A mistake that undoubtedly happens even more frequently than it is discovered is the fouled board. Hands are returned to the board in differing compositions, the board is turned, or there is a misduplication. If a deal is played differently at one table, a 60%-60% artificial adjusted score is usually the result. When it has been played more than once, a fouled board is the result.
Two or more different deals will now have been played. Part of the pairs have played one deal, the rest have played another.
It does not matter how great the difference between the differing deals is. The switching of the club two and three is enough. The Laws do not impose on the director to rule on whether the change is significant or not. The differing vulnerability or even a different dealer is enough to be of possible importance to the result on the board, so a fouled board is ruled according to Law 87.
Law 87B also states what the director should now do : He rates each
group separately.
This much is clear, but after that, there are certain adjustments to be made.
In the Mitchell system, these corrections are the most important. Scoring a subset of the total group yields result on a lower Top. This should always be upgraded to the normal top. The Mitchell 1 system dealt with this problem in exactly the same way as they handled the artificial score problem as described in the next chapter :
To the scores awarded in each subset is added a number, equal to the number
of tables in all other groups.
Suppose that on the next board 4 tables play one version, 7 another version
:
620 | - | 6 | 13 |
80 | - | 4 | 11 |
- | 90 | 2 | 9 |
- | 100 | 0 | 7 |
110 | - | 12 | 16 |
90 | - | 9 | 13 |
- | 50 | 2 | 6 |
0 | 0 | 4 | 8 |
90 | - | 9 | 13 |
- | 90 | 0 | 4 |
80 | - | 6 | 10 |
110 |
The first group get seven matchpoints extra, the second group 4 MP. As you can see, the average within both groups is 10 MP, the total is the correct 110.
The Mitchell (1) method of dealing with this problem is unsatisfactory. The first pair, scoring 620, deserves a top or near-top, no matter what. In stead, they score a meagre 65%.
A better way of solving the problem could be to use percentages straight away :
620 | - | 6 | 20 |
80 | - | 4 | 13.3 |
- | 90 | 2 | 6.7 |
- | 100 | 0 | 0 |
110 | - | 12 | 20 |
90 | - | 9 | 15 |
- | 50 | 2 | 3.3 |
0 | 0 | 4 | 6.7 |
90 | - | 9 | 15 |
- | 90 | 0 | 0 |
80 | - | 6 | 10 |
110 |
This is not the approach chosen by Gerard Neuberg however, and for a good
reason :
Although the first pair now does score near to the absolute top, this is
probably too high a score. Since 7 tables did NOT play the board, nobody
can be absolutely sure that their score of +620 will not be matched or even
bettered. And the pair scoring 110 have proved themselves slightly better
: they have outclassed a larger field.
In this discussion, we do not take into consideration that the first pair
have played superbly, perhaps executing a double squeeze, any more than we
would subtract points if it were their opponents who by revoking allowed
them to win an unmakeable contract.
In the calculations above, we have divided by the original top and multiplied by the new top : x10/3 and x10/6.
Gerard Neuberg chose in stead to base the multiplication on the number of results in each group and in total : x11/4 and x11/7.
If one does this calculation on the raw scores, the averages do not remain average. In the first group, the average 3 would become 8.25, which is less than the average 10 in the total field. In order to have averages return to average, in Mr Neubergs formula, one MP is added before the multiplication, and subtracted again afterwards :
620 | - | 6 | 18.25 |
80 | - | 4 | 12.75 |
- | 90 | 2 | 7.25 |
- | 100 | 0 | 1.75 |
110 | - | 12 | 19.43 |
90 | - | 9 | 14.71 |
- | 50 | 2 | 3.71 |
0 | 0 | 4 | 6.86 |
90 | - | 9 | 14.71 |
- | 90 | 0 | 0.57 |
80 | - | 6 | 10 |
110 |
The score for the first pair is calculated as : 6+1 = 7; 7/4x11 = 19.25;
19.25-1 = 18.25.
Although they score near the top (91.25%), they have a lower score than the
best in the other group.
