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Gordon Bower | 1 Sep 2002 04:19

Re: Lower tops (long)


On Sat, 31 Aug 2002, Marvin L. French wrote:

> It has some more inelegance. I don't know how it is handled elsewhere, but in
> the ACBL one, two, or three scores are not Neuberged. A single score gets 60%,
> two get 55% and 65%; and three get 50%, 60%, and 70%, with no regard to the
> larger group's top.

To clarify, this regulation applies specifically to fouled boards. This
makes some sense: if Table 1 plays a board and then Table 2 promptly
reshuffles it, the Table 1 players have been deprived of a meaningful
matchpoint result through no fault of their own and have an honest claim
to an "average-plus-like" adjustment. The 55-65 and 50-60-70 never did
make sense to me as being 'too narrow', but then, I didn't write the
regulation. They certainly don't represent any attempt at estimating what
really would have happened on the board, and are a completely separate
issue from whether Neuberg is a good idea or not.

Even in the ACBL, if you have a series of boards some with a 3+ top and
some with a 2 top, the boards with a 2 top are factored to the usual
17-50-83, not to 50-60-70. (Does this ever happen? Sure it does... my unit
ran its North American Open Pairs qualifying game last week, a two-session
game. Had four tables enter and one pair withdrew at the half.)

As to the original comment about the "inelegance" of Neuberg, I was more
inclined to see independence from the number of boards as a feature, not a
flaw. I am accustomed to "elegant" being associated with simplicity.

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Marvin L. French | 1 Sep 2002 07:58
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Re: Lower tops (long)


From: "Gordon Bower" >
>
>  Marvin L. French wrote:

> > > > [Jeremy Rickard:]
> > > > However, I think that it's clear that we're never going to decide from
> > > > first principles what the appropriate scaling method is. This depends
>
> Yes. Worded differently, we've seen people pick 4 different sets of first
> principles, and seen them derive the method appropriate to each.  Not
> surprisingly, as long as we disagree on the starting point we will
> disagree on the best solution.

I see only three, Ascherman and Neuberg having the same "first principle."
>
> > And, presumably, include the variance of results. -100, -50, -50, +100
should
> > not be equal to -800, 0, 0, +800, one would think.
>
> If we are playing matchpoints, we've explicitly agreed in advance that
> these *are* equal.

I didn't mean to imply otherwise. Most people cannot understand why they should
be treated equally.
>
>
> > Henry's method somehow accounts for the variance difference by shrinking the
> > smaller group's scores toward 50% a bit more than Neuberg does, which  looks
> > right.
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Marvin L. French | 1 Sep 2002 08:15
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Re: Dummy asks defender about revoke


From: "Todd Zimnoch"

> From: Marvin L. French
>
> I thought we were talking about actions permitted by
> the Laws, whether the
> converse of a law is legitimate because not forbidden.

> That's what I'm talking about and if that is not the
> issue, sorry. The
> reference to Item 7 of the Lille WBFLC meeting gave me
> that impression.

     Item 8.  Here it is again.

goddammit, it's item 7!!

I'm outta this thread, tired of the sophistry.

Marv
Marvin L. French
San Diego, California

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Jurgen Rennenkampff | 1 Sep 2002 11:16
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RE: Lower tops

Finally, at the end of this muddled discussion, some clarity...

There are two distinctly different problems being confounded:
(1) How to average two or more scores obtained by the same player on boards
not played the same number of times.
(2) How to obtain a fair ranking of scores obtained by two or more players
in different sections when
(a) One played more boards than the other
(b) The boards were not played the same number of times

Case (1) is not difficult if, as Jeremy points out, what is known about the
statistics of the situation suffices to make reasonable simplifying
assumptions. Suppose you have two measurements - two bridge scores, the
position of a satellite in the sky, or whatever - and one is known to be
more reliable than the other. How do you estimate the mean? You form a
weighted average, giving more weight to the more reliable measurement. The
correct weights are (under further assumptions concerning the distributions)
inversely proportional to the variances. This is not equivalent to the
various correction schemes in use, nor to the new proposal. The weighted
average deflates the difference between the 2 scores, while  Newberg et al.
deflate the difference between a score and the global mean.

The resolution of Case (2) is impossible unless there is extraneous
information and a useful definition of what is meant by the 'better
performance'. Consider two baseball players; one bats .300 and breaks his
leg in late August. The other also bats .300 but plays until the end of the
season. Who is the better hitter? The answer is that you cannot know, based
on this information alone. If you rank one higher than the other and wonder
whether this is fair you will unavoidably have to quantify what you mean by
'fairness', and any such definition will be somewhat arbitrary. - In bridge
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Jeremy Rickard | 1 Sep 2002 10:31
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Re: Lower tops (long)

"Marvin L. French" <mfrench1 <at> san.rr.com> writes:

> Neuberg assumes that the six scores with a 5 top would be duplicated in a game
> with 23 top, each of the six scores becoming four identical scores, hence the
> same variance. We know that larger tops have smaller variances, making good
> scores more difficult to achieve. A 5 top is very unlikely to be a top in a
> much larger field (factoring), and not likely to be a tie for top (Neuberg).
> We are dealing with probabilities, aren't we?
> 
> Henry's method somehow accounts for the variance difference by shrinking the
> smaller group's scores toward 50% a bit more than Neuberg does, which  looks
> right.

