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As I mentioned in a previous article on ratings systems, the log5 estimate for participant 1 beating participant 2 given respective success probabilities p1,p2 is
The limit extreme case is log5; the n=2 extreme case I call tanh5. We compute
p=p1q2p1q2+q1p2=p1/q1p1/q1+p2/q2pq=p1q1⋅q2p2
where q1=1−p1,q2=1−p2,q=1−p.
Where does this come from? Assume that both participants each played average opposition. In a Bradley-Terry setting, this means
p1=R1R1+1p2=R2R2+1,
where R1 and R2 are the (latent) Bradley-Terry ratings; the 1 in the denominators is an estimate for the average rating of the participants they've played en route to achieving their respective success probabilities.
In a Bradley-Terry setting, it's true that the product of the ratings in the entire pool is taken to equal 1. But participants don't play themselves! Thus, if participant 1 played every participant but itself, the average opponent would have rating R, where R1⋅Rn−1=1. Here n is the number of participants in the pool.
Our strength estimate for the average opponent faced is then
Our strength estimate for the average opponent faced is then
R=R1−1n−1.
There are two extreme cases. If n=2, then R=1R1; as n→+∞, R→1.
The limit extreme case is log5; the n=2 extreme case I call tanh5. We compute
p1=R1R1+1/R1p2=R2R2+1/R2tanh5=p=√p1q2√p1q2+√q1p2.
Why tanh5? We can think of log5 as derived from the logistic function by setting log(R)=0 for the opponent's rating; analogously, tanh5 is derived from the hyperbolic tangent function by setting log(R)=0 for the opponent's rating.
Note that we have a spectrum of estimates corresponding to each value for n, but these are the two extremes. This also gives a new spectrum of activation functions for neural networks, but I'll explore this application later.
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