This task can be quite daunting.  For the most part, the difficulty
depends on the number of units involved.  To correctly determine the
odds of a particular result, you multiply the chance of hitting for
each unit involved.  For example, take two battleships (keeping the 
odds balanced for now).  The attacking battleship has a 4/6
(everything will be in sixths here) chance of hitting.  No matter
that roll, the defending battleship also has a 4/6 chance of hitting.
With two results per die (hit or miss, though the chances are
weighted), there are a total of four possible outcomes.  They are:

   Attacker  Defender                Probability  Result

 1   Hit      Hit   ==> (4/6) * (4/6) =  16/36    Mutual Annihilation
 2   Hit      Miss  ==> (4/6) * (2/6) =   8/36    Attacker Wins
 3   Miss     Hit   ==> (2/6) * (4/6) =   8/36    Defender Wins
 4   Miss     Miss  ==> (2/6) * (2/6) =   4/36    Draw; continue

Note that the sum of the probabilities is 1.  This makes for a
simple means of verifying your work.  Adding more units makes this
calculation much more complicated.  To calculate such odds, you must
take into consideration all rolls.  This task can be simplified by
creating a binary tree.

To generate a binary tree, draw a branch for each possible outcome
(hit or miss) for each unit involved for each round of combat.  Keep
them all in a "path" within the tree (e.g., you should be able to go
from the left (top) of the tree all the way to the right (bottom)
and have a representation of every die rolled during a specific
battle).  So, for example, if there were two attacking units and one
defending, you would show both attackers one after the other, then
the defender (by now, there are eight outcomes), then continue based
on the outcomes of those rolls.  Be sure to label each branch
(Mutual Annihilation, Attacker Wins, etc.).  I always work sideways
and make the upward branch the hit branch and label it by placing
the number of ways (out of six!) that a hit could be rolled.
Likewise, the downward branch is for misses and is labelled with the
number of ways (out of six!) to miss. For the battleship example
above:

                     Result                Probability

           Defender

                4 /  Mutual Annihilation    (16/36)
                /
   Attacker   <
            /   \
       4  /     2 \  Attacker Wins           (8/36)
        /
Start <
        \
       2  \     4 /  Defender Wins           (8/36)
            \   /
              <
                \
                2 \  Draw                    (4/36)


To get the probabilities, divide each number by 6 and multiply all
numbers on a path from left (top) to right (bottom).  Write this
result at the left (bottom) of the tree next to the label of the
outcome at the end of that path.  When you have all those numbers,
add all fractions of similar labels (like Mutual Annihilation or
Attacker Wins), no matter where in the tree it is.  This sum is the
probability of that outcome.  The sum of all outcomes is 1 (or a
mistake has been made).  But, what to do at "Draw"?  By the rules,
you continue the battle (as we are concerned about the odds of
possible outcomes we will not consider withdraw options).  If units
have been lost, they will no longer appear in the tree.  Each "Draw"
could actually be considered the "Start" of a new battle with only
the units that survived to get to that "Draw".  In this example, the
entire tree could be copied and placed where the "Draw" is located,
though this would become a recursive loop (which would never end),
making calculations quite difficult.

There is a simplification!  Recall that to find the odds for an
outcome when the tree is complete, you will add the fractions of all
occurances of that outcome. Under the "Draw" node, you will find the
same ratios of results as in the parent branch immediately above it,
thus as you add the fractions that occur under it, you do so in the
same proportions as the top tree (so, as the number of recursive
branches approaches infinity, you reach a limit for the other
outcomes and the "Draw" probability becomes zero).  With that, you
can ignore the "Draw" branch with one provision:  Instead of
counting the other outcomes out of 36, count them out of 36 minus
the 4 occurances of draw.  This is of course 32.  This simplification
will work at each "Draw", but take care to normalize each "Draw" node
separately.  This is done so that the sum of probabilities is still 1
(called normalization).  So, the odds of a mutual annihilation with
two warring battleships is 0.50, and of either combatant winning is
0.25.  While this is how you can calculate the exact odds of a
result, it is obviously no easy task when there are more than a
few units!

While all you have to do is separate between the three results
above, you can distinguish more.  For example, Attaker wins but
loses two units, Defender wins but loses three units, or Mutual
Annihilation.  With more categories you can get a more detailed
breakdown of the outcomes.  This obviously takes more time, but is
quite feasible.

For a more in-depth discussion, or if more explanations of
the odds mechanism is needed, please email Dewey Barich
(barich@tam2000.tamu.edu).