What are the Elo formulas when assuming performance to be logistically distributed?

Question

I'm struggling to find a definite source on the Elo rating system but from what I understand:

• it originally assumes players performance to be normally distributed with $\sigma = 1$
• the expected score of player A is $E_A = \frac1{1+10^{(R_B - R_A)/c}}$
• the update rule is $R_A = R_A + K (\text{actual score} - E_A)$

I also read that the chess people use the logistic distribution instead of the normal distribution in the assumption of how performance is distributed. But I can't find any second set of formulas for when a logistic distribution is assumed. (When the formulas stay the same there shouldn't be a need for a different distribution because nothing would change.)

Answer 1

Here is one model giving rise to the Elo formula in which the difference in the random components of performance is logistically distributed.

Suppose player a's skill is $R_a$

Player a's performance is $\alpha_a = R_a + \epsilon_a$ , which we can interpret as performance being the sum of skill and other factors summarised by $\epsilon_a$.

If he plays against b, he'll win if he performs better, $\alpha_a - \alpha_b > 0$, which is equivalent to $\epsilon_a - \epsilon_b > R_b -R_a $

Assume that the random components are independent and distributed according to the Gumbel distribution $\text{Gumbel}(\mu, \beta)$ and $\text{Gumbel}(\mu, \beta)$. Then $u = \epsilon_a - \epsilon_b$ follows a mean-zero logistic distribution $\text{Logistic}(0, \beta)$ with scale parameter $\beta$ and

$$\Pr(\text{a wins}) = \Pr(u > R_b -R_a) = \Pr(u < - (R_b - R_a)) = \frac{1}{1 + \exp\left( \frac{R_b - R_a}{\beta}\right)}$$

Which is the same as the Elo function you've written down for $\beta = \frac{c}{\log 10}$