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  1. #1
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    Unplayed games and Sonneborn-Berger tie-break

    Hello,

    I have two questions. Here are the FIDE rules regarding tie-breaks
    http://www.fide.com/component/handbo...&view=category

    The virtual opponent seems to appear in section F

    F. Handling Unplayed Games for Calculation of Buchholz (Congress 2009)

    If one specifies "for Calculation of Buchholz" I guess that the system should be used ONLY for Buchholz.
    The question: how to consider the unplayed games in a Swiss tournament if I use the Sonneborn-Berger tie-break? Am I missing something?

    When the virtual opponent was proposed it appeared in a preliminary FIDE report and I had the opportunity to understand how to perform the calculations. Perhaps they copied and pasted the explanation of his inventor and it was very clear. For some reason in the final official version the same system and, I hope, calculations have been reworded in a way that I cannot recover the same calculations. I would like to know from you if this depend by my poor understanding of English or if the text is a bit cryptic.

    Thanks,
    Luigi

  2. #2
    CC FIDE Master Jesper Norgaard's Avatar
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    Quote Originally Posted by forlano
    Hello,

    I have two questions. Here are the FIDE rules regarding tie-breaks
    http://www.fide.com/component/handbo...&view=category

    The virtual opponent seems to appear in section F

    F. Handling Unplayed Games for Calculation of Buchholz (Congress 2009)

    If one specifies "for Calculation of Buchholz" I guess that the system should be used ONLY for Buchholz.
    The question: how to consider the unplayed games in a Swiss tournament if I use the Sonneborn-Berger tie-break? Am I missing something?

    When the virtual opponent was proposed it appeared in a preliminary FIDE report and I had the opportunity to understand how to perform the calculations. Perhaps they copied and pasted the explanation of his inventor and it was very clear. For some reason in the final official version the same system and, I hope, calculations have been reworded in a way that I cannot recover the same calculations. I would like to know from you if this depend by my poor understanding of English or if the text is a bit cryptic.

    Thanks,
    Luigi
    In the November 2012 column Geurt Gijssen affirms that the inventor of the virtual opponent tie break score is Mr. Wim van Beersum.

    http://www.chesscafe.com/geurt/geurt174.htm

    Stewart Reuben claims that the only reason this new approach was accepted was that they couldn't understand Reuben's suggestion that the average tiebreak of the games actually played be used.

    http://www.ecforum.org.uk/viewtopic....nplayed#p76551

    I don't know if there is any on-line definition of Stewart's suggestion. However, it might be close to my own definition of using the average score of players in the same score group, of the same round as the unplayed game, seen in the thread http://www.chesschat.org/showthread....ge+score+group

    Question 1: "Should the unplayed games tie break score by virtual opponent ONLY apply in Buchholz, not in other tie breaks like Berger?"
    Most other people that have implemented the virtual opponent, do this also in the Berger for instance http://www.chess-results.com has an Excel file with definitions of both using the virtual opponent. In fact it makes little sense to define a very specific way to handle unplayed games when it is completely optional to use whatever tie break method you want.

    I would like to point out that as I have documented on the tie break threat in post #154 http://www.chesschat.org/showpost.ph...&postcount=154
    there is a positive effect on predictiveness in Buchholz, Berger, Norgaard tie breaks when using the average score in score group instead of virtual opponent. Progressive and several other tie break methods are unaffected.

    Question 2: "For some reason in the final official version the same system and, I hope, calculations have been reworded in a way that I cannot recover the same calculations. I would like to know from you if this depend by my poor understanding of English or if the text is a bit cryptic."
    I think that it is impossible to answer unless you will give some example on what you would have expected the calculations to be. The example in the FIDE page looks right to me.
    Chess well played is imagination, calculation, observation, experience and memorization in order of importance.

