Shooting Squad Probability Problem?

Question

There are two squad playing airsoft gun.
Squad A have 5 members and shooting accuracy of 70%. Squad B have 3 members and shooting accuracy of 85%. Given that both squad have unlimited ammunition and always shoot in the same time, how much is the probability that squad A win, squad B win or draw ?
Of course shooted squad member may not shoot anymore.

The winning condition is by wiping out the other squad. All squad member always shoot in the same time and free to choose their target.

All squad members shoot until they died or the other squad is wiped out. But they always shoot in the same time, just like musketter combat without meele.

Answer 1

This is the best I could do so far. It is a very complicated problem, not that it is difficult, but that there are so many cumbersome calculations that need to be performed.

Here are a few points. Initially A has 5 shots, B has 3 shots. That's in the first round. Clearly both are binomial distributions. But there is a further twist. A cannot make 4 or 5 hits, because B only has 3 members. So what is the P(A = 3). As far as the binomial formula goes, it is equal to
P(A = 3,4,5) because A making 4 hits is equivalent to a member of team B getting shot twice. It's a possibility that cannot be discounted.

So for the first round.
P(A = 3) = 0.84
P(A = 2) = 0.13
P(A = 1) = 0.03
P(A = 0) is too small to consider.
If we don't ignore the small probs, the problem will go on indefinitely.
P(B = 3) = 0.61
P(B = 2) = 0.33
P(B = 1) = 0.06

From those numbers we then need to find likely combinations after the first round. The unlikely combinations will be those involving small probs, eg
P(A = 1 and B = 1) = 0.06*0.03

After we determine the likely scenarios, for each one we need to determine how many members each team has left for round 2, then go at it again. I did this for 3 rounds, assuming that because of the high accuracy, the game should finish fast.

So after doing a probability tree of the likely scenarios, I ended up with

P(A winning) =~ 0.93
P(B winning) =~ 0.03
P(tie) = ~ 0.04

In fact, the strongest likelihood is that A would finish off B in the very first round. There is a 0.84 chance of that happening.

Answer 2

The problem is not well-defined. For one thing, you say nothing about the aiming policy of the shooters. Does each shooter pick a target at random?