To handle hands with a different number of scores, in a pairs contest with match point scoring, the Neuberg method is recommended to combine the scores. This method is sometimes considered difficult or even incomprehensible. However a few examples show that this solution is quite obvious. We present a simple derivation.

### Introduction

On a certain club evening only 6 pairs were present, such that each hand was played only 3 times. The next week, there were 12 pairs and every hand was played 6 times. Let us consider one frequency table from each of those two session. Only the NS results are presented.

 session 1        3 tables session 2        6 tables score number MP percent score number MP percent 430 1 4 100 430 2 9 90 400 1 2 50 400 2 5 50 ‑50 1 0 0 ‑50 2 1 10

A notable difference! Left and right there are 3 different scores that each occur equally often. Yet the results left and right are different. This happens to be a property of the way we score pairs events, and it would be no reason for concern if these were two single drives. But suppose they were part of a competition over several sessions. Then a pair that scores 430 on the first night and ‑50 on the second would have an average of 55% and a pair that scores ‑50 the first time and the second time 430 has an average of 45%. We see very different rewards for comparable performances.

We should be able to do better then that. Of course we want every hand to be equally important, but we can do something about the percentages. A natural and obvious way is to award the same percentages in session 1 as in a comparable session with 12 pairs, such as session 2. Also if the distribution of scores is more complicated the recipe is simple. Let 50% remain 50%, and, in session 1, multiply the difference from 50% by a factor 0.8.

The above is an example of the method of Neuberg for combining results for hands that do not all have the same number of scores. It is often said the Neuberg is difficult and complicated, but in fact it is nothing more that a consequent application of the idea outlined above.

Or, to give another example, the wish that a top shared by 3 pairs in a field of 14, is the same as a top shared by 30 pairs in a field of 140, automatically leads to the method of Neuberg.

### Derivation

In the following we use
n = the number of scores actually present
N = the number of scores that should be present
Pn = the percentage before Neuberg, from a normal calculation with n scores
PN= the percentage after converting to N scores.
The formula we seek should have the following form:
 PN = (Pn − 50) ·A + 50
(1)
Indeed, it satisfies the demand: if Pn = 50, then also PN = 50.
All that is left to do is to determine the number A, for any given N and n.

In the special case that N is 2n, as in the introduction, we want a full top for a hand with n scores (Pn=100) to be equivalent with a top shared by 2 pairs for a hand with N scores. More generally, if N/n is an integer number, say m, then Pn=100 should correspond to a top shared by m pairs for a hand with N scores.

A full top is 2N−2 matchpoints. A top shared by 2 pairs 2N−3. In general: a top shared by m pairs is 2N−1−m matchpoints.
Thus a top shared by m pairs yields a percentage of
 PNtop(m)  =  100 · 2N − 1 − m 2N −2
(2)
To obtain our desired PN we put m=N/n
 PNtop(N/n)  =  100 · 2N − 1 − (N/n) 2N −2
(3)
Furthermore we know from (1) that for Pn = 100
 PNtop = 50 ·A + 50
(4)
Equating (3) en (4) and solving for A we find
 A = n−1 n · N N −1
(5)
What to do if N/n is not an integer? The simplest solution is to generalize our result and also apply it in that case!
Inserting this in (1)
 PN = (Pn − 50) · n−1 n · N N −1 + 50
(6)
This is the Neuberg formula.
Usually this relation is expressed in terms of matchpoints. When we write PN and Pn in terms of the corresponding matchpoints SN en Sn,
 PN = 100 · SN 2N − 2
(7)

 Pn = 100 · Sn 2n − 2
(8)
and substitute this in (6) we obtain after some algebra:
 SN = (Sn + 1) · N n −1
(9)
This form of the Neuberg formula is better known. But, what we want when presenting the results are the percentage scores anyways. When we apply the Neuberg formula, expressed in percentages, the intermediate step via matchpoints is not necessary. Formula (6) is also very useful to "Neuberg" a session as a whole when combining sessions with different numbers of participants, or, more generally, different tops.

### Summary

We have derived the Neuberg formula on the basis of the following postulates:
The relation between PN and Pn is linear (not involving squares or worse)
At the value of 50% PN and Pn are equal
If N/n is integer the top at n scores corresponds to a shared top at N scores, shared by N/n pairs.
Generalising this idea, we use the same dependency on N and n, when N/n is not integer.

In the literature the Neuberg formula is often given without any clarification, as something that is difficult to understand. We see here there is a simple and logical explanation for this formula.

Formulae translated from TEX by TTH, version 4.03.