[computer-go] Is skill transitive? No.

[Nick Apperson]

I feel that what we need essentially is a set of functions that tell us
expected winning percentages with certain matchups.  In an extreme example,
we could imagine 3 rock, paper, scissors players.  One always plays rock,
one always scissor and one always plays paper.  In this case, we would be
able to define a ranking function as a relative ranking, but absolutre
ranking would not exist.  And this is consistent with our whole approach
here that skill is not transitive.  A simple 2D ranking system could work
like this:

let W = chance that player 1 will beat player 2
1-W = chance that player 2 will beat player 1

our skill is expressed in R and T (for theta)

ELO = (R1-R2)+k*sin(T1-T2)   where k is some constant, R1, R2, T1, T2 are R
of player 1 and 2, theta of player 1 and 2 respectively


This would result in a  nontransitive 2D skill map.  There are many, more
compex functions that could be worked out and this has the nice property
that it easily collapses into a 1D map by merely setting everyone's theta to
the same value.  Essentially R is "general skill" and theta is a rock paper
scissor type thing where one strategy is better against certain other types
of strategies.


On 1/31/07, Vlad Dumitrescu <vladdu55 at gmail.com> wrote:
>
> Hi,
>
> On 1/30/07, Don Dailey <drd at mit.edu> wrote:
> > It would be interesting if it would be possible to construct a 2
> > dimensional
> > model statistically.   A 2 dimensional system would not be a perfect fit
> > either,
> > but would simply be a better approximation.    So in some way a players
> > "strength" could be expressed by 2 numbers instead of 1,  and the 2
> > numbers
> > together would predict your chances of beating another (2 dim) player
> > more accurately that a 1 dimension system could.   And of course you
> > could
> > extend this.   But I don't have a clue how one would construct such a
> > system
> > or if it's even possible - but it seems like more information should be
> > better
> > than less.
>
> Unfortunately, having more than one dimensions makes comparisons
> impossible - if an ordering relation is defined over the domain, then
> this domain is "one-dimensional" with regard to that relation.
>
> In other words, one can't compare vectors, just scalars. So the
> multi-dimensional "strength vector" has to be turned into a scalar (by
> for example a weighted sum) and we're back where we started...
>
> best regards,
> Vlad (master of the obvious :-)
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