Ok, so you might know the game of Risk. If you don't here is a summary of the relevant rules.

The attacking player may, if attacking with more than 3 armies, throw 3 dice to descide the results of a battle. The defender may, if the defending territory holds 2 armies, defend with 2 dice. The higher 2 of the Attacker's dice are matched against the 2 of the defender.
The result of the rolls descide the amount of downsizing in the military force of the battleing territories.

For example, these rolls, will result as follows.
6, 6, 6 vs 6, 6 will result in a win of 2 armies as the defence has the favor in any tie.
If the highest of the attacker is 3 and the highest of the defender 3 the attacker will lose at least 1 army.

So only if the highest of the attacker and it's 2nd are both higher than the matching defender's dice. (5 3 1 vs 4, 2) (the 4 doesnt beat the 3 because it's matched vs the other highest, the 5)
So there you have it, the relevant rules of Risk. Now for the actual problem.

I want to calculate the odds of the result of a single 3 vs 2 battle of the dice in Risk. 5 dice, it's n/7776 chance for the attacker making the defender lose 2 armies. p/7776 for a tie, (each makes the other lose 1 army, for example 6 4 1 vs 6 3) and an q/7776 chance the defender makes the attacker lose 2 armies.

I already found a promising way of getting my answer.
I wanted to use the following, I actually NEED to use the following, because if I can find out the correct use of the figures I figured out I can use it in a decision algorhythm in case the defender gets to choose wether to use 1 or 2 dice.

So here it is,
I found out that the odds for the higest of a roll by the defender being 6 is 11/36
5: 9/36
4: 7/36
3: 5/36
2: 3/36
1: 1/36

And exactly the opposite for the odds of the lower of a defenders roll being:
6: 1/36
5: 3/36
4: 5/36
3: 7/36
2: 9/36
1: 11/36

The Highest for the attacker has it's distribution as follows.
6: 91/216
5: 61/216
4: 37/216
3: 19/216
2: 7/216
1: 1/216

The second highest for the attacker has the following odds.
6: 16/216
5: 40/216
4: 52/216
3: 52/216
2: 40/216
1: 16/216

Now does anyone know how I can easily pair these odds to get to n/7776: p/7666 and q/7666???? With n, p and q all adding up to 7666 so theres no hidden variable. These are all possible outcomes of my question.