jmc

There are 36 possible outcomes for the two dice. Of these, there is 1 where both dice roll a six, 5 where the first die rolls a six and the other rolls something less than a six, and 5 more where the second die rolls a six and the first die rolls something less than a six. So, there are a total of $1+5+5=11$ ways the larger number rolled can be a six. Similarly, there are $1+4+4=9$ ways the larger number rolled can be a five, $1+3+3=7$ ways the larger number rolled can be a four, $1+2+2=5$ ways the larger number rolled can be a three, $1+1+1=3$ ways the larger number rolled can be a two, and $1$ way the larger number rolled can be a one. The expected value of the larger number is $$\begin{array}{rl}\frac{1}{36}(11(6)+9(5)& +7(4)+5(3)+3(2)+1(1))\\ & =\frac{1}{36}(66+45+28+15+6+1)\\ & =\boxed{\frac{161}{36}}\end{array}$$