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