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Weighted scores and 12c1(c) stories
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Law 12c1(c) reads as:

In order to do equity, and unless the regulating Authority forbids it, an assigned adjusted score may be weighted to reflect the probabilities of a number of potential results.

What does this mean? For example, after an irregularity, the TD decision is that the result on the hand without the irregularity would be 60% of the time 4= (420) and 40% of the time 4-1 (-50).

One of the problems with this deceptively simple paragraph is of a technical nature: "How to calculate the final result of a match, or a tournament"? With tournaments sometimes decided by fractions of a point (be it a MP, or a VP, or an IMP), to correctly calculate the final result is essential.

When we see a pairs tournament with the winner 0.1 matchpoints ahead of second place and 0.15 matchpoints ahead of third place, for example, did the winner really win? If we accept that any difference, no matter how small, is enough for a win, then yes, of course... As long as the results areaccurate. There are ways through whichextra undeserved (fractions of) matchpointscan creep inside a final ranking. The best we can do is to try to make sure that we minimize chances of this happening.

Given the conditions of contest, and a set of results for a tournament, calculation of the final ranking should be straightforward and not questionable. If it is somehow "approximated", or erroneous, something is terribly wrong and must be corrected. Of course, CoC may state that the "erroneous" method is the used one, and players tacitly accept it when entering the event...

Writing this, a story almost 22 years old came to my mind. On October 1994, a mathematics teacher, Prof. Thomas Nicely, discovered and reported a bug in the floating-point unit of the early Pentium processors, producing inaccurate results in some circumstances. The wholething was perpetuated through jokes like:

"What's another name for the "Intel Inside" sticker they put on Pentiums? Answer:The warning label."

Those days are long gone,but the initial reaction from Intel,the public outcry and the damages that were made to Intel, both financially and in terms of corporate image, at the time, are well worth studying, even today.

My main concern with this article is to show how 12c1(c) calculations should be performed so that final results are as accurate as possible.

For the rest of the article we will assume a TD decision giving 60% of 4 = (420) and 40% of 4-1 (-50), and when talking about matchpoints a scoring sheet for the other pairs reading:

420 1 time

400 2 times

170 1 time

-50 1 time

A basic confusion, which has long been erradicated, was to consider 60% of 420 and 40% of -50 (232). This is wrong, wrong, wrong.

When the TD decides for a weighted outcome, it means that some percentage of the time the contestant would be assigned an outcome, and some other percentage of the time another one (weighted scores may have as many items as judged appropriate, not necessarily only two).

So, if a contestant receives 60% of the time 420, he receives 60% of the time the "consequences" of 420.

Playing in a team match (KO), this is relatively straightforward. Suppose that on the other table the result on the board was 400.

Team A scores 60% of the time 420 (winning 20) and 40% of the time -50 (losing 450).

Here again one does NOT calculate 0.6*20+.4*(-450).

60% of the time team A would win 1 IMP and 40% of the time lose 10 IMPs. The calculations should be performed directly in IMPs.

0.6*1 IMP + 0.4*(-10) = 0.6 - 4.0 = -3.4 IMPs

This often results in fractional IMPs, rounded according to regulations to the nearest integer. So the result on the board would be -3 IMPs for Team A.

In case of a .5 result, different organizations use different rules. I personally like a rule where the rounding is in favor of the non-offending side, and if both are offendersor non-offenders, to the lowest integer. But it´s just an opinion.

What to do with rounding when there are multiple adjustments like this on a match? IMHO, rounding should occur just for the final addition.

Playing in a Swiss, there is one further argument, and a very valid one, namely now that we are using continuous VP scales... Lets proceed with the same example, 60% of 420 and 40% of -50, closed room has 400, and now team A wins by8 IMPs without the weighted comparison.

Score on the board would be, like we saw in the previous page, -3 IMPs, so team A wins by 5 in total (12.18 VPs) according to

http://www.worldbridge.org/Data/Sites/1/media/documents/regulations/WBFVPscales.pdf

However, if we take the reasoning one step further, we could conclude that 60% of the time team A would score 420 on this board, win 1 on the boardand win by (8 + 1 = 9), scoring 13.65 VPs, and 40% of the time team A would score -50, lose10 on the board and lose by (8-10 = 2), scoring 9.08.

60% of 13.65 + 40% of 9.08 would be 11.82 VPs.

Note that the procedure, also in Swiss, and in all the countries that I am aware of, according with the regulations, is to calculate the weighted scores in IMPs. It is a procedure that works, is relatively easy to understand by the players, and is accepted worldwide.

But, IMHO,the additional step of performing the calculations directly in VPs should be discussed. It is thought provoking, and different. Maybe it is better?

Of course, this approach of calculating weighted scores directly in VPs entails the problem that we cannot report an IMP result of a match, and probably it was one of the reasons why the approach was never adopted.

But doesn´t it sound better?

Playing board a match the solution for 12C1(c) adjustments is simple and similar to the one used for KO play.

In our example, 60% of the time team A wins the board and 40% of the time team A loses. Score on the board would then be 0.6*(1)+0.4*(-1) = 0.2 boards for team A.

And what about pairs tournaments?...

This is where it gets more challenging and fascinating (if you are the type of person that finds such things fascinating)...

