Let $D$ and $A$ be the number of defenders and attackers respectively killed during a roll.  One quantity of interest is $\mathbb{E}(D-A)$.  

According to the [wikipedia page](http://en.wikipedia.org/wiki/Risk_%28game%29#Dice_probabilities), if the attacker rolls 3 dice and the defender rolls 1 dice we have $\mathbb{E}(D)=0.66$ and $\mathbb{E}(A)=0.34$. By linearity of expectation, $\mathbb{E}(D-A)=0.32.$

On the other hand, if the attacker rolls 3 dice and the defender rolls 2 dice we have $\mathbb{E}(D)=1.08$ and $\mathbb{E}(A)=0.92$.  It is reasonable to normalize so that the expected number of death is one (to match the first case).  In this case, $\mathbb{E}((D-A)/2)=0.08.$

So, clearly it is better to defend with two dice instead of one (in some sense four times better).