I think you're mixing up two different variances. The "Unfairness of
Variance" arises from the fact that in a small field, the variance of
a particular pair's percentage score is greater than in a large
field. But here you are talking about the variance of the distribution
of the scores achieved (by all players) on a particular board, which
may be almost completely unrelated.

    Jeremy.

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Email: j.rickard <at> bristol.ac.uk
WWW: http://www.maths.bris.ac.uk/~pure/staff/majcr/
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Jeremy Rickard | 1 Sep 2002 11:33
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Re: Lower tops (long)

"Marvin L. French" <mfrench1 <at> san.rr.com> writes:

> From: "Gordon Bower" >
> > All the methods posted here, and a bunch more methods not posted here, all
> > involve making certain assumptions.
> 
> All assumptions are not equal.
> 
> Factoring. Assume a lower number of results would each have an identical
> percentage score in a larger field. A top remains a top, a zero remains a zero.
> So obviously wrong it isn't necessary to prove it so.

Factoring gives:

(a) an unbiased estimate of the matchpoints a particular pair would
get in a larger field. (I.e., "a 60% pair will average 60%".)

and

(b) an unbiased estimate of the matchpoints the pair would get in a
larger field, given the at-the-table score they have got (e.g.,
+1440) and the opponents they are playing against.

The only assumption is that the smaller field is chosen at random from
the larger field. The proof is simple undergraduate level probability
theory, and makes no assumption about what the distribution of scores
would be in a large field.

Neither Neuberg nor Bethe have properties (a) or (b).
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vitold | 1 Sep 2002 14:42
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Minutes from Paris-2001, (was L25B in KO play)

Hi all:)
Some days before we (Sergey and Vitold) had discussed privately a similar
problem, therefore this letter reflects our agreed
opinion.

>From David Stevenson
Sent: Friday, August 30, 2002 11:30 PM
Subject: Re: [BLML] L25B in KO play

<snip>

>   Let us suppose that you reach 4S on minimum values, which is cold, and
> will get you an 80% score.  Let us also suppose that an overtrick would
> make it a 90% score.  At this point your RHO leads a club out of turn.
>
>   You decide, correctly, that if you can stop a club lead you will make
> an overtrick.  However, the TD gets the OLOOT wrong, and while he allows
> you to ban a club lead he fails to say that the ban continues for as
> long as LHO has the lead.  LHO cashes an ace and then switches to a
> killing club.  The UI is irrelevant: the club switch is evident.
>
>   So, you get an 80% board, but if the TD had not made an error you
> would have got a 90% board.  Half an hour later, the TD proudly returns
> to your table, says he has made a mistake, and so gives you A+/A+, ie
> you now get a 60% score.  Are you happy?
>
>   At least your opponents are happy.  They got 20%: they should have got
> 10%: and the TD has now given them 60%.
>
>   Can this be right?  Fortunately, it is not.  Reading the Law carefully
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Adam Wildavsky | 1 Sep 2002 16:19

Re: 12C2 Interpretation (was Las Vegas)

At 1:08 PM +0100 8/29/02, David Stevenson wrote:
>Adam Wildavsky writes
>  >My sincere apologies to all. I sent a message to David which was
>>intended to be private. I neglected to indicate to him that he ought
>>to reply to me only, and not the BLML.
>
>   As a matter of good Netiquette I do not quote other people's emails
>without permission.  However, the email you sent to me you *also* copied
>to BLML.

I see now that you are correct, I did. I committed the cardinal sin 
of email, not knowing whether I was sending publicly or privately. 
Just to make this clear:

David is right.
Joan is right.
Grattan is right.
I was wrong.

Mea culpa. Mea maxima culpa.

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adam <at> tameware.com         http://www.tameware.com
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Jeremy Rickard | 1 Sep 2002 19:23
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Re: Lower tops (long)

Jeremy Rickard <j.rickard <at> bristol.ac.uk> writes:

> Steve Willner <willner <at> cfa.harvard.edu> writes:

> > Absent a true mathematical basis, Neuberg strikes me as being as good as
> > anything.
> 
> You may be right. However, I just don't know whether the size of the
> Neuberg adjustment is roughly appropriate or not, and, as far as I
> know, nobody else knows either. Some empirical investigation could be
> enlightening.

I performed the following computer experiment.

I had 20 pairs, of four different ranks:

A: 3 very good pairs.
B: 7 good pairs.
C: 7 bad pairs.
D: 3 very bad pairs.

I assumed that every board that was played had five possible results,
numbered from 0 (bad) to 4 (good). 

When two pairs of the same rank play, each result 0-4 is equally
likely.
When two pairs whose ranks differ by one play, the better pair gets a
result in the range 1-4 (each with equal probability).
When two pairs whose ranks differ by two or more play, the better pair
gets a result in the range 2-4 (each with equal probability).
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Todd Zimnoch | 1 Sep 2002 19:41
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RE: Lower tops (long)

> -----Original Message-----
> From: Jeremy Rickard
> Sent: Sunday, September 01, 2002 1:24 PM
> To: bridge-laws <at> rgb.anu.edu.au
> Subject: Re: [BLML] Lower tops (long)
> 
> I performed the following computer experiment.

	Thank you for running this experiment and sharing your results.

-Todd

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Gmane