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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.

2007-05-11 19:33:53 · 2 answers · asked by seed of eternity 6 in Science & Mathematics Mathematics

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.

2007-05-12 23:59:48 · update #1

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.

2007-05-13 00:01:44 · update #2

2 answers

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.

2007-05-12 06:12:18 · answer #1 · answered by Dr D 7 · 1 0

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?

2007-05-17 22:48:14 · answer #2 · answered by Anonymous · 0 0

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