# Degrees of freedom for asymptotic likelihood ratio test?

When testing:

$$H_{0}: p_{1}=p_{2}=p_{3}$$ vs. $$H_{1}$$: at least one is different

With the asymptotic likelihood ratio test, by Wilk's theorem, how many degrees of freedom would the Chi-square distribution have? I'm having a hard time understanding how many free parameters the alternate hypothesis has.

For the asymptotic generalized likelihood ratio test, the guideline for degrees of freedom is $$df = \text{dim } \Omega - \text{dim } \omega_0$$. Notice I'm being vague about what I'm referring to by "dimension" intentionally (it's a bit long). $$\Omega$$ and $$\omega_0$$ are subsets of $$\mathbb{R}^3$$ that are allowable values for the vector $$(p_1, p_2, p_3)^T$$ under the alternative and null hypotheses, respectively.
The parameter space $$\omega_0$$ assuming the null hypothesis is true is "dimension" 1. Intuitively, this is because we have three parameters, but we want all of them to be equal. So, fixing any one of them means that they all are fixed to the same value; we have only one "free choice".
In the alternative, we are dealing with the "full" parameter space. Namely, we have three parameters that can take on any value they want to. So the parameter space $$\Omega$$ under the alternative hypothesis is of "dimension" 3. This is because we are not constraining the value of any of our parameters; all three are free to vary.
So, from the degrees of freedom formula above, we get $$df = 3-1 = 2$$.