Suppose that Jon has a coin with a probability of landing on heads p and Ann has a coin with a probability of q landing on heads.
Each time they flip a coin and they write the sum of the outcome according to the following table:
| Head | Tail | |
|---|---|---|
| Jon | 3 | 4 |
| Ann | 2 | 3 |
i.e. If Jon gets Head, Ann gets Tail, the sum is $3+3=6$. So, the possible the sum are $\{{5,6,7}\}$
From this information, we know that the probability of getting $5$ is: $p_{5}=p\times q$
Getting $6$ is: $p_6=p\times(1-q)+q\times(1-p)$
Getting $7$ is: $p_7=(1-p)(1-q)$.
Now, let's say that we have given the $Y$ list, which is the outcome of the experiment (data) containing the values of $\{5,6,6,7,5,6,6,7\}$ i.e. two 5-s, four 6-s and two 7s.
My question is
How can I find the likelihood of $p$ and $q$ given $Y=\{ 5,6,6,7,5,6,6,7\}$?
I guess that from the binomial likelihood that the number of $5,6,7$ in the $Y$ is required in the first place.
The likelihood function for a binomial random variable with parameters $n$ and $p$ is: $P=(^n_k)p^k(1-p)^{n-k}$
My guess is that the likelihood for the case will be something like below, but I could not figure it out the combination part (i.e.$ (^n_k)$):
$ L(p\ \text{ and }\ q \mid Y) = (p_5)^2 (p_6)^4 (p_7)^2 =(pq)^2 \times[p(1-q)+q(1-p)]^4 \times[(1-p)(1-q)]^2$
Your likelihood function is completely right. As you have $Y=\{5,6,6,7,5,6,6,7\}$ the order matters, so there is no binomial. Otherwise if you had only the information $2\times 5$, $4\times 5$ and $2\times 7$ given, then you would have to multiply by the number of possible combinations, i.e. $\frac{9!}{2!\cdot 4! \cdot 5!}$.