Then, the highest-scoring attacking die is compared with the highest-scoring defending die: if the defender's score is greater than or equal to the attacker's, the defender wins the pair, and so 1 army is removed from the attacker (i.e. \(n\) drops to \(n-1\)). If the attacker's score is greater, the attacker wins and 1 army is removed from the defender (\(m\) becomes \(m-1\)). If there are 2 (or 3) attacking dice

**and**2 defending dice being rolled, then there will be a second pair as well, where the second-highest scoring attacking die is compared with the second-highest scoring defending die, with a loss for either side costing that side 1 army.

The probabilities of every 1 turn outcome are known, having been ascertained largely with the help of BGM in this thread:

http://www.talkstats.com/showthread.php/43951-Dice-probability-problem

We can write the probability that, when 1 attacking die is rolled and 2 defending die are rolled, the attacker will win as P(Attacker wins 1,1,2). The probability that each side will win one pair when 3 attacking dice are rolled and 2 defending dice are rolled is P(Both win 1,3,2). In general P(Outcome,Attacking Dice,Defending Dice) is my suggested notation.

Then, in terms of these different probabilities (suggest your own notation if you want, but I'd like it algebraic) - all their values are known to me, I can provide them if necessary - what is the probability of an attacking side with \(n\) attackers finally winning over the defending side with \(m\) defenders? And in such a battle, what are the expected number of losses the side with more armies would suffer?