revised May 2017

# The Bussemaker model.

The Bussemaker model is described in an article by Pieter Bussemaker in WEKOwijzer-58 (A journal published by the Nederlandse Bridge Bond NBB). It is assumed that the pairs are ordered according to strength, and the difference in strength between consecutive pairs is equal. The score on a board is simply taken as the difference in pair numbers times 100.

For instance, if pair 4 plays a board against pair 8, pair 4 obtains a score of +400, and pair 8 obtains -400. After all, 8 - 4 = 4, and multiplied by 100 is 400. If pair 5 would play the same board against 1, the score would therefore also be 400.

The scores are converted to MP scores in the usual way.

In the ideal case pair 1 would end up first, followed by pair 2, then pair 3, et cetera, and the difference between consecutive pairs should be equal. The results show differently.

I have written a program to calculate the variation of the results when the same field of pairs is assigned arbitrary pair numbers (i.e. the pairs keep their original strength but their pair numbers in the movement scheme are randomized. The score is now: difference in strength times 100). A similar study was done before by Gerrit van der Velde, see WEKOwijzer-68.

The program calculates the MP scores for a large number of random sets of pair assignments. From these the standard deviation from the ideal value is derived. The term standard deviation should be taken with a grain of salt. Actual differences are quite often larger than 3 standard deviations.

Furthermore a N x N - matrix is formed of the percentage probability that pair i ends up at place j in the final ranking. (N is here the number of pairs, i and j are numbers from 1 to N). In the ideal case this is a diagonal matrix with all diagonal elements equal to 100.

It turns out that there are very few movements for which this ideal is obtained. Also for a movement with perfect balance there is a considerable chance that the best playing pair finishes second. The results show that even with perfect movements and pairs that score according to their strength in the completely deterministic way, as described above, there remains a considerable "luck factor" that is correlated to the way the pairs are assigned.

Note. The probability of shared places is not represented in this matrix. Instead, when 2 or more pairs have the same score, their order in the ranking is assigned randomly; e.g. the case that pairs 1 and 2 share places 1 and 2 is treated half the time as if pair 1 comes first and pair 2 comes second, and half the time the other way around. In early versions of the program (before version 5.8) pair 1 would always be first and pair 2 second in such a case.

For a description of the options and use of the program click here.

### Some Examples

As an example we will test the superperfect Howell for 8 pairs. For this movement the "2 strong pairs" model gave an ideal score. For every possible choice of pair numbers of the two strong pairs, both ended up equal, and also all the weak pairs had the same total score. But alas, the Bussemaker model offers no guarantee for an ideal result, even for this movement. Depending on how we assign the pairs with strength 1, 2, . . . 8 large variations occur. For instance, for the pair with strength 1 the ideal result within the model is 83.3%, but values found range from 76.2 tot 90.5%. The following table gives a survey of this spread. For every pair the average result is shown, together with the standard deviation and the extreme values.

```Pair average   s.d.  min.  max
1     83.3   3.3  76.2  90.5
2     73.8   3.9  64.3  81.0
3     63.8   4.7  52.4  73.8
4     54.3   7.3  38.1  66.7
5     45.7   7.3  33.3  61.9
6     36.2   4.7  26.2  47.6
7     26.2   3.9  19.0  35.7
8     16.7   3.3   9.5  23.8
```
There is also no guarantee that the strongest pair ends up first, although it will always end up in first or second place. A complete survey of the chances that pair i ends at place j is given in the following probability table.
```
Pair  1  2  3  4  5  6  7  8
Rank
1   92  8
2    8 82 10
3      10 73  8  8
4         17 72 12
5            12 72 17
6             8  8 73 10
7                  10 82  8
8                      8 92

```
How is it possible that the best playing pair does not win?
 Paar NZ score MP NZ OW NZ OW 1 2 100 0 6 5 7 200 2 4 3 6 300 4 2 4 8 400 6 0
In our model the score on a board is proportional to the difference in strength of the pairs. As a result the strongest pair does not score a top on every board. As an example, a score card where pair 1 even scores a bottom.

Needless to say that such a model is far from realistic. But the trend that appears from this score card may occur in actual life. For instance, pair 2 finds a superior play, that is not found at the other tables. Then pair 1 can do nothing to prevent a 0 score.