  3. #3
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    Hello Jesper,

    Quote Originally Posted by Jesper Norgaard
    In the November 2012 column Geurt Gijssen affirms that the inventor of the virtual opponent tie break score is Mr. Wim van Beersum.

    http://www.chesscafe.com/geurt/geurt174.htm
    I was unaware of the latest Gijssen's article. However that answer does not remove my doubts. By the way, why Mr. Gijssen needed to ask to the inventor of the method? As arbiter he should be able to perform by hands this easy task. For me this is a sign that the words used in the rules are a bit confusing and a hole has appeared in the document.
    Moreover, Mr. Wim van Beersum say, with the permission of Mr. Gijssen,

    However, when you win a normal game in the first round, and your opponent leaves the tournament immediately thereafter, all your opponent's losses by default (losses by default indirectly) are considered to be draws for Sonneborn-Berger calculation. So this player will give you 0.5 * (N – 1) Sonneborn-Berger points where N is the number of played rounds.
    and extend a rule that FIDE consider valid for Buchholz to another tie break such Sonneborn-Berger. Is it so easy to emend the rules? What I want to say is that the current document needs a very serious revision to make easy the arbiter task and avoid complains from the players.

    Question 1: "Should the unplayed games tie break score by virtual opponent ONLY apply in Buchholz, not in other tie breaks like Berger?"
    Most other people that have implemented the virtual opponent, do this also in the Berger for instance http://www.chess-results.com has an Excel file with definitions of both using the virtual opponent. In fact it makes little sense to define a very specific way to handle unplayed games when it is completely optional to use whatever tie break method you want.
    I know that the tie-breaks are arbitrary, but I am asking what to do if I want to adhere to the FIDE reccommendation to let the players understand something of general use. Moreover I do not want to use in important FIDE tournament something that FIDE consider no longer valid.

    Question 2: "For some reason in the final official version the same system and, I hope, calculations have been reworded in a way that I cannot recover the same calculations. I would like to know from you if this depend by my poor understanding of English or if the text is a bit cryptic."
    I think that it is impossible to answer unless you will give some example on what you would have expected the calculations to be. The example in the FIDE page looks right to me.
    I learned how to use the virtual opponent from the minute of Kallithea congress :
    7. Handling of unplayed games in Swiss tournaments
    The committee agreed unanimously to handle unplayed games in Swiss tournaments as follow:
    There are two points of view:
    a. For the player himself who gets a result by default or is absent
    b. For the opponents in other rounds of the player who gets a reult by default
    a. The new style Buchhloz uses a virtual opponent to calculate the Buchholz score for a result by default. A virtual opponent has the same points at the beginning of the round and the result by default of a player is treated as a normal result, so a loss by default (by absence) is a win for the virtual opponent and vice versa. For each next round the virtual opponent gains half a point.
    b. For reducing the consequence for the opponents when calculating Buchholz, each result by default of a player is counted as a half point (draw) for the buchholz of the player’s opponents
    Examples:
    1 In a 9 round swiss the player A achieves 6 points including a default win in round 3.
    After round 2 A had 2 points score.
    The contribution of round 3 for A is 2 + 0 + 6 x 0.5 = 5 points Buchholz
    The contribution of A for his opponents’ Buchholz is 5.5
    2 In a 9 round swiss the player B was absent in round 7 and scored 6 points after round
    9. After round 6 B had 4 points.
    The contribution of round 7 for B is 4 + 1 + 2 x 0.5 = 6 points Buchholz
    The contribution of B for his opponents’ Buchholz is 6.5
    This was much, much, better. First they speak about unplayed games in Swiss tournaments that make clear that even Sonneborn-Berger and other can be treated in the same way (moreover Round Robin are something else). Second, it was explicit a two step procedure.
    Perhaps you knew the same document. If I try to forget what I knew, the new words do not permit me to perform the calculations with the necessary two steps. They say first that the unplayed game is drawn and then introduces the virtual opponent. How to mix both info? I would like to know what may think an arbiter that was unaware of the previous Kallithea document.
    Anyway I do not want to insist on this point as I recognise that English is a language that i do not master.

    One last thing. What to do in a Round Robin with the unplayed game for tie break purpose? I hope we agree the the virtual opponent should be banned. In a Swiss tournament at each round I face an opponent with my same score. This is not the case of a RR. So, what to do, now that the old way, "draw against himself", is disappeared?

    Regards,
    Luigi
    Last edited by forlano; 02-12-2012 at 07:58 PM.

