Laws & Ethics Committee Section- Articles
We have seen three different solutions proposed over the years - and they all generate a slightly different final result and therefore potentially a different overall winner as well!
| (a) | Insert an average into the results, so the top becomes 19 and the bottom becomes 1 (and the average is still 10). The ten actual scores are match-pointed in the normal way 19, 17, 15, 13, 11, 9, 7, 5, 3, 1. This has the merit of recognising a very important principle in pairs play, which is that all boards should be equally significant (I'll give an extreme example later of why this is important). It also has the merit of simplicity and is the method which has been used by experienced club scorers for years in the pre-computer age. However, it has the considerable de-merit of being blatantly unfair! Our poor pair 'X' now get only 19 points on the board, which we have already demonstrated is not enough |
| (b) | Because there are only 10 results, score this board with a top of 18 and express everyone's final result as a percentage of the maximum score which was available to them. Superficially, this sounds fine, as pair 'X' are getting 100% on the board, which is certainly closer to the mark than the 95% they got under method (i). However, consider the following rather extreme case. Take a tournament where there are 26-boards in play, of which everyone plays exactly 25 (this just happens to be a convenient number to use). Say it was a very large event - several sections with 51 tables overall and a top of 100. Pair 'Y' score 50% on all their boards bar one. On one board they have achieved a 65% score. So, (24 x 50) + (1 x 65) = 1265 out of 2500 = 50.6% However there is something strange about the movement and board 26 is only played in one of the sections and only has 11 results on it. Pair 'Y' have never played this board, so their final score is not affected. Of course, our poor pair 'X' have played the board. Pair 'X' have also scored 50% on all their boards bar one. On one board they have scored a complete top - and wouldn't you know it, it's board 26: the only one with 11 results! So method (ii) would give them (24 x 50) + (1 x 20) = 1220 out of 2420 = 50.41%. Oops! Pair 'Y' have beaten pair 'X'. Is this fair? Is this what you thought or expected would happen? Their results are identical save that pair 'Y' have a 65% score on one board and pair 'X' have a 100% on one board. Yet pair 'Y' are the winners? So, what has gone wrong? Well, what's happened is that the significance of board 26 has become almost irrelevant because it has been played less often that all the other boards. This is why the basic philosophy contained in approach (ii) is that all boards should be equally significant is very important |
| (c) | So, there are serious flaws in both the above approaches. Moreover, we need to find a solution to the problem as it really is most unsatisfactory that two (or even three) different and perfectly competent scores might produce two (or even three) different winners given the same set of data. |
To the rescue comes Gerard Neuberg of France