[BLML] Scoring 2 and 3 Fouled Boards

Roger Eymard roger-eymard at orange.fr
Tue Mar 1 16:51:29 CET 2016


In order to assign MP scores to a group of results from a fouled board, 
Neuberg’s formula proceeds from the assumption that without the full number 
of real results, the best we can do for a fouled board is to assume that all 
the other tables would have shared in the same proportion the same results 
as those obtained by the tables playing it (easy proof below).

This implies that the results on the fouled board are supposed to be 
distributed in the same way as if played the total number of times.
It is obviously not true for a board played 1 or 2 times, and highly dubious 
for 3 times.
It is apparently why Neuberg’s formula is used for 4 or more results.

But what must be the size of a subset of N results (on the fouled board) for 
it to be representative of N hypothetical real results? More than one third 
seems IMO to be a rule of thumb without rationale. A minimum size to apply 
Neuberg’s formula must be related to N in some way. May be sqr(N), but I 
know no evidence for that.

Thanks for your comments

Roger

Proof:
Within n scores, p equal scores are topped by q scores.
Within those n scores, the MPs for these equal scores are:
(1) x = 2(n-1) – 2q – (p – 1)
We repeat those n scores as many times as necessary to reach a total of N 
scores.
Within these N scores, we obtain (N/n)p equal scores topped by (N/n)q 
scores.
The MPs for these repeated equal scores are:
(2) X = 2(N - 1) - 2(N/n)q – [(N/n)p – 1]
>From (1) and (2) :
2q + p = 2n - 1 – x = (n/N)(2N - 1 - X)
And then Neuberg’s formula: X + 1 = (N/n)(x + 1), extended to the cases 
where N/n is not an integer

----- Original Message ----- 
From: Gordon Rainsford
To: Bridge Laws Mailing List
Sent: Tuesday, March 01, 2016 9:20 AM
Subject: Re: [BLML] Scoring 2 and 3 Fouled Boards


The EBU, which used to Neuberg all fouled boards, has recently changed to 
have a regulation like the ACBL's (and the WBF & EBL I think) except that we 
Neuberg whenever a group is more than one third of the field we Neuberg even 
if it's a small number. So a five table movement played in one form at two 
tables and another at three tables would be Neuberged in both subfields.


On 01/03/2016 04:16, David Grabiner wrote:

The ACBL policy is that Neuberg is used for a group of four or more, and is 
used for a group of three if it is the larger group.  Most groups of three 
score 70/60/50, and all groups of two score 65/55.


On 2/29/2016 4:43 PM, Jan Peach wrote:


Can anyone direct me to learned papers on the above please?
I favour 70% 60% 50% to Neuberg for 3 different scores. To me, Neuberg looks 
over generous/harsh to the Highest/Lowest scores.
I would like to read something authorative on the subject.
Jan






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