  4. #4
    CC FIDE Master Jesper Norgaard's Avatar
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    Quote Originally Posted by forlano
    ...
    One last thing. What to do in a Round Robin with the unplayed game for tie break purpose? I hope we agree the the virtual opponent should be banned. In a Swiss tournament at each round I face an opponent with my same score. This is not the case of a RR. So, what to do, now that the old way, "draw against himself", is disappeared?

    Regards,
    Luigi
    Hi Luigi
    I will answer the last bit first. In Round Robin the notion of the virtual opponent does not make much sense, because the rounds before and after the unplayed games are not depending on the result of the unplayed game, indeed are just a permutation of the same opponents, so this rule would work out unfairly if applied literally.

    For the average score of scoregroup (which I invented) the logical equivalent would be not the average result of the player's score group *in that round* but rather *against that same player* irrespective of which round the other players in the score groups met the player that had an unplayed game against the player we are calculating. For example if players A, B and C ended up on 5.5 points each, and A had a forfeit result against X, while B drew X and C won against X. Suppose X ended up with 4 points. Now the contribution of X to B will be 2 points (drawing), and 4 points (winning) to C, so the contribution to A would be (4+2)/2 = 3 points. However, since the actual result is very important in Round Robin, it should be included in the calculation. So if A won on forfeit against X he would get 3 points , if he drew he would get 1.5 points, and if he lost on forfeit he would get 0 points.

    Another view would be that if X withdraws after 60% of the games, he could have scored more in the last rounds and therefore contribute less to each player. However, that is the same effect for all players. Therefore normal Berger can be applied, e.g. if the forfeit is a win against X, then all points of X, if the forfeit is a draw then half the points of X, and if the forfeit was a loss, no points.

    You are right that the tie break section of the FIDE page is deficient in many ways to say the least. Recommending a tie break method as default for Swiss that reduces predictiveness to around 52% is of course a disaster. Luckily most tournaments are not following that recommendation, but use some form of Buchholz or Berger directly, which have much better predictiveness. Also the definition of contribution to a players Buchholz (in the Kallithea document) seem to have vanished in the FIDE page.

    The definition "The contribution of A for his opponents’ Buchholz is 5.5" has been somewhat hidden. However, the FIDE page says "For tie-break purposes all unplayed games in which players are indirectly involved (results by forfeit of opponents) are considered to have been drawn." I consider that equivalent, but not easy to understand. If it was a default win or default loss the contribution to his score is 1 or 0 respectively. But his contribution of Buchholz for that round for another player that played him in another round, will be 0.5. I don't like the formulation, and players seeking an explanation with the arbiter, will be hard to convince with this vague definition. Anyhow, with good will it is possible to interpret it like it appears in the Kallithea document.
    Chess well played is imagination, calculation, observation, experience and memorization in order of importance.

  5. #5
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    Here is the Gijessen's answer to the question

    http://www.chesscafe.com/geurt/geurt176.htm.

    It seems that there is no longer an official way to handle unplayed games in round robin.

    Luigi

  6. #6
    CC FIDE Master Jesper Norgaard's Avatar
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    In http://www.chesscafe.com/geurt/geurt176.htm Geurt Gijssen shows an example of a RR tournament that in fact shows the problems of applying "unplayed game considered a draw":

    Code:
    A.	*  1  1  0
    B.	0  *  1  + 
    C.	0  0  *  0 
    D.	1  -  1  *
    As we can see C gets 4.th. place with 0 points, the other 3 players get 2 points. If the forfeited game had been a normal win for B against D, in fact between these three players (A,B,D) each win one and lose the other game. It should be clear that unless using something artificial as average rating (that has nothing to do with the actual result) there is no way to break the tie. Returning to the tournament with the forfeited game, Geurt arguments that using Berger calculation when the forfeited game is calculated gives D the better tie break score. That is correct (at least if counting the forfeit as a draw in the tie break scores) but this produces a very unfortunate effect.

    Suppose B playes D and A playes C in the last round. If we use Geurt's way of calculating, then we can calculate the options for D who has won 2 games to have the only sure winning strategy (of the tournament) is to forfeit his game against B. As we can see above, if he does that, he will get the better Berger tie break, no matter if A wins against C, A draws C or A loses against C. If he plays and wins or draws, he will also win the tournament outright. However, if he plays and loses, if A wins against C there is an unbreakable tie between A, B and D, and if A draws or loses against C, B will win the tournament.

