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    Hey....Nice Marmot BorisTheFrugal
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    Can someone explain for us less statistically savvy players what reaching equilibrium means?
    I understand what it is in theory: the score that one's skill level should gradually move towards as the number of games goes to infinity.

    But there have been a number of references to the calculation of an equilibrium in some threads, and I haven't the slightest clue how you're doing those calculations (I'm looking at you, Hugh).


  2. #2 / 9
    Standard Member Hugh
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    I derived it in post 5 of

    http://www.wargear.net/forum/showthread/474/Ratings_Equilibrium

    I naturally assumed that everyone here had read that post ;)

    I was being a bit winded, but equilibrium is reached when the amount you expect to lose playing equals the amount you expect to win. Once reached, variation will keep you going above and below it, but generally you shouldn't stray too far.

    e^ix=cos x + i*sin x. Tell your friends.

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    Standard Member Hugh
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    A condensed version of the  calculation (b/c man can my explanations be winded!!!):

    You win at rate p, lose at rate 1-p against a person rated R. You've hit Equilibrium E when

    20*(1-p)*(E/R) = 20*p*(R/E)

    Cancel the 20's divide by 1-p, reciprocal R/E to the other side and we get

    E^2/R^2 = p/(1-p), or E = Sqrt(p/1-p) * R.

    e^ix=cos x + i*sin x. Tell your friends.

  4. #4 / 9
    Hey....Nice Marmot BorisTheFrugal
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    Exactly what I was looking for.  Thanks for the help!!


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    Shelley, not Moore Ozyman
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    I always meant to post this in that thread, but since it is revisited...

     

    I think from your formulas you are saying you can estimate what a players equilibrium score would be based upon current games.

    If that is true, I think it would be really cool to have that 'predicted equilibrium' on the rankings page somewhere.   It would be neat to see a new player with only 50 games played, but they've won a lot and you can see that it is expected they'll end up at 2000. 

    Edited Thu 19th Apr 16:25 [history]

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    Standard Member Hugh
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    Agree, it would be cool. The prediction requires a win rate and an opponent rating, so one way to make the formula simple would be to use H-Rating and average opponent rating.

    e^ix=cos x + i*sin x. Tell your friends.
    Edited Thu 19th Apr 16:32 [history]

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    Standard Member Luieuil
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    To be move precise I think your current H-rating compared to the average opponent rating you're losing to is the best predictor of your equilibrium rating.

    For example, Player A has an H-rating of 80% and plays to opponents with an average rating of 1100 his ER is about 2200. This falsely assumes his losses are equally distributed between good and bad players. Now if we state player A loses to opponents with an average rating of 1500, then the ER moves to 3000. 

     

    "Battles are won by slaughter and maneuver. The greater the general, the more he contributes in maneuver, the less he demands in slaughter" - Winston Churchill

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    Brigadier General M57 M57 is offline now
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    Might it be possible to include with such a rating a "confidence interval" or "margin of error" type of number based on number of games played?  ..or the number of games opponents have played.

    I.e., is there any concern that the H-ratings of opponents may not be based on a statistically significant games in the first place?  ..or am I over-thinking the problem? (not unlikely)

    It should be possible to play WG boards in real-time ..without the wait, regardless of how many are playing.
    https://sites.google.com/site/m57sengine/home

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    Standard Member AttilaTheHun
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    M57 wrote:

    ...  ..or am I over-thinking the problem? (not unlikely)

    yes :D

    "If an incompetent chieftain is removed, seldom do we appoint his highest-ranking subordinate to his place" - Attila the Hun

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