A good illustration why the Neuberg formula is correct is provided by the following example : In a tournament of 15 tables, everybody plays 4H. This contract is made if one can find the CQ. 60% of the players manage this. However, at the end it becomes clear that 5 tables have played a board in which the D2 an D3 have been switched. Law 87 comes into operation. In the different Mitchell systems, this would give the following results :
NS EW MP/g mpM1 mpM mpM2 620 6 16 18 20 620 6 16 18 20 620 6 16 18 20 100 1 11 3 5 100 1 11 3 5 620 13 18 19,5 20 620 13 18 19,5 20 620 13 18 19,5 20 620 13 18 19,5 20 620 13 18 19,5 20 620 13 18 19,5 20 100 3 8 4,5 5 100 3 8 4,5 5 100 3 8 4,5 5 100 3 8 4,5 5
Only when using the Neuberg formula are the results in both groups the same.
In the Ascherman system, the average is equal to the number of comparable scores, so the first and second methods above coincide to give :
620 | - | 7 | 19.25 |
80 | - | 5 | 13.75 |
- | 90 | 3 | 8.25 |
- | 100 | 1 | 2.75 |
110 | - | 13 | 20.43 |
90 | - | 10 | 15.71 |
- | 50 | 3 | 4.71 |
0 | 0 | 5 | 7.86 |
90 | - | 10 | 15.71 |
- | 90 | 1 | 1.57 |
80 | - | 7 | 11 |
121 |
All remarks concerning the correctness of these results that
applied to the Neuberg formula, also apply here.
Please remark that the Ascherman scores are of course 1 MP higher than those
obtained in the Mitchell-Neuberg system. This explains why the Ascherman
system will still give exactly the same ranking as the Mitchell system. It
will not surprise you to know that in the second example above, the two groups
of the fouled board produce the same results when using the Ascherman scoring
(21 and 6 MP).
Normally, after calculating all cross-differences and converting them to IMPs, an average is made, although this is usually only done to obtain more manageable numbers.
This immediately solves the problem of the fouled board, but it leaves us with one question : should one divide by the number of results on the board, or by the number of comparisons ? It is my opinion that you should divide by the number of results, for two reasons :
To illustrate the point we calculate our special example, using the division by number of comparisons (A), and by the number of scores (B) :
NS EW totIMPs CI(A) CI(B) 620 +24 +6 +4.8 620 +24 +6 +4.8 620 +24 +6 +4.8 100 -36 -9 -7.2 100 -36 -9 -7.2 620 +48 +5.4 +4.8 620 +48 +5.4 +4.8 620 +48 +5.4 +4.8 620 +48 +5.4 +4.8 620 +48 +5.4 +4.8 620 +48 +5.4 +4.8 100 -72 -8 -7.2 100 -72 -8 -7.2 100 -72 -8 -7.2 100 -72 -8 -7.2
It is clear that method B should be preferred.
The Butler system does not really pose any problems in calculating fouled boards.
Nor does the Bastille system, but it does present us with an interesting
question, when we check our special case as above. In the calculation
of the mean score, one score at each end is not used. This causes the mean
score to be different within both groups.
Without a foul, over the 15 results, the mean would be : (9x620-4x100)/13
= 398.5
But within the two groups, these averages become :
small group : (2x620-1x100)/3 = 380
large group : (5x620-3x100)/8 = 350
And the results would change accordingly.
There is a problem with the Butler and Bastille systems. When the groups
become too small, it gets less and less interesting to drop the extreme scores
when calculating the average. This poses the problem when one should, and
when not to drop scores. Perhaps it would be best never to drop any scores.
Or we could use another approach : always drop, either side, a same percentage
of scores, say 10%. With 10 results on a board, this is one score either
side, so we revert to the normal Butler or Bastille method. With 20 results,
two results are dropped from either side. With any other number of results,
the appropriate last result is counted in part. Example : our
special example above :
In the big group, with ten results, the average is calculated over the normal
8 scores : 350
In the smaller group, we drop half the extreme scores, so we have :
(620x50%+620x2-100-100x50%)/4 = 350
Or without the foul, we drop one 620 completely and one 620 by 50%, at the
other end the same, so we are left with : (7.5x620-4.5x100)/12 = 350
This way of calculating the average gives us a result consistent with our
wishes that events that occur with the same relative frequency should produce
the same results.
This alternate way of calculating the average was first used at a second
Revolutionary Tournament at the Royal Squeeze Bridge Club on
14 july 1995.
Last Modified : 1996-09-12