Calculating Bridge Tournaments

Problem 6. Different number of scores

Description of the problem

Sometimes a movement is used which leads to different boards being played a different number of times. This is of course a thing to be avoided, but sometimes this is not always possible. Here are a few instances where I have encountered this problem :

When playing Multiple Mitchell movements in several rows of unequal length. There is one set of boards that is not in use in all series.
When playing an appendix tournament, using the first round to duplicate the boards. Those boards not in use in the first round, will play all scheduled rounds = one more than the boards duplicated in the first round.
In tournaments in which there are more boards than are actually played, when a bye occurs. The bye can be considered as a ‘phantom’, a pair with its own movement. Since this phantom does not play all boards, some boards contain more results than others. In my club we always play a Mitchell movement, with between 26 and 32 boards. With 23 pairs we have a problem : we have to curtail a 12 table Mitchell, and there are then two sets of boards that have been played 10 times, all the others having only 9 results.

I am only dealing here with this problem occurring within the one session. The next chapter deals with the similar problem, when it occurs over two different sessions.

Solutions

1) Mitchell

In the Mitchell (1) system, it would be most logical to deal with this problem by using percentages. Since the system was never fully written down, and since it is no longer of practical use, we should not go further into this.

One mistake should however be cited at this moment, so as to make clear how not to go about tackling this problem. It is the solution we used when I started calculating my club tournaments, by hand, some 15 years ago. We used to keep track, for every pair, of the total of the tops of the boards they did play. At the end, we divided the total of their matchpoints by what they could have got.
This breaks a fundamental principle I wish to cite now : every board should count equally heavy. Every board should be brought to the same top, and only then added. The ways in which the boards are brought to the same top are perhaps still under discussion, but it must be done.

2) Mitchell-Neuberg

The WBF has not specifically stated how this problem should be dealt with. Law 78A does not give a solution, Law 87B, which deals with fouled boards is not applicable.
I think most Tournament Directors would use percentages to solve this problem. All boards with the same top would be taken together, and by using percentages and a formula such as the one described in the next chapter, they would get a result.
This system is actually equal to one in which the scores are handled board per board. Their results are recalculated on a ‘normal’ top using a simple formula : (most other boards contain only 10 results = top 18)


NS	EW	MP	MP-M2
Mitchell 2
620	-	20	18
80	-	11	9.9
-	90	3	2.7
-	100	0	0
110	-	18	16.2
90	-	15	13.5
-	50	6	5.4
0	0	8	7.2
90	-	15	13.5
-	90	3	2.7
80	-	11	9.9
			99

The average is indeed 9, the top 18.

There is however not one good reason why the Neuberg formula should not also be used in this case.
As in fact it is in European Pairs’ Championships. I was first made aware of this during the European Ladies’ Pairs Championships in Killarney in 1991, when this did prove necessary. Claude Dadoun, the French TD in charge had no doubt that it had to be done this way, even if no regulation stipulated this.

This is where I draw the line between the Mitchell 2 and 3 systems. If you do not use the Neuberg formula in this case, you are working with the Mitchell 2 system. In the Mitchell 3 system, the Neuberg formula is used for this problem.

Everyone is completely entitled to use the Mitchell 2 system, since no WBF guidelines have been given concerning this problem. In contrast, the Mitchell 1 system should not be used.

Let’s use the Neuberg formula on the previous example :


NS	EW	MP	MP-M3
Mitchell 3
620	-	20	18.09
80	-	11	9.91
-	90	3	2.64
-	100	0	-0.09
110	-	18	16.27
90	-	15	13.55
-	50	6	5.36
0	0	8	7.18
90	-	15	13.55
-	90	3	2.64
80	-	11	9.91
			99

Two remarks :

Don’t say that it doesn’t change much. What did you expect, a different winner ? I once got that. See the example.
Negative scores ? A score higher than the top ? Yes indeed. This follows logically out of our concern to be ‘correct’. You don’t like it : tough. Gerard Neuberg suggests one solution : always take the highest number of scores as basis for your ‘normal’ top.
Indeed we can prove that it does not matter which ‘top’ you choose, as long as you score all boards on the same top.

3) Ascherman

Since in the Ascherman system percentages and the (adapted) Neuberg formula yield the same result, there is no problem here :


NS	EW	MP	MP-As
Ascherman
620	-	21	19.09
80	-	12	10.91
-	90	4	3.64
-	100	1	0.91
110	-	19	17.27
90	-	16	14.55
-	50	7	6.36
0	0	9	8.18
90	-	16	14.55
-	90	4	3.64
80	-	12	10.91
			99

4) Cross-IMPs, Butler, Bastille

These problems do not pose any more problems in IMP-style tournaments than those already dealt with.

Last Modified : 1996-09-13