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**log5**estimate for participant 1 beating participant 2 given respective success probabilities \( p_1, p_2 \) is

\begin{align}

p &= \frac{p_1 q_2}{p_1 q_2+q_1 p_2}\\

&= \frac{p_1/q_1}{p_1/q_1+p_2/q_2}\\

\frac{p}{q} &= \frac{p_1}{q_1} \cdot \frac{q_2}{p_2}

\end{align} where \( q_1=1-p_1, q_2=1-p_2, q=1-p \).

Where does this come from? Assume that both participants each played average opposition. In a

**Bradley-Terry**setting, this means \begin{align}

p_1 &= \frac{R_1}{R_1 + 1}\\

p_2 &= \frac{R_2}{R_2 + 1},

\end{align} where \( R_1 \) and \( R_2 \) 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! T…