(Reproduced by kind permission of the author)
1. Introduction
   Consider the following situation. The last table to finish play for the evening are just about to start their
final board. Everyone else has finished, and the scoring has been completed save for the final board.
   Pair 'X' currently have a top on the final board - but there is still one result to come.
What do you think the chances are that Pair "X" will still have a top once the last table have finished
   Well, there are three possibilities. Let us say that it is an 11-table complete movement, so a top on a
board is 20 match-points.
   The three possibilities are:-
∙ the final table will beat  'X's result, so pair 'X' will score only 18 points out of 20.
∙ the final table will get the same result, so pair 'X' will score 19 points.
∙ the final table will get a worse result, so pair 'X' will indeed score their complete top and get 20 points.
   So, if all three outcomes are equally likely, pair 'X' would have a normal expectancy of 19 points out of 
20 in such a situation (the average of 18, 19 and 20).
   However, demonstrably, not all three outcomes  are equally likely. After all, pair 'X' have already
beaten 9 out of 9 other results, so it must be heavily odds-on that they will beat the 10th and final result
as well - not certain, but very likely. So their normal expectancy in such a situation must be closer to 20
points than it is to 19. We will return to this question later.
2. Boards with unequal tops.
Of course, in the actual example quoted above, pair 'X' must simply wait until the last table have finished
before they know their real score on the board.
  But what happens if the tournament director tells the last table that they cannot play the last board
because they are too slow? "Take an average" he says to the last table.
Now we have a real problem. What score to give pair 'X', or indeed all the other pairs who have already
played the board.
3. Three possible approaches to the problem.
  We have seen the three different solutions proposed over the years - and they all generate a slightly
different final result and, therefore, potentially a different overall winner as well!
      a) Insert an average into the results, so the top becomes 19 and the bottom becomes 1 (and the
average is still 10).The ten actual scores are match-pointed in the normal way 19, 17, 15, 13, 11, 9, 7, 5,
3, 1.
  This has the means of recognising a very important principle in pairs play, which is that all boards
should be equally significant (I'll give an extreme example later of why this is important). It also has the
merit of simplicity and is the method which has been used by experienced clubs for years in the
pre-computer age. However it has the considerable de-merit of being blatantly unfair. Our poor pair 'X'
now only get 19 points on the board, which we have already demonstrated is not enough.
      b) Because there are only 10 results, score this board with a top of 18 and express everyone's 
final result as a percentage of the maximumscore which was available to them. Superficially, this
sounds fair, as pair 'X' are getting 100% on the board, which is certainly closer to the mark than the
95% they got under method (a).
  However consider the following rather extreme case. Take a tournament where there are 26 boards in
play, of which everyone plays exactly 25 (this just happens to be a convenient number to use). Say it
was a very large event - several sections with 51 tables overall and a top of 100.
  Pair 'Y' score 50% on all their boards bar one. On the one board they have achieved a 65% score. So
(24 x 50%) + (1 x 65%) = 1265% out of 2500% = 50.6%.
  However there is something strange about the movement, board 26 is only played in one of the
sections and only has 11 results on it. Pair 'Y' have never played this board so their final score is not
affected. Of course, our poor pair 'X' have played the board.
Pair 'X' have also scored 50% on all their boards bar one. On one board they have scored a complete
top - and, wouldn't you know it, it is board 26; the only one with 11 results.
  So method (b) would give them (24 x 50pts) + (1 x 20pts) = 1220pts out of 2420pts = 50.41%. Oops!
  Pair 'Y' have beaten pair 'X'. Is this fair? Is this what you thought or expected would happen? Their
results are identical save that pair 'Y' have a 65% score on one board and pair 'X' have a 100% on one
board. Yet pair 'Y' are the winners.
  So, what has gone wrong? Well, what has happened is that the significance of board 26 has become
almost irrelevant because it has been played less often than all the other boards. This is why the basic 
philosophy contained in the approach (a) " that all boards should be equally significant" is very
      c) So, there are serious flaws in both the above approaches. Moreover, we need to find a solution
to the problem as it really is most unsatisfactory that two (or even three) different and perfectly
competent scores might produce two (or even three) different winners given the same set of data.
  To the rescue comes GERARD NEUBERG of France
4. The Neuberg Principle
  I'm afraid that this is where life begins to get complicated in which case I would urge the reader to just
trust us and leap to section 6 and beyond.
