Answers.Body: <p>According to the FIDE handbook (<a href="https://handbook.fide.com/chapter/C0401" rel="nofollow noreferrer">C.04.1</a> Basic rules for Swiss Systems):</p>
  <blockquote>
  <p>For each player the difference between the number of black and the number of white games shall not be greater than 2 or less than –2.<br />
  Each system may have exceptions to this rule in the last round of a tournament.<br />
  No player shall receive the same colour three times in a row.<br />
  Each system may have exceptions to this rule in the last round of a tournament.</p>
  </blockquote>
  <p>And according to the FIDE handbook (<a href="https://handbook.fide.com/chapter/C0403" rel="nofollow noreferrer">C.04.3</a> FIDE (Dutch) System):</p>
  <blockquote>
  <p>non-topscorers (see A.7) with the same absolute colour preference (see A6.a) shall not meet (see C.04.1.f and C.04.1.g).</p>
  </blockquote>
  <blockquote>
  <p>An absolute colour preference occurs when a player’s colour difference is greater than +1 or less than -1, or when a player had the same colour in the two latest rounds he played. The preference is white when the colour difference is less than -1 or when the last two games were played with black. The preference is black when the colour difference is greater than +1, or when the last two games were played with white.</p>
  </blockquote>
  <p>So how can you legally get a better ratio than what you propose? Longer tournaments are out; the longer you go the more times you'll have to play White in order to keep it within 2. The obvious way would be a 3 round tournament, but that seems awfully short and on the edge of your &quot;unnatural&quot; 1 or 2 round tournaments.</p>
  <p>But a 3 round tournament isn't the only way you could play 3 games. You could have a player in a 4 round Swiss with a first-round bye. The colors could then go XBBB. This would seem to be somewhat improbable, however, since any players who had played all the games would have a weaker color preference in the last round. You'd need multiple people at XBB going into the last round that <em>have</em> to play each other.</p>
  
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Answers.CreationDate: 2022-10-13T20:14:48.340Z
Answers.Text: Would be (peripherically :-) interesting to actually construct such a tournament in a Swiss programm. (Unfortunately the FIDE rules only state that color check *may* overruled in the last round, but not *what* can overrule. Surely not whim of the organizer :-) The other 15 pairing rules? But that is stuff for a new question...)
Answers.UserDisplayName: Hauke Reddmann
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Body: <p>Is the worst B/W ratio you can get in a &quot;natural&quot; tournament a Swiss 5 round with lifted color check in the last round, BBWBB? (Did that actually happen in real life?)</p>
  <p>(Obviously, you can get 100% Black with &quot;unnatural&quot; events - all White opponents forfeiting, a tournament just with one or two rounds, feel free to comment...)</p>
  
Tags: <swiss-pairing>
Title: Worst B/W ratio
ThreadText: # Worst B/W ratio
  
  (Asked by: username_0 on 2022-10-13)
  
  Is the worst B/W ratio you can get in a "natural" tournament a Swiss 5 round with lifted color check in the last round, BBWBB? (Did that actually happen in real life?)
  
  (Obviously, you can get 100% Black with "unnatural" events - all White opponents forfeiting, a tournament just with one or two rounds, feel free to comment...)
  
  **Tags:** swiss-pairing
  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_1 on 2022-10-13. Score: 1 | Accepted answer
  
  According to the FIDE handbook ([C.04.1](https://handbook.fide.com/chapter/C0401) Basic rules for Swiss Systems):
  
  > For each player the difference between the number of black and the number of white games shall not be greater than 2 or less than –2.  
  > Each system may have exceptions to this rule in the last round of a tournament.  
  > No player shall receive the same colour three times in a row.  
  > Each system may have exceptions to this rule in the last round of a tournament.
  
  And according to the FIDE handbook ([C.04.3](https://handbook.fide.com/chapter/C0403) FIDE (Dutch) System):
  
  > non-topscorers (see A.7) with the same absolute colour preference (see A6.a) shall not meet (see C.04.1.f and C.04.1.g).
  
  > An absolute colour preference occurs when a player’s colour difference is greater than +1 or less than -1, or when a player had the same colour in the two latest rounds he played. The preference is white when the colour difference is less than -1 or when the last two games were played with black. The preference is black when the colour difference is greater than +1, or when the last two games were played with white.
  
  So how can you legally get a better ratio than what you propose? Longer tournaments are out; the longer you go the more times you'll have to play White in order to keep it within 2. The obvious way would be a 3 round tournament, but that seems awfully short and on the edge of your "unnatural" 1 or 2 round tournaments.
  
  But a 3 round tournament isn't the only way you could play 3 games. You could have a player in a 4 round Swiss with a first-round bye. The colors could then go XBBB. This would seem to be somewhat improbable, however, since any players who had played all the games would have a weaker color preference in the last round. You'd need multiple people at XBB going into the last round that *have* to play each other.
  
  > **Comments:**
  > **username_0**: Would be (peripherically :-) interesting to actually construct such a tournament in a Swiss programm. (Unfortunately the FIDE rules only state that color check *may* overruled in the last round, but not *what* can overrule. Surely not whim of the organizer :-) The other 15 pairing rules? But that is stuff for a new question...)  
  
  ---
  
  