This is clearly maximal when nnn is the smallest value possible, which here is 4 (since it’s not possible to draw a 4 with a 3-faced die). So far this is quite easy, but the confidence interval is another affair, and illustrates quite well the idea of “add-on”. One way to find it is to find all the values of nnn for which P(Xmax≤4∣n)≥α/2P(X_{\mathrm{max}} \leq 4 | n) \geq \alpha/2P(Xmax≤4∣n)≥α/2, where α\alphaα is the confidence level (usually chosen to be 5%). For a given nnn, this probability is equal to (4n)8\left(\frac{4}{n}\right)^8(n4)8 which yields a CI of the form [4,6][4,6][4,6], so there we have it!2
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