As a comparison the probability table for the best known movement for 8 pairs and 6 rounds, discussed on the vacancy quality page. Here the factor luck is quite a bit larger.

```Pair  1  2  3  4  5  6  7  8
Rank
1   83 15  2  .
2   16 63 16  4  2
3    1 21 55 16  9
4       2 26 56 14  2
5          2 14 56 26  2
6             9 16 55 21  1
7             2  4 16 63 16
8                .  2 15 83
Qd 64.1
```
As above, numbers have been rounded to integer values. The '.'-s stand for values larger than 0, but less than 0.5.

For another example we return to the subject "Arrow Switching". Again we look at the complete Mitchell for 14 pairs. Arrow switching one complete round we obtain:

```Pair  1  2  3  4  5  6  7  8  9 10 11 12 13 14
Rank
1   73 22  4  .  .  .  .
2   22 54 20  4  .  .  .  .
3    4 20 54 19  3  .  .  .  .
4    .  4 19 55 18  3  .  .  .
5    .  .  3 19 56 18  3  .  .  .  .
6    .  .  .  3 19 56 18  3  .  .  .
7       .  .  .  3 19 55 19  3  .  .  .
8       .  .  .  .  3 19 55 19  3  .  .  .
9          .  .  .  .  3 18 56 19  3  .  .  .
10                .  .  .  3 18 56 19  3  .  .
11                   .  .  .  3 18 55 19  4  .
12                   .  .  .  .  3 19 54 20  4
13                      .  .  .  .  4 20 54 22
14                            .  .  .  4 22 73
Qd 57.8
```

Switching 2 rounds gives:

```Pair  1  2  3  4  5  6  7  8  9 10 11 12 13 14
Rank
1   64 26  8  2  .  .  .  .  .
2   24 41 23  9  2  .  .  .  .
3    8 20 36 23  9  3  1  .  .  .
4    3  8 19 34 22 10  3  1  .  .  .
5    1  3  9 19 33 21 10  3  1  .  .  .
6    .  1  3  9 20 32 20 10  4  1  .  .  .
7    .  .  1  3  9 20 32 20 10  4  1  .  .  .
8    .  .  .  1  4 10 20 32 20  9  3  1  .  .
9       .  .  .  1  4 10 20 32 20  9  3  1  .
10          .  .  .  1  3 10 21 32 19  9  3  1
11             .  .  .  1  3 10 22 34 19  8  3
12                .  .  .  1  3  9 23 36 20  8
13                   .  .  .  .  2  9 23 41 24
14                      .  .  .  .  2  8 26 64
Qd 38.8
```
Another proof that too many switching rounds deteriorate the quality of the movement.

## Variations on the Bussemaker model

A variation on the Bussemaker model is the following. We assign to every pair a strength that does not necessarily have to be different for all pairs. Again strenght 1 is very strong, 2 a little less strong, et cetera. The score is determined again by the difference in strength. These strengths may be given on the command line.
Strength ordering

`1 2 3 ...`
yields the original Bussemaker model, while
`1 1 2 2 2 2 ...`
is the same case as the "2 strong pairs" model.

This option has proven to be useful in the judgement of the quality of methods for seeding

In another variation the pairs are classified in groups. In each group the pairs are assigned random pair numbers, but they always stay in their own group. One may think of movements with separate NS and EW scoring, and also of schemes with scoring across the field

### An example

We use again the complete Mitchell for 14 pairs but this time without arrow switches. The field is separated into NS (pair 1 - 7) and OW (pair 8 - 14) groups. The strengths are chosen as
1 2 3 4 5 6 7 1 2 3 4 5 6 7.
The probability table for this case looks as follows:
```Pair  1  2  3  4  5  6  7  8  9 10 11 12 13 14
Rank
1   98  2  .  .          98  2  .  .
2    2 96  2  .  .        2 96  2  .  .
3    .  2 96  2  .        .  2 96  2  .
4       .  2 96  2  .        .  2 96  2  .
5          .  2 96  2  .        .  2 96  2  .
6          .  .  2 96  2        .  .  2 96  2
7             .  .  2 98           .  .  2 98

```
We see that the strongest pairs have 98% probability to finish first in their group. For the other pairs the chances to obtain their ideal rank are also much larger than in the case of that perfect Howell.