    This is similar to the infamous tie break of the European Championship where some players would qualify if they resigned in the last round, but if they played and won, they would not qualify. This is the worst scenario you can possibly present for a tie break.

    Instead if we count forfeited games as normal games, and D forfeits his game against B, he has exactly the same odds as if he plays and loses, e.g. in 2 scenarios B is the winner, and in the last there is an unbreakable tie between A, B and D. In that case of course the more intelligent approach would be to play and hope for a draw or a win, when D will still win the tournament, as it gives no worse chances of winning the tournament than forfeiting the last game.

    Code:
    Unplayed considered a draw, 6 outcomes:
    
    ----- D plays and loses -----
    A. * 1 ˝ 0  1˝   2 + 0.25 + 0 = 2.25
    B. 0 * 1 1  2    0 + 0.25 + 2 = 2.25  {B winner}
    C. ˝ 0 * 0  ˝    0.75 + 0 + 0 = 0.75
    D. 1 0 1 *  2    1.5 + 0 + 0.5 = 2
    
    A. * 1 0 0  1    2 + 0 + 0 = 2
    B. 0 * 1 1  2    0 + 1 + 2 = 3  {B winner}
    C. 1 0 * 0  1    1 + 0 + 0 = 1
    D. 1 0 1 *  2    1 + 0 + 1 = 2
    
    A. * 1 1 0  2    2 + 0 + 0 = 2  {tie A, B, D}
    B. 0 * 1 1  2    0 + 0 + 2 = 2
    C. 0 0 * 0  0    0 + 0 + 0 = 0
    D. 1 0 1 *  2    2 + 0 + 0 = 2
    
    ----- D forfeits ---------
    A. * 1 ˝ 0  1˝   1.5 + 0.25 + 0 = 1.75
    B. 0 * 1 +  2    0 + 0.5 + 1.25 = 1.75
    C. ˝ 0 * 0  ˝    0.75 + 0 + 0 = 0.75
    D. 1 - 1 *  2    1.5 + 0.75 + 0.5 = 2.75  {D winner}
    
    A. * 1 0 0  1    1.5 + 0 + 0 = 1.5
    B. 0 * 1 +  2    0 + 1 + 1.25 = 2.25
    C. 1 0 * 0  1    1 + 0 + 0 = 1
    D. 1 - 1 *  2    1 + 0.75 + 1 = 2.75  {D winner}
    
    A. * 1 1 0  2    1.5 + 0 + 0 = 1.5
    B. 0 * 1 +  2    0 + 0 + 1.25 = 1.25
    C. 0 0 * 0  0    0 + 0 + 0 = 0
    D. 1 - 1 *  2    2 + 0.75 + 0 = 2.75  {D winner}
    So my conclusion is, it is better to consider forfeited games as normal results when dealing with Round Robin or Double Round Robin tournaments. Comments?
    Chess well played is imagination, calculation, observation, experience and memorization in order of importance.

  7. #7
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    Quote Originally Posted by Jesper Norgaard
    So my conclusion is, it is better to consider forfeited games as normal results when dealing with Round Robin or Double Round Robin tournaments. Comments?
    I totally agree with you.
    I always felt that any averaging score of the unplayed game in round robin is unfair based on the fact that
    1) I can no longer met the forfeited opponent (in contrast to what happen in swiss tournament). Because I cannot be punished for what is not my responsability, and is unfair to to get more than what may bring me a normal victory given that in RR all met all and the pairing is not based on probability (as in swiss tournament);
    2) Drastically avoid any cheating effect connected with the different evaluation of the forfeited games.

    Today you gave me a mathematical demonstration.
    Thank you!

  8. #8
    CC FIDE Master Jesper Norgaard's Avatar
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    Quote Originally Posted by forlano
    Today you gave me a mathematical demonstration.
    Thank you!
    Thank you, Luigi!
    Chess well played is imagination, calculation, observation, experience and memorization in order of importance.

  9. #9
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    Quote Originally Posted by forlano
    Hello,

    I have two questions. Here are the FIDE rules regarding tie-breaks
    http://www.fide.com/component/handbo...&view=category

    The virtual opponent seems to appear in section F

    F. Handling Unplayed Games for Calculation of Buchholz (Congress 2009)

    If one specifies "for Calculation of Buchholz" I guess that the system should be used ONLY for Buchholz.
    The question: how to consider the unplayed games in a Swiss tournament if I use the Sonneborn-Berger tie-break? Am I missing something?

