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The FIBS rating formula : why sqrt(n) ?

Posted By: Kevin Whyte
Date: Tuesday, 13 September 2011, at 8:58 p.m.

In Response To: The FIBS rating formula : why sqrt(n) ? (Fabrice Liardet)

I think the idea goes something like this : the number of games in the match should be roughly linear with the length (call it n). Over that span the "noise" from luck will have the score deviate from the long term expected win rates of the players by something like the random walk ( so like sqrt(n) ).

So, for a fixed skill difference, as the length grows the expected margin of victory grows like n while the luck only like sqrt(n) - thus the amount of luck needed for the weaker player to win is growing like n/sqrt(n) which is also sqrt(n). Thus the amount of evidence provided about who is stronger is growing like sqrt(n).

This is not just a single point games thing. In general the sum of scores from repeating some fixed event will have mean growing linearly and standard deviation growing like sqrt and be quite close to the corresponding normal distribution in the long run. This doesn't quite apply to backgammon matches for a few reasons. First, matches are finite and not "for n large". Second, play is affected by the score so the individual game results aren't independent. Third, and relatedly, the skills to use a MET and handle match score are different for different match scores so trying to get a rating that accurately reflects in one number a player's ability at different match lengths just isn't possible with people free to play whatever lengths they like.

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