THE NEUBERG FORMULA | ||||||

by MAX BAVIN E.B.U. CHIEF TOURNAMENT DIRECTOR | ||||||

(Reproduced by kind permission of the author) | ||||||

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MATCHPOINTING BOARDS WITH AN UNEQUAL NUMBER OF SCORES | ||||||

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 | ||||||

important | ||||||

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) | ||||||

A | ||||||

Where | ||||||

∙ M is the match points considering the scores in isolation (14, 11, 6, 2, 0) in the previous | ||||||

example. | ||||||

∙ 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 | ||||||

computer. | ||||||

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 | ||||||

night. | ||||||

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. | ||||||

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