Risk and Die-Rolling Probabilities If you aren't familiar with Risk, here is a short description of how the game works:

In the game of Risk, players control countries by occupying them with a variable number of “armies.”  The object is to gain more territory by conducting battles between neighboring countries.  A battle consists of the armies of a single country going against an opponent’s armies occupying a neighboring country.  The battle progresses by each player rolling a prescribed number of dice, and applying rules to determine from the outcome how many armies are lost for each player.  The dice are rolled repeatedly until either the defender loses all his armies (in which case the attacker can occupy the disputed country), or until the attacker is reduced to a single army (in which case there is no army to spare for occupying the opponent’s country).  An attacker can never lose control of his own country.  The battle may also be stopped at any prior time at the attacker’s discretion.
The rules for determining how many dice a player may shake are as follows:
  
  
*
  
*The attacker may shake one less die than the number of armies on his country, to a maximum of three.
  
*The defender may shake as many dice as the number of armies on his country, to a maximum of two.
  
  
  The rules for deciding the outcome of a particular throw of the dice are as follows:
  
  
*
  
*The highest attacker die is compared against the highest defender die.  Whoever has the lower number loses one army.  Ties go to the defender.
  
*The procedure is repeated for the second-highest dice.
  
  
  In cases where either the attacker or the defender only rolls a single die, a total of only one army will be lost; in all other cases a total of two armies will be lost.

For the 1-1 die rolling situation, it is easy to determine that the probability of winning one battle is 15/16.  
How can I determine probabilities for the other scenarios? Do I have to brute force the calculation (which wouldn't be hard if I were to use python), or does there exist a mathematical way to determine the probabilities of the attacking side winning?
 A: No need to apply brute force. Let's calculate the probability that the highest die is $k$ when $n$ dice are thrown. Let's name this probability $P_n(h=k)$
We know that $P_1(h=k) = \frac16, \;\forall\ k \in\{1,2,3,4,5,6\}$
What if we throw two dice? To find $P_2(h=6)$ it's easier to calculate the complementary probability that the highest die is not $6$. 
$$P_2(h<6) = \left({\frac56}\right)^2 \implies \\
P_2(h=6) = 1- \left({\frac56}\right)^2$$
In general for any $n$: $$P_n(h=6) = 1-\left({\frac56}\right)^n$$
Similarly: $$P_n(h<5) = \left({\frac46}\right)^n \implies\\
P_n(h=5) = 1- \left({\frac46}\right)^n - P_n(h=6)$$
In general for any $n$ and any $i$:
$$P_n(h<k) = \left({\frac{k-1}6}\right)^n$$ and 
$$\begin{aligned}
P_n(h=k) &= 1 - P_n(h<k) -P_n(h>k)\\
&=1-P_n(h<k) -P_n(h \ge k+1)\\
&=1-P_n(h<k) - (1- P_n(h < k+1))\\
&= P_n(h < k+1) - P_n(h < k)\\
&= \left({\frac{k}6}\right)^n - \left({\frac{k-1}6}\right)^n = \frac{k^n - (k-1)^n}{6^n}\\
\end{aligned}$$
Just to get a numerical sense, here are the values for $P_3(h=i)$:
$$\begin{aligned}
P_3(h = 6) &= \frac{91}{216} \approx 0.421 \\
P_3(h = 5) &= \frac{61}{216} \approx 0.282 \\
P_3(h = 4) &= \frac{37}{216} \approx 0.171 \\
P_3(h = 3) &= \frac{19}{216} \approx 0.088 \\
P_3(h = 2) &= \frac{7}{216} \approx 0.032 \\
P_3(h = 1) &= \frac{1}{216} \approx 0.004 \\
\end{aligned}
$$
So let's compute the probability of the attacker winning if the attacker throws $3$ dice and the defender $2$. Let's denote the attacker's highest value as $A$  and the defender's highest value as $D$. We want to compute $P(A>D)$.
$$\begin{aligned}
P(A>D) =& \sum_{k=1}^6 P_3(A = k) \cap P_2(D<k)  \\
=&\sum_{k=1}^6 P_3(A = k) \cdot P_2(D<k) &&\text{(because throws are independent)}\\
=& \sum_{k=1}^6 \frac{k^3 - (k-1)^3}{6^3} \cdot \frac{(k-1)^2}{6^2} 
=& \frac{3667}{7776} \approx 0.471
\end{aligned}$$
Sum computed with WolframAlpha
Interesting to know that the attacker is slightly more probably to lose, even with three dice against the defender's two. Ties going to the defender does the job :)  
If the attacker throws three dice and the defender only one then we get: $P(A>D) =\frac{95}{144} \approx 0.66$
