Expected number of sides of a dice I have two dice, one with m sides (labeled $1,2,...,m$) and one with $n$ sides (labeled $1,2,...,n$). I roll both three times. The $m$-sided one comes up $1, 2, 9$ and the $n$-sided one comes up $7, 7, 8$. Which is higher: the expected value of $m$ or the expected value of $n$? Now compute both expected values and give their approximate value with a $95\%$ confidence interval.
Part (a) of the question seems pretty straightforward and I tried approaching part (b) using MLEs but that didn't turn out too well because the likelihood function of this involves a $n!$ term, namely
$$P(x_1,x_2,...x_n|n)=\frac{n!}{x_1!x_2!...x_n!}\Big(\frac{1}{n}\Big)^{x_1+x_2+...x_n}$$
Any suggestions? Thanks!
 A: From a Bayesian statistics point of view, this is an ill-posed question because you don't have a proper prior distribution on the number of sides of the dice. For example suppose that you know the number of sides is between 2 and 9. Then clearly the first die has higher expected number of sides because it must equal 9, which is the maximum. Since you have to use Bayes rule to appropriately write down the probabilities for this problem, the only "non-Bayesian" way to do it is to basically assume a "flat" prior on the number of sides of the dice, i.e. before you observe the data then each number of sides is equally likely. However that is an improper prior because you would get that each number of sides has arbitrarily small prior probability (basically 0) because there are infinite possibilities and yet the infinite number of prior probabilities must sum to 1. There is no proper flat prior that can accomplish this.
A: With a maximum likelihood approach:
for the first dice, the probability of every 3 throw event is $1/m^3$, with $m\ge 9$. Hence the value of $m$ that maximizes the probability is $m^{ML}=9$. The same way you get $n^{ML}=8$.
To get a confidence interval you need a Bayesian approach. But you are not given any prior.
A: The other answers have already pointed out the missing prior. However, as both dice were thrown the same number of times and the highest roll is a sufficient statistic, for any prior that has non-zero probability for the number of sides to be $8$ the expected number of sides will be higher for the die that rolled a $9$ than for the die that rolled an $8$.
We can also use an improper flat prior (or equivalently use a finite uniform prior and then take the upper bound to infinity). In that case the expected number of sides for the die that rolled a $9$ is
$$
\frac{\sum_{m=9}^\infty\frac m{m^3}}{\sum_{m=9}^\infty\frac1{m^3}}=\frac{\zeta(2)-\sum_{m=1}^8\frac1{m^2}}{\zeta(3)-\sum_{m=1}^8\frac1{m^3}}=\frac{\frac{\pi^2}6-\frac{1077749}{705600}}{\zeta(3)-\frac{78708473}{65856000}}\approx17\;,
$$
and for the die that rolled an $8$,
$$
\frac{\sum_{m=8}^\infty\frac m{m^3}}{\sum_{m=8}^\infty\frac1{m^3}}=\frac{\zeta(2)-\sum_{m=1}^7\frac1{m^2}}{\zeta(3)-\sum_{m=1}^7\frac1{m^3}}=\frac{\frac{\pi^2}6-\frac{266681}{176400}}{\zeta(3)-\frac{9822481}{8232000}}\approx15\;.
$$
See also the German tank problem.
