Related to Probability of failing to decode a biased die
Say we are given a biased die that shows up M of its sides with a probability of $p = 0.8$ and the rest of the sides (6 - M) with a probability of $ 1 - p = 0.2$. Note we only know the count of favoured sides (M) and not their identity.
Our objective is to determine $R$ -- the number of times that we should throw the die to determine the favoured sides. The user is allowed to provide a confidence threshold $\delta \in [0,1)$ around the result.
Additionally if we say throw the die a fixed number of times $R1$, then can we determine the probability of failure.
Proposed Solution: Let $X_i$ be the number of times a number from the favoured sides has shown up. Let $Y_i$ be the number of times a number from the remaining sides has shown up.
Say the die is thrown $R$ times and $\delta = 0.9$.
Say we are using the approach in which we repeatedly throw the dice, create a frequency histogram, and then try to cluster the output into two groups.
Now, let's first translate the confidence $\delta$ into a number ($D$) that will ensure that if the difference in the number of times side A and side B show up is greater than $D$, these sides will be counted in different groups.
Now, in $R$ repetitions, each favoured side is expected to be picked up $\frac{p \cdot R}{M}$ times, while any remaining side will be picked up $\frac{(1-p) \cdot R}{6-M}$ times. Ideally, the difference between their frequencies should be around $ T = \frac{p \cdot R}{M} - \frac{(1-p) \cdot R}{6-M}$.
Now, let $D = \delta \cdot T$
Then the quantities $P(X_i - X_j < D)$, $P(X_i-Y_j < D)$ should be the part of the failure probabilities.
I am not sure if the above formulation will lead me to the result—any suggestions or maybe any other direction to approach the solution.