help,practical,interval measure of ranks
From: Robert Frick (junk_at_rfrick.info)
Date: 10/28/04
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Date: Thu, 28 Oct 2004 15:34:54 GMT
In bridge, they have tournaments, and the top finishers are given
"masterpoints". The current formulas for computing masterpoints are
garbage. I want to change them, and it is unlikely that anyone will
listen, but I still am at least curious what they should be.
Suppose, for the sake of example, that coming in 1st out of 40
contestants receives 10 masterpoints. It makes sense that second place
would be some proportion of that, and third place would be some
proportion. Say the top 10% receive awards.
Currently, the formula is to give second place 75% of the first place
award, and third place 75% of the second place award, etc.
Given some scheme like that, we can calculate what the first place award
should be for any number of contestants, using the assumption:
Axiom 1: The number of contestants should not influence the total number
of points awarded per contestant.
Sorry this is taking so long. The problem is determining the
proportions. I think that the percentage drop from first to second
should be smaller when there are more people. I am more confident that
the drop from 2 to 3 should be not only a smaller amount (which it
alrady is), but even a smaller percentage.
And I am fairly sure that
Axiom 2: The award for 3rd out of 40 should be about the same as 30th
out of 400. To do that, the percentage drop has to be less for more
contestants.
This essentially is the same as converting rank scores to a linear
measure. Are there accepted ways of doing this?
I am not looking for "the correct" answer. I am looking for a defensible
mathematical procedure that does not violate either common sense or
correct axioms. For example, it seems allowable to assume that ability
is normally distributed (a reasonable assumption) and that the observed
rank was a perfect ordinal measure of ability (it is not). However, I
then can't do the math to get at an answer. Or probably understand the
answer either, but I thought I would ask.
Bob Frick
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