To demonstrate the effect of the balance some authors use the example of one pair of very good players, who always score a top, in a field of normal players, who always get average results. Sometimes the strong pair is replaced by a pair of very bad players, but that does not change the idea. In a movement with perfect balance only 2 scores will occur, the good scores and the bad scores. With a not so perfectly balanced movement the good and bad scores will vary around the average.
The next model is of course a contest with two strong pairs amidst average pairs. The conditions are the same. Average pairs always get average results when they meet, strong pairs always get the top. But now a top may also be a shared top and a bottom may be a shared bottom, so there is more variation in the score cards. We will see how this works out for a perfectly balanced Howell.
To be specific let us consider a 4-table Howell, 8 pairs, 7 rounds of 1 board. The balance is perfect: each pair of pairs play 3 boards in the same compass direction, 3 boards in opposite directions, and 1 board against each other. Every board is played 4 times, so the top is 6. If this board was played by the 2 strong pairs in different directions, they will each have a 6, their opponents a 0, and all other pairs a 3. If the strong pairs where playing in the same direction, they have a shared top of 5, and their opponents a 1. But it is more interesting to look at the scores of the average pairs that did not meet the strong pairs directly. The 2 pairs that played in the same direction as the strong pairs are unlucky. They are left with 1 point each, since the strong pairs already took 10 out of the 12 available MP's. Consequently their opponents get 5, just as much as the strong pairs! A beautiful illustration of the fact that pairs that play a board in opposite direction from you are your allies. Finally, there is one board where the 2 strong pairs meet, and on this board everyone gets 3.
So the score of the strong pairs will be 3 x 6 + 3 x 5 + 1 x 3 = 36
out of the possible 7 x 6 = 42, and therefore good for 85.71%.
And the average pairs? If we sort out the scores for them we find they have to
divide between them:
6 scores of 0 12 scores of 1 18 scores of 3 6 scores of 5In the ideal case each of them gets 1 x 0 + 2 x 1 + 3 x 3 + 1 x 5 = 16.
A) 8- 1 A 3- 6 D 2- 7 F 5- 4 G 8- 2 B 4- 7 E 3- 1 G 6- 5 A 8- 3 C 5- 1 F 4- 2 A 7- 6 B 8- 4 D 6- 2 G 5- 3 B 1- 7 C 8- 5 E 7- 3 A 6- 4 C 2- 1 D 8- 6 F 1- 4 B 7- 5 D 3- 2 E 8- 7 G 2- 5 C 1- 6 E 4- 3 FHere '8- 1 A' means: pair 8 (NS) plays against pair 1 (EW), board A.
B) 8- 1 A 6- 5 B 4- 2 C 7- 3 E 8- 2 B 7- 6 C 5- 3 D 1- 4 F 8- 3 C 1- 7 D 6- 4 E 2- 5 G 8- 4 D 2- 1 E 7- 5 F 3- 6 A 8- 5 E 3- 2 F 1- 6 G 4- 7 B 8- 6 F 4- 3 G 2- 7 A 5- 1 C 8- 7 G 5- 4 A 3- 1 B 6- 2 DHere some pairs score a point more or less than the average. Let us assume, for instance, the strong pairs are 1 and 2, and look at the score for pair 5. It is 3 x 1 + 4 x 3 = 15. Even though they never had a full bottom, they had no shared top either, ending up 1 below the score of 16 they were entitled to.
Notice that both movements have a perfect balance as defined before. The score matrix is perfectly regular, the quality factor is 100, and the standard deviation is 0. Yet, for the present model of the 2 strong pairs we come to the conclusion that scheme A is better balanced than scheme B. This is difficult to swallow if you find it impossible to be better than perfect. We propose to name perfect movements of type A "superperfect".
It is an open question whether such movements also exist for perfect Howells with a larger number of tables. We have not yet been able to spot one for 6 tables.