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hi all,

on the matter of scoring i would like to propose the following in case a ruling breaks a result in pieces
example 1: (result broken down)
ns score -50 in 40% of the cases
ns score +110 in 60% of the cases
this means that the frequencie of -50 will increase with 0.4
and the frequencie of +110 will increase with 0.6
example 2: (split score)
ns score -50 in 100% of the cases
ew score +110 (ns score) in 100% of the cases
here herman de wael advocates the method of counting these scores for 0.5 frequencie each, what makes sense, because that's their real frequencie. when calculated,
ns should recieve only the points for -50 and ew for +110, as if the frequencie is 1 (100%) for them, which is the point of a split score.
example 3: (result broken down and split)
ns score -50 in 40% of the cases
ns score +110 in 60% of the cases
ew score -50 in 20% of the cases
ew score 0 in 30% of the cases
ew score +110 in 50% of the cases
allthough this is very unlikely to occur, it is not impossible.
in fact the possibility to calculate such a result may lead to frequent use of this type of ruling.
the point i am trying to make is that each ns score should add half of its frequencie, as the ew scores should do as well.
a simple way of calculate all the rulings is then to weigh to scores:
in example 3:
line ns ew score weightns weightew
x xx yy -50 2 2
x+1 xx yy 0 0 3
x+2 xx yy +110 3 5
added frequencie for -50: 0.5 * 2/5 + 0.5 * 2/10 = 0.3
added frequencie for 0: 0.5 * 0 + 0.5 * 3/10 = 0.15
added frequencie for +110: 0.5 * 3/5 + 0.5 * 5/10 = 0.55
total added frequencie: 1 (to be expected)
each score should than be awarded a number of points according their frequencie.
each pair should than be awarded a number of points
according the weight of their scores.
this method should be used in any form of calculation:
matchpoints, imps, total points.
e.g. matchpoints in example 3:
-50: 7.7 points for ns
0: 10 points for ns
+110: 20.3 points for ns
score ns: 2/5 * 7.7 + 3/5 * 20.3 = 15.26
score ew: 2/10 * 7.7 + 3/10 * 10.0 + 5/10 * 20.3 = 14.69 (ns points!)
also by calculating according the frequencie, it's easier to neubergize. the method of comparing all results in ns, and in ew after that should be made obsolete. it is only debatable if on a certain table ns and ew have different scores. in that case, each score should be accounted for. what i mean is that if on a certain board, ns recieves score A and ew score B, all ns pairs should be compared to score B (and A) and all ew pairs to score A (and B), simply because these scores occurred. a more intuitive reason would be that it seems wrong that a certain table will receive more or less points on that board than another, only because of the ruling at some table.

best regards,

tom cornelis.

International Bridge Laws Forum

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International Bridge Laws Forum

If you need help with the Laws or rulings from any country in the world, this is the place!Hosted by David StevensonSenior Consultant DirectorEnglish Bridge Union

If you need help with the Laws or rulings from
any country in the world, this is the place!

Hosted by David Stevenson
Senior Consultant Director
English Bridge Union