Consider two football teams $V$ and $L$ with strengths $W_V$ and $W_L$, respectively. Let's assume that the draw probability $\mathbb{P}(Draw)$ is known.
Then this model is supposed to give estimates for the winning probability of the two teams:
$\mathbb{P}(V_{wins}) = \displaystyle\frac{1-\mathbb{P}(Draw)}{1+e^{K_W(W_V-W_L)}}$
$\mathbb{P}(L_{wins}) = \displaystyle\frac{1-\mathbb{P}(Draw)}{1+e^{K_W(W_L-W_V)}}$
Assume that there exists $n$ historic games, all of which team $V$ won, that can be considered in the process of estimating $K_W$.
The likelihood function would be:
$L=\prod_{i=1}^n\displaystyle\frac{1-\mathbb{P}_i(Draw)}{1+e^{K_W(Wi_V-Wi_L)}}$
$\ln L=\sum_{i=1}^n(\ln(1-\mathbb{P}_i(Draw))-ln(1+e^{K_W(Wi_V-Wi_L)}))$
$\displaystyle\frac{\partial \ln L}{\partial k_W}=-\sum_{i=1}^n(\frac{(Wi_V-Wi_L)e^{k_W(Wi_V-Wi_L)}}{e^{k_W(Wi_V-Wi_L)} + 1})$
There seems to be no analytic solution to $\displaystyle\frac{\partial \ln L}{\partial k_W}=0$.
My question is then: How can $k_W$ be estimated through a method such as Newton-Raphson or something else?
I would first transform the variables. $1-P(\texttt{Draw})=a$ and $W_V-W_L=b$
$PV=\frac{a}{1+e^{b\cdot K_w}}$
$\frac{1}{PV}=\frac{1+e^{b\cdot K_w}}{a}$
$\frac{a}{PV}=1+e^{b\cdot K_w}$
$\frac{a}{PV}-1=e^{b\cdot K_w}$
$\ln\left( \frac{a}{PV}-1\right)=b\cdot K_w$
Let $\ln\left( \frac{a}{PV}-1\right)=y$ we get
$y=b\cdot K_w$
Now you can apply linear regression.
$$\min \sum_{i=1}^n (y_i- b\cdot (K_w)_i)^2$$