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

It now turns out that it is possible to devise Howell movements that have this property, but also Howell movements that don't, but still are perfectly balanced movements in the customary sense of the word.

Let us take an example of each:

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.

No matter which way you choose the 2 strong pairs from the numbers 1 to 8,

the average pairs will have a score of 16, i.e.

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.