I have two discrete distributions. The second one is a simple distribution where all values from 1 to 6 (think rolling dice) have equal probability (so, that must be $\frac{1}{6}$). The first one is defined like this: $f(6) = 0.18$, $f(5) = 0.14$, $\ f(4) = f(3) = f(2) = f(1) = 0.17$. There are two observations. I need to find the most powerful test for a given significance level (0.0196, so this clearly hints at something like getting two 5s in a row, since $f(5)^2 = 0.0196$).
I only know how to apply the Neyman-Pearson Lemma. First I need to get the likelihood ratio. I can't even do this because I don't understand how to construct the joint probability mass function for the first distribution since it's not a closed-form expression. Maybe it's supposed to be a sort of a multinomial distribution? But the number of observations (2) is less than the number of possible values (6), so how would this work...
Then, I need to construct a test based on this likelihood ratio: $$ \phi(x_1, x_2) = \begin{cases} 1, \lambda(x_1, x_2) > c \\ 0, \lambda(x_1, x_2) < c \\ \alpha, \lambda(x_1, x_2) = c \end{cases} $$
And I do not have a good grasp on how to derive $\alpha$ and $c$. For $c$ I need to construct an inequality based on the likelihood ratio and use the significance level somehow, but I don't really understand how it's supposed to be done.
If you have not confused anything and the distribution corresponding to the main hypothesis is really symmetric, and the distribution corresponding to the alternative hypothesis is asymmetric, then the test will be the following.
For a single observation, $$ \lambda(x_1)=\frac{f_2(x_1)}{f_1(x_1)}=\begin{cases}6\cdot 0.18, & x_1=6 \cr 6\cdot 0.14, & x_1=5, \cr 6\cdot 0.17, & x_1=1,2,3,4\end{cases} $$ And for two observations we have $$ \lambda(x_1,x_2)=\frac{f_2(x_1)f_2(x_2)}{f_1(x_1)f_1(x_2)} $$ $$=\begin{cases}36\cdot 0.18^2, & x_1=x_2=6 \cr 36\cdot 0.18\cdot 0.14, & x_1=6,x_2=5 \text{ or } x_1=5,x_2=6 \cr 36\cdot 0.14^2, & x_1=x_2=5, \cr 36\cdot 0.18\cdot 0.17, & x_1=6, x_2\in\{1,2,3,4\} \text{ or } x_2=6, x_1\in\{1,2,3,4\} \cr 36\cdot 0.17\cdot 0.14, & x_1=5, x_2\in\{1,2,3,4\} \text{ or } x_2=5, x_1\in\{1,2,3,4\}\cr 36\cdot 0.17^2, & x_1, x_2\in\{1,2,3,4\}\end{cases} $$ The greatest value of $\lambda(x_1,x_2)$ is $36\cdot 0.18^2$ if two sixes observed. If we consider critical region as $x_1=x_2=6$, then the significance level of this MP test is $$ \mathbb P_1 (x_1=x_2=6)=\frac{1}{36}=0.02(7) > 0.0196. $$ So, we should consider randomized test where $\alpha$ is such that the significance level of the test is exactly $0.0196$: $$ \mathbb P_1(\lambda(x_1,x_2)>36\cdot 0.18^2)+\alpha\cdot \mathbb P_1(\lambda(x_1,x_2)=36\cdot 0.18^2) = 0+\alpha\cdot \frac{1}{36} = 0.0196 $$ So, $\alpha=36\cdot 0.0196=0.7056$
Finally, we get most powerful test for a significance level $0.0196$ $$ \phi(x_1,x_2)=\begin{cases} 0.7056 & x_1=x_2=6\cr 0 & \text{ other cases}\end{cases} $$
And I really sure that you swapped null and alternative hypotheses. In this case swap numerator and denominator of $\lambda(x_1,x_2)$ and you'll get that the largest value will be at $x_1=x_2=5$. If we take it as critical region then $$ \mathbb P_1(x_1=x_2=5)=0.0196 $$ as required. Note that I swap numbers of distributions and now the first one is asymmetric. Here $\alpha$ in $\phi(x_1,x_2)$ is zero and $c$ is any number lies strictly between $\frac{1}{36\cdot 0.14^2}$ and $\frac{1}{36\cdot 0.14\cdot 0.17}$. The last value is the next greatest value of $\lambda(x_1,x_2)$.