Having to do something other than command 99 other bots to press F1 would be healthy for this game.

LOL

I jsut reread what you said.

Your fleet member random choose 2 targets each in 100v100 and puts 2 damps on them.

Law of Probability gives a 95% chance that no less than 90 hostile ships would have 2 damps on them.

Law of Probability gives 74% chance that no less than 68% hostile ships would have 4 damps on them.

You use the nCr combinatoric probability distribution.

If each Megathron random chooses 4 ships and puts one damp on them, you now have a 99.95% chance that at least each hostile ship has at least one damp on them.

87% chance that any hostile has two damps.

21% chance any hostile has three damps

2% chance any hostile get hit with 4.

1% for 5

sub 1% (cumulative) to get hit with 6 to 100 damps.

Thus the Fleet commander chooses if he wants more random breadth (each mega chooses 4 different ships) or more random focus (each mega chooses 2 different ships).

Higher **breadth** increases the total amount of hostiles damped (at least one damp); higher **focus** increases the amount of individual hostiles with at least two damps, but leaves more hostiles possibly untouched.

Higher breadth is to force range (reliable), higher focus is to stop damage (reliable).

You can use the below diagram for a quick estimation of random choice, although it’s not as accurate.

Let us now show how the real formula would work:

Let G = number of allied megas

Let E = number of hostile Muninns.

Let a = number of damps on Mega

Let b = number of damps used on each target.

Let n = number of targets randomly chosen by a single mega, such that:

n = a/b, where n is a whole number.

For the mega we have 4 mid slots, one is reserved for Prop mod.

We can fit 3 damps to a mega, so:

a = 3

Since n must be a whole number, b cannot be equal to 2, therefore:

b = 1 or 3, therefore

n = 3 or 1.

We will assume the case of b = 3, such that each Mega chooses only one target and puts all 3 damps on them, therefore n = 1, this is the FOCUS model.

Now say we have 100 Megas and 100 Munnins, then:

G = 100

E = 100, set E = {e1, e2, e3…e100}

n = 1

The first Mega chooses his target, {e1}

The second Mega chooses his target, with (E - 1)/probability of NOT choosing {e1}. This would be a (99/100) probability. Call this target {e2}.

Before we move on, you must understand the following combination chart for the binomial coefficient:

.

The amount of distinct pair combinations that the first two Mega pilots can choose is given by the formula nCr; or 2 C 100 = 4950, let 4950 = Y

The amount of same choice combinations is simply 100, where the first and second pilot chose the same target. Let 100 = Z.

let X = (Y + Z) = 5050. The chance that both pilots choose different targets is TRULY (Y/X) = 98.02%, the TRUE chance they both chose the same target is the compliment (Z/X) = 01.98%

Now let us proceed to the Third Mega pilot, he chooses enemy {e3}. The number of distinct combination that {e3} is not {e1} or {e2} is given by the formula 100 C 3 = 161700, let this be equal to T.

T = 161700

Since 100 C 2 = 4950 distinct {e1, e2} pairs, and there are 100 same choice {e1,e2} pairs (Y and Z).

There are still 100 same choice pairs where {e1} = {e2} = {e3}.

There are 161700 triplets where {e1} ~= {e2} ~= {e3}.

We know need to know the chance of {e1} = {e2} but ~= {e3}.

We set {e1} = {e2} = {h}. Then for each {h} there are 99 choices of {e3}. The number of choices for {h} is 100. Therefore there are 99(100) combination for this scenario, which is 9900 choices.

We repeat this again for the case of: {e1} = {e3} but ~= {e2}.

and we repeat again for the case of: {e2} = {e3} but ~= {e1}.

Both of the above yield 9900 separate combinations. Let 9900 = Y

Now we can proceed to obtain the true probability the Mega Pilot 3 chooses a third distinct enemy.

Let X = (T+Z+3(Y)) = (161700 + 100 + 3(9900)) = 191,500.

Then there is a (T/X) chance, 84%, that the third mega pilot does not choose a Muninn to damp that was not already damped.

As we proceed up the chain from the third mega pilot to the last Mega pilot, it becomes far easier to count the probability that any particular subset of Muninns is NOT damped.