    When the virtual opponent was proposed it appeared in a preliminary FIDE report and I had the opportunity to understand how to perform the calculations. Perhaps they copied and pasted the explanation of his inventor and it was very clear. For some reason in the final official version the same system and, I hope, calculations have been reworded in a way that I cannot recover the same calculations. I would like to know from you if this depend by my poor understanding of English or if the text is a bit cryptic.

    Thanks,
    Luigi
    This question remains unanswered...
    In a swiss system tournament, when there are unplayed games, how is the Sonneborn-Berger tiebreak computed?
    In particular, is the virtual opponent used for Sonneborn-Berger?
    Yes or no?
    I hope somebody asks the technical commission in the next FIDE Congress or General Assembly.

  10. #10
    CC FIDE Master Jesper Norgaard's Avatar
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    Quote Originally Posted by Pepechuy
    This question remains unanswered...
    In a swiss system tournament, when there are unplayed games, how is the Sonneborn-Berger tiebreak computed?
    In particular, is the virtual opponent used for Sonneborn-Berger?
    Yes or no?
    I hope somebody asks the technical commission in the next FIDE Congress or General Assembly.
    Out of sheer logic of the dynamics of Swiss, it seems that handling unplayed games with the virtual opponent works well here whether the specific tiebreak used is Buchholz or Sonneborn-Berger or Zermelo or ...
    At least that is the logic I have been applying in the Norgaard tiebreak program.

    As stated above, in a Round Robin, virtual opponent makes no sense, but the actual result makes sense.

    It would be nice to clarify these things in FIDE recommendations about tiebreaks, e.g. how to apply unplayed games in all variations of tournament type and tiebreak. In Progressive of course it doesn't matter if there was an unplayed game, as it is just calculated with the actual result.

    In Average Rating in Swiss, I suppose you can just omit the game because it is not accumulative, so an average of 9 games is just as good as an average of 8 games, without having to speculate what the rating of the unplayed games opponent would have been.

    For Most Blacks in Swiss it seems that the default has been counting it as a white game, but if a player would have 2 unplayed games, it seems unfair to count them all as white games, making that player lose any possible tiebreak against all other players, so perhaps 1 white and 1 black would be more reasonable.

    For Most Blacks in Round Robin I presume that the color should be retained as specified by the expected opponent, for one just to avoid a possible speculation by someone to get an advantage for someone else by sandbagging a game, for the detriment of the player receiving the forfeit win (hope that makes sense).
    Chess well played is imagination, calculation, observation, experience and memorization in order of importance.

  11. #11
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    Quote Originally Posted by Pepechuy
    This question remains unanswered...
    In a swiss system tournament, when there are unplayed games, how is the Sonneborn-Berger tiebreak computed?
    In particular, is the virtual opponent used for Sonneborn-Berger?
    Yes or no?
    Hi,

    here, http://www.chesscafe.com/text/geurt177.pdf page 4, Geurt say yes and it make sense in Swiss tournament.
    Of course all his suggestions spreaded in chesscafe should appear in some FIDE document one day. At the moment the Handbook alone does not help the arbiter.

    One more question. Does the player's score need to be adjousted before to apply Buchholz and Sonneborn? I mean, should we consider the unplayed game as draw before to proceede with calculations?
    Jesper clearly demonstrated that the answer is definitely NOT in Round Robin, while Geurt seems to suggest the opposite. Jesper, please consider sending your example to Geurt.
    Who knows if this way to proceedes backfire even in Swiss system. After all it produces less points for the tied players that met who won a forfeited game. Why this player needs this advantage if we already gave him a suitable virtual opponent? I am afraid that the adjusted score could make sense in conjunction with the old way (draw againsta himself), but make no sense at all with virtual opponent.

    Anyway, in the next version of my program will appear the option to adjust the score (default) or disable it. Virtual opponent is the default in Swiss system but not apply to Round Robin (who want to test the prerealise can pm me).

    Best regards,
    Luigi

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