Either provide a reference to a definition that establishes an objective difference in usage so that I may understand and use that definition, or accept this as the proper usage of terminology.
You quoted a definition, and it didn't reflect what you are asking for. A confidence interval is a set of values of a population
parameter, such that that there is a certain probability (confidence level) that the
true parameter is in that interval, based on
observation of a sample. You are not talking about estimating a parameter of your dice based on observation of throws. So this is not the right term.
But as I said, I'm not interested in arguing about the meaning of words; I just want to figure out what you actually want, and see if I can help.
There's probably a more succinct way to express the limitations of k in terms of number of sides of the dice, such that k = aX where X is the set {1,2,3,4,5,6} and aX<=36.
I'm not sure what you mean by aX. How are you multiplying a set by a number to get a number?
But that specificity narrows the problem space down to just the discrete dice model.
I would rather pursue an avenue that leads to a continuous model of probabilities where p is any real number in the range [0..1], and the confidence interval is a range across a normalized set of possible outcomes [0..x], where 0 < x <=1.
Your question WAS about dice. If you're taking that as just an example of what you want, state what the real goal is, rather than wasting people's time with other problems that may not need the same technique. There is a huge difference between the dice question we've been talking about and a continuous distribution.
You haven't responded to the most important thing I said. The question as I stated it, which you say is accurate,
does not have a solution! For some values of p there will be
no intervals (unless you allow some latitude), and for others there will be
two very different intervals.
Now, the continuous situation you are talking about appears to have infinitely many solutions, unless you can clarify it. As I picture it, suppose you have a normal distribution. You want to find an interval such that the probability in this interval is a given number p. You can pick just about any value of x for the left end of the interval and find a right end that will give the desired p. But I'm probably completely misinterpreting what you said.