We will bring again the scoring sheet that we are using as an example:

TD decision giving 60% of 4 = (420) and 40% of 4-1 (-50), other pairs getting:

420 1 time

400 2 times

170 1 time

-50 1 time

There should be six results on the scoring sheet, but because of the adjustment only 5 were obtained.

How should we proceed?

One simplified and approximated procedure (remember the story of the Pentium bug?) would be to matchpoint the scoresheet without the weighted score resulting in

420 (4 MPs), 400 (2.5), 170 (1), -50 (0)

and then apply theNeuberg formula (https://www.ebu.co.uk/documents/laws-and-ethics/articles/neuberg-formula.pdf)to the results, giving

420 (5 MPs), 400 (3.2), 170 (1.4), -50 (0.2)

and now, for the weighted score, use 60% of the MPs corresponding to 420 (5) and 40% of the MPs corresponding to -50 (0.2) = 3.08 MPs

However, this doesn´t sound right (and it is definitely not right). The board was played 6 times, and 100% of the time there are six results on the scoring sheet (60% of the time a 420 and 40% a -50 on that sixth line).

The correct principle, described very briefly inthe February Bridge Bulletin, is to matchpoint each of the weighted scores against the rest of the field and then average the result. So, 420 would get 4.5, -50 would get 0.5, and the final result would be 0.6*4.5+0.4*0.5 = 2.9. However, the article omits an important point: What to do with the other results on the scoring sheet?

The main point is that for each of the possibilities in the weighted score there should be a separate matchpointing on the scoring sheet, and then the different outcomes are weighted. This affects ALL the scores on the scoresheet, and not just the weighted score.

So let´s see, step by step, how to perform an accurate calculation.

60% of the time the score sheet would look like this (third column is the MPs corresponding to the result):

420 2 4.5

400 2 2.5

170 1 1

-50 1 0

and 40% of the time it would look like this:

420 1 5

400 2 3.5

170 1 2

-50 2 0.5

Let´s see a score that is not part of the possibilities in the weighting, but is affected indirectly by them: 170. 60% of the time 170 gets 1 MP, and 40% of the time 2 MPs, meaning in total 1.4 (0.6*1+0.4*2)

For the other scores, it would be:

420 4.7

400 2.9

170 1.4

-50 0.2

This is an extremely important principle, when applying 12C1(c) in a Pairs tournament: The weighting of possibilitiesfor one pair of contestantsmay affect the calculations for all the other pairs in the same board. The scoresheet must be matchpointed taking into account each possibility and the outcomes weighted for all the contestants, not only for the two contestants involved at the table.

Let´s go further in this example...

Suppose that there were by fortune (or misfortune, as the case may be) two weighted scores,the one we have been using, and another with 50% of the time 170 and 50% of the time 420, and the remaining scores are:

420 1

400 2

-50 1

Now, Pair A is 60% of the time 420. Of this, Pair B is 50% of the time 170 and 50% 420,

so the sheet becomes 30% of the time (60%*50%)

420 3 4

400 2 1.5

-50 1 0

and 30% of the time

420 2 4.5

400 2 2.5

170 1 1

-50 1 0

And Pair A is 40% of the time -50. Of this Pair B is again 50% of the time 170 and 50% 420, so the sheet becomes 20% of the time (40%*50%)

420 1 5

400 2 3.5

170 1 2

-50 2 0.5

and 20% of the time

420 2 4.5

400 2 2.5

-50 2 0.5

Multiple weighted scores on the same sheet result in the multiplication of possibilities and the correct MP calculation requires that all of these are taken into account

In this example, for the pairs not involved in the weighted adjustments, the MPs would be:

420 4.45 (0.3*4 + 0.3*4.5 + 0.2*5 + 0.2*4.5)

400 2.40

-50 0.2

For pair A, 60% of the time 420 results in (0.3*4+0.3*4.5) because pair B is half the time 420 and half the time 170, and 40% of the time -50 results, for the same reasons, in (0.2*0.5+0.2*0.5) = 2.63. For Pair B, similarly, it would be, from each of the four scoresheets in succession: 0.3*4+0.3*1+0.2*2+0.2*4.5 = 2.80

Sounds complicated, but with basic spreadsheet proficiency practice, once one gets a graspof the basic principles,it isn´t that hard.

How about other forms of the game?

Same principle, be it playing Cross Imps, IMPs against a Butler,etc. As long as the event produces a scoresheet with a number of results to be compared as a whole unit, each weight (or combination of weights in different adjustments on the same board) produces a separate"provisional"scoresheet. The weights are then applied to each provisional scoresheet and the results combined to produce a "weighted scoresheet", valid for all the pairs not involved in the adjustments. For the pairs involved in the adjustments, the weights (or combination of weights) are used in combination with each result in each "provisional" scoresheet to calculate their result on the board.

A funny circumstance with Butler scoring is that for each of the weighted possibilities, theaverage that all scores are compared against (the "butler") will be different. According the the February Bridge Bulletin ACBLScore was not ready to do this math yet. Will it be in Reno? Calculations are feasible by hand and relatively easy for small tournaments, but for bigger tournaments they can quickly become a nightmare of routine clerical work...

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