Here's a question from an old exam paper, I am struggling to understand how to construct an answer:
"A six-sided die has an unknown number of faces marked with a
six. Let $k$ be this unknown number, which we would like to estimate.When the die is thrown each face has an equal chance of showing. The observed data is that
the die was thrown twice, and it showed a six exactly once. Write down the likelihood for the observed data. What is the maximum likelihood
estimate for $k$?"
I really just don't get how to start this. I think I need to decide what distribution the observed data comes from, but when I try a binomial distribution, I end up with only being able to estimate the probability of getting a six given the data, not the number of sixes on the die.
EDIT: With thanks to user Mees de Vries, I believe I may have found a way to model the problem correctly. Say the die has $k$ faces with a six on them, $k \in \{0,1,2,3,4,5,6\}$. Then, say we have rolled the die $n$ times and observed $m$ successes, where success is rolling a six. Then we could treat this as $$m \sim \text{Bin}(n,k/6) $$
Then of course, the MLE is $$\frac{\hat{k}}{6}=\frac{m}{n} \implies \hat{k}=\frac{6m}{n}$$ Our observed data is that $m=1,n=2$ so here $\hat{k}=6(1)/2=3$
Which matches my intuition in the comments. Now, is my logic correct?
Two long for a comment. I didn't unerstand what die is.
My approach to do your question without "Binom":
We have two expectations. The first one is fake expectation. In your problem, fake expectation is $E_f=\sum_{i=1}^n(\frac{k}{6})6=\sum_{i=1}^nk=kn$. On the other hand we have the reality, real expactation. Here, it is $E_r=6m$ because we are $m$ times succesful. From the conversation of expactations, $E_f=E_r\implies kn=6m\implies k=\frac{6m}{n}$.