The larger of two dice

Idea. The event “the maximum is at most $k$” is a product of two independent events, so its probability is a simple square. The distribution of $M$ follows by taking differences.

Step 1. $M\le k$ exactly when both dice are at most $k$. The dice are independent, so $$\mathbb{P}(M\le k)={\left(\frac{k}{6}\right)}^2=\frac{k^2}{36}.$$

Step 2. Taking differences, $$\mathbb{P}(M=k)=\frac{k^2-(k-1{)}^2}{36}=\frac{2k-1}{36}$$ for $k=1,\dots ,6$. As a check, these are $1,3,5,7,9,11$ out of $36$: the outcomes with maximum $k$ form an L-shape in the $6\times 6$ grid, of size $2k-1$.

Step 3. $$\begin{array}{rl}\mathbb{E}[M]& =\frac{1}{36}(1\cdot 1+2\cdot 3+3\cdot 5\\ & +4\cdot 7+5\cdot 9+6\cdot 11)\\ & =\frac{161}{36}\approx \mathrm{4.47.}\end{array}$$

Another route. For a variable with values in $\{1,\dots ,6\}$, $\mathbb{E}[M]={\sum }_{k=0}^5\mathbb{P}(M>k)={\sum }_{k=0}^5\left(1-\frac{k^2}{36}\right)=6-\frac{55}{36}=\frac{161}{36}$.