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The FIBS rating formula : why sqrt(n) ?

Posted By: Timothy Chow
Date: Wednesday, 14 September 2011, at 5:24 p.m.

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

I haven't looked at this issue before, but I agree with you; it doesn't seem that the cited theorem from random-walk theory is relevant here.

Suppose we fix the rating difference between two players, and ask for the functional relationship between the match length N and the probability P that the weaker player wins. If we simplify and assume that all games are worth one point, then we're asking for the probability that a biased random walk from the origin (i.e., which steps to the right with probability p and to the left with probability 1–p, where p > 1/2) ends up to the left of the origin after 2N–1 steps. This is a large-deviation inequality, which if I remember correctly means that P should decrease like exp(–N) rather than exp(–sqrt(N)).

Now, with the cube in play and with gammons and backgammons counting, maybe this simplified model breaks down. But the cited passages don't seem to give any hint that the sqrt(N) comes from modeling the doubling cube.

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