Consider one game between two opponents (I am actually thinking about chess but it doesn't have to be that). The ELO rating system gives the expected score $E_A$ and $E_B$ for both players. That is
$$E_A=Pr(A\ wins)+\frac12Pr(draw)+0Pr(A\ loses)=Pr(A\ wins)+\frac12Pr(draw)$$ $$E_B=Pr(B\ wins)+\frac12Pr(draw)+0Pr(B\ loses)=Pr(B\ wins)+\frac12Pr(draw)$$
From Elo's logistic formulas
$$E_A=\frac{Q_A}{Q_A+Q_B}, E_B=\frac{Q_B}{Q_A+Q_B}$$
where $Q_A=10^{R_A/\alpha},Q_B=10^{R_B/\alpha}$, $R_A,R_B$ are the ratings of the opponents and $\alpha$ is a normalization constant. As expected
$$E_A+E_B=Pr(A\ wins)+Pr(draw)+Pr(B\ wins)=1$$
which agrees with the formula.
Prof. Elo originally made his system for two outcomes only, win and loss. However if we have three outcomes and want explicitly the probability of win lose or draw for the game we need more data, right? We need one more equation. Anybody have an idea how to remedy this?
Thanks in advance, I appreciate.