  This said, let us return to our original example - the board with 10 results where 11 results were
expected. The correct (i.e. mathematically sound) approach to the problem is as follows:-
(a) We were expecting 11 results, but in fact we have only 10. The ratio of 11/10 is 1.1.
(b) So, assume that each of the 10 results have occurred 1.1 times (rather than only once) and
match-point accordingly.
(c) Example
N/S Score Frequency Factored Actual frequency Match Points
+ 520 1 1.1 19.9
+ 500 1 1.1 17.7
+ 490 1 1.1 15.5
+ 480 1 1.1 13.3
+ 460 1 1.1 11.1
+ 450 1 1.1 8.9
+ 430 1 1.1 6.7
+ 420 1 1.1 4.5
+ 400 1 1.1 2.3
-50 1 1.1 0.1
  Total 10 Total 11  
(d) The Match points quoted are out of a theoretical top of 20. The top score of 19.9 is arrived at by 
subtracting the frequency (1.1 in this case) from 21 (i.e. 1 more than the theoretical top). This is exactly
the same process you use already, possibly without realising it, when match pointing normally. If the
best score had occurred only once, you would give it 20 out of 20 (i.e. subtract 1 from 21 - had it
occurred twice (a shared top) you would give it 19, and so on. The match points for subsequent scores
go down in units of 2.2 rather than 2.0. 2.2 is the sum of the previous frequency (1.1 in this case) and
the frequencyof the result you are trying to calculate (also 1.1 in this case).
(e) A second example may help to illustrate the principle, this time using whole numbers only. Say a
board should have been played 16 times (top = 30), but has actually only been played 8 times. So, the 
ratio of 16/8 is exactly 2 this time. So, imagine that each of the eight actual results had occurred
exactly twice and match point accordingly 
Let us say that our results are as follows:-
  Number Normal   Factored  
N/S score of match points Factored match points Calculation
  occurrences (top of 14) frequency (top of 30)  
+ 490 1 14 2 29 31 - 2
+ 460 2 11 4 23 29 - 2 - 4
+ 430 3 6 6 13 23 - 4 - 6
+ 400 1 2 2 5 13 -6 - 2
- 50 1 0 2 1 5 - 2- 2
  Total 8   Total 16    
You can verify this by match pointing a board yourself in the usual way with two scores of + 490, four
of + 460, six of + 430, two of + 400 and two of - 50. You will end up with match point scores of 29, 23,
13, 5 and 1 for the five different results.
5. The Neuberg Formula.
  The actual formula is as follows:-
  Match Points = (M x E) + (E-A)
∙ M is the match points considering the scores  in isolation (14, 11, 6, 2, 0) in the previous
∙ E is the expected number of scores (16).
∙ A is the actual number of scores (8).
So, the + 490 above scores (14 x 16) + (16 - 8) = 29
6. Application of the formula.
  Boards with a different number of results can arise in a variety of different ways,such as the nature of
the movement itself or a table being unable to play a board for whatever reason.
  In all such circumstances, and in an ideal world, the Neuberg formula should be used to calculate the
final result.
  However it is acknowleged that there are difficulties, not the least being:-
(a) it is difficult (very difficult) to understand; and
(b) it is even more difficult (though not impossible) to perform such calculations without the aid of a
  For this reason, and for this reason alone, the formula has not been written into the laws of Bridge as
a 'must do it this way'.Indeed, for clubs who score manually we would strongly recommend that you do
not even attempt to score this way.
  The important thing is to have a rule about how you will score such boards, and then keep to it.
What ever you do though, please don't have three different scorers all scoring it a different way each
All World Bridge federation, European Bridge League and English Bridge Union events are scored using
the Neuberg formula, and have been for the last couple of decades. All -or nearly all - modern computer
scoring software has the Neuberg formula already built-in, so if your club uses a computer it is likely
that you too are already using the Neuberg formula without even realising it.
7. Final example
  We have seen how the formula generates a score of 19.9 out of 20 for our pair 'X' in the question at the
very start of this paper. And for the case of the 50.6% (1265 match points) pair who somehow managed
to beat the pair with 24 averages and 1 top? Well pair 'X' would score
[(20 x 51) + (51 - 11)] / 11=96.4% on board 26 (out of 100%). So their final score would be
(24 x 50) + 96.4 = 1296.4 out of 2500 = 51.86%. A clear victory for them, which is precisely how it
should be.