NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/rec/games/go/2299

Text File | 1992-11-17 | 3.8 KB | 80 lines

Newsgroups: rec.games.go
Path: sparky!uunet!charon.amdahl.com!pacbell.com!ames!saimiri.primate.wisc.edu!hp9000.csc.cuhk.hk!uxmail!uxmail.ust.hk!schmidt
From: [e-mail] (DR. ROY SCHMIDT)
Subject: Re: winning ratio
Message-ID: <[e-mail]>
Sender: [e-mail] (usenet account)
Organization: Hong Kong University of Science and Technology
References: <[e-mail]> <[e-mail]> <[e-mail]>
Date: Wed, 18 Nov 1992 02:10:06 GMT
Lines: 68

In article <[e-mail]> [e-mail] (Bernd Wechner) writes:
>[e-mail] (Fred R Stearns) writes:
>
>>Somewhere (I can't remember where now) I saw the following
>>formula:
>>
>>P = 1/2 * (2/3)^(2*d)
>>
>>where d is the absolute rank difference and P is the probability
>>that the weaker player will win.
>
>Is this with, or without handicap stones? Presumably without.
>
>--
>Bernd Wechner, Research Engineer                       ([e-mail])
>BHP Sheet and Coil Products Division, Research and Technology Centre
>Port Kembla, New South Wales, Australia.

There are several problems with such a "formula" approach.

1.  If we assume no difference in ranks, then the formula yields a
    probability of 0.333 of a given player winning.  Since the game is
    not three-handed, we would have to assign the remaining 1/3 to draws
    Having played not a few games in my life, draws just don't seem to
    come up often enough :-).

2.  Such a formula would have to be strength-dependent.  That is, the
    probability that a 16-kyu would beat a 12-kyu is likely much higher
    than the probability that a ten-kyu would beat a six-kyu in an even
    game, and so forth.

3.  By the formula, the probability runs to a very small number too
    quickly.  For instance, for a one-rank difference (weaker player
    holds Black), p=0.2222.  For a two-rank difference, p=0.0195.  In
    other words, beyond a one-rank difference, the formula has no
    practical significance, since we are giving the weaker side less
    than 2% of the games.  A rating system based on this formula would
    not discriminate well enough beyond two ranks difference.

4.  The formula lacks an empirical basis.  First, you have to find a
    bunch of players and establish, through extended handicap play, the
    absolute difference in their ranks.  Next, you have them play each
    other a sufficient number of games with no handicap to build a
    sample of results for each absolute difference range.  During all
    this, you must have the players promise not to improve :-).  The
    games must be between various combinations of players at each
    absolute difference, since we can't build a formula based on the
    quirks of the difference between just two players.  The way I see
    it, such an experiment would take a good while to complete, and
    the results would probably indicate that because of factors like
    the one example given in paragraph 2, a blanket formula could not
    be set out as a linear expression in simple terms.

Before you flame, dear readers, let me entreat you not to take all my
assertions too seriously.  This is just my opinion, not a scientific
thesis.  For the thesis, call me back in about 300^30 games -- by then
I should have some preliminary results and know how to proceed with the
other 95% of the sample yet to be taken :-).

My point is, we can't be so precise with a ranking system that is
subjective in nature and ranks that wax and wane through experience and
age.

--
Roy Schmidt        [e-mail]        [e-mail]
Business Information Systems Dept, School of Business and Management
         The Hong Kong University of Science and Technology
                Clearwater Bay,  Sai Kung,  HONG KONG