TITLE: Closed form equation for win percentage of two battling armies
QUESTION [6 upvotes]: I was pondering a battle mechanic for a board game that is similar to, but simpler than battling armies in Risk.  Consider one army of size X and a second army of size Y.  The battle occurs by creating pairs of one-on-one fights where the two combatants each have 50% probability of winning.  The pairs of fights continue until one army is defeated.
For Example, a 2 vs 3 battle could have the following outcomes for the first army, when the first army wins:
1. W W W
2. L W W W
3. W L W W
4. W W L W

Note that the final fight must be a win, all outcomes must have 3 wins total (the size of the second army, Y), and there cannot be more than 1 loss (the size of the first army, X, minus 1).
I'm having trouble finding a closed form equation that gives the probability of the first army winning the battle where the sizes of the armies are X and Y.
If that's easy, how about if the probability of winning the battle if an individual fight is not 50% but rather probability P?
Edit: just wanted to add a 3 vs 2 example as well:
1. W W
2. L W W
3. W L W
4. L L W W
5. L W L W
6. W L L W

Here's the 2 vs 2 case, which produces a 50% probability:
1. W W
2. L W W
3. W L W

REPLY [3 votes]: You succeed if and only if you win at least $Y$ battles out of $N = X + Y - 1$ attempts.
Indeed, if you win less than $Y$, then you lose at least $X$, and you are dead before your enemy. Conversely, if you win at least $Y$, then your enemy will be dead before you.
The number of battles won is given by the binomial distribution, so the probability of succeeding is
$$\sum_{k=Y}^{N}{N\choose k}P^k(1-P)^{N-k}$$
For large $N$, we can approximate the binomial distribution with a normal one, and thus approximate the above formula to
$$ \Phi\left(\frac{PN - Y}{\sqrt{NP(1-P)}}\right) $$