I'm having a hard time solving the following problem :
We have $n$ uniform dices, each with $m$ different faces. We throw the dices simultaneously and see what comes out. Then we order the results in increasing order to get a number read in base $m$. Call this $r$. What is the average of $r$?
I thought of representing $r$ as a random variable and calculate the mean, but I'm not sure how. The problem is that some faces may appear more than one time, and that's what makes the problem hard. There's an intuitive trick that I failed to see here.
Here is how far I was able to simplify, perhaps there is an easier way to do this:
Let $X_k$ denote the outcome of dice $k$. Then $X_1,\dots,X_n$ are IID with uniform distribution, and $P(X_k=j) = m^{-1}$ for each $k=1,\dots,n$ and $j=0,\dots,m-1$. Let $Y_1,\dots,Y_n$ be the order statistics of $X_1,\dots,X_n$ (i.e., $Y_1$ is the smallest of $X_1,\dots,X_n$, and $Y_2$ is the next smallest, all the way to $Y_n$ being the largest). Then it holds $$R = Y_nm^{n-1} + \cdots + Y_2m + Y_1 \implies \mathbb E[R] = \sum_{k=1}^n\mathbb E[Y_k]m^{k-1}.$$ From the wikipedia page on order statistics, the CDF of $Y_k$ is given by $$P(Y_k \leq x) = \sum_{j=k}^n\binom{n}{j}\left(\frac{x}{m}\right)^j\left(1-\frac{x}{m}\right)^{n-j}, \quad x=0,\dots,m.$$ If $Z_x\sim\text{Binom}(n,x/m)$, then we see $P(Y_k \leq x) = P(Z_x \geq k).$ Thus we have $$\mathbb E[Y_k] = \sum_{x=1}^mP(Y_k \geq x) = \sum_{x=1}^m P(Z_x \leq k).$$ So, $$\mathbb E[R] = \sum_{x=1}^m \sum_{k=1}^n P(Z_x \leq k)m^{k-1}.$$