position because White and Black can't force a checkmate (including self-inflicted) nor a stalemate. Thus, if no draw is agreed and FIDE rules (dead-position, three-fold and fifty-move) don't apply (edited thanks to comment below), both sides would keep playing for ever and without any possible outcome.
  The position below is the minimal example of a dead position:
  [fen "k7/8/8/p2p2p1/P2P2P1/8/8/K7 w - - 0 1"]
  
  Example
  White to play and force a dead position:
  [fen "1k4n1/1p2p3/1P1pPp1p/pP1P1PpP/P5P1/8/2Q5/K6B w - - 0 1"]
  
  Where the solution is 1. Qc8+
  IMO, the example I show is overly easy and has no "wow" factor. So, I am curious to see a good composition that arises from your observation.
  
  > **Comments:**
  > **username_4**: If no mate is possible, even with total co-operation, then the position is a draw by dead position, under FIDE rules, and the game ends immediately  
  > **username_5**: Isn't the position in the OP "Position where neither player can force a win and neither player can force a draw" a **dead position** by definition? If not, how is it different?  
  > **username_0**: No, that was not my intent. In a dead position, both players can force a draw; in fact, they have no choice but to force a draw. Please study carefully the examples I provided. The point is that if White has a forced selfmate then Black cannot force a draw, because White can force Black to win the game, and a win is not a draw.  
  > **username_5**: Thanks for the clarification. I guess I jumped to conclusions too quickly from the title and I will need to re-read the text at least 10 more times and focus on the diagrams. Leaving the answer here in case someone else misinterprets the OP like me.  
  
  ---
  
  ### Answer by username_6 on 2022-09-18. Score: -8
  
  One sometimes sees the claim that in every chess position, either White has a forced win, or Black has a forced win, or both players can force a draw.
  
  One doesn't. The claim is patently nonsense. Here is a position, which I am sure we are all familiar with, which is neither a forced win nor a forced draw for anyone. All 3 results are possible and everything is to play for:
  [fen ""]
  
  
  While this claim is "morally" correct
  
  No, it's not. It's not correct in any sense.
  
  My question is, has this observation been made before (most likely, by chess problem composers)?
  
  Only if they've had far too much to drink.
  The argument starting "but if I had a 32-piece tablebase then ..." is also nonsense. Since there are more possible chess positions than there are atoms in the universe a 32-piece tablebase is a physical impossibility. If you could store one position per atom the whole universe would still be too small to hold your tablebase.
  
  > **Comments:**
  > **username_2**: I have never seen the claim "in every chess position, either White has a forced win, or Black has a forced win, or both players can force a draw.", and although somewhat opinion based, the claim seems to be completely bogus. There are many positions where one color is playing for two results, while the other is playing for just one.  
  
  ---
  
  