Consider the guessing milk tea example
$$\begin{array}{c|c|c|} & \text{Guess Milk} & \text{Guess Tea} \\ \hline \text{Milk} & 3 & 1 \\ \hline \text{Tea} & 1 & 3 \\ \hline \end{array}$$
I want to test that $H_0: \theta = 1$ (independent) vs. $H_a: \theta \neq 1$ (associated)
The formula for $P(n_{11} = t) = \frac{{n_{1+} \choose t} {n_{2+} \choose n_{+1}-t}}{{n \choose n_{+1}}}$
It's fixed on $n_{11}$
The range is given by $m_- = max(0, n_{11} - n_{22}) = max(0,0) = 0$ and $m_+ = min(n_{11}+n_{12}, n_{11}+n_{21}) = min(4,4) = 4$.
$$0 \leq n_{11} \leq 4$$
This is what my notes did after that
$P(n_{11} = 0) = 0.0143, P(n_{11} = 1) = 0.2285, P(n_{11} = 2) = 0.5143, P(n_{11} = 3) = 0.2285, P(n_{11} = 4) = 0.0143$
Thus, the two sided p-value is $$P(n_{11} = 0) + P(n_{11} = 1) + P(n_{11} = 3) + P(n_{11} = 4) = 0.4857$$.
Why was $P(n_{11} = 2)$ excluded ?