Akmatov, I have been thinking about the situation you presented and it appears to be the case that you probably can't do what you want.

 

The problem is that (I am guessing here, someone correct me if I am wrong) the probability of a group assigned to a base actually appearing is independent of the probability of another group appearing, whereas what you really need is a conditional probability. For example, you want a 99% chance of something being there, and IF something is there, you want a 1/3 chance that it is of type A, B, or C. Alternatively, you want a 33% chance of group A being there, and IF group A is not there, you want a 49.5% chance that group B is there, and IF group B is not there, you want a 97% chance of group C being there.

 

That way you would get the following result: "a 99% chance of one of the three types of a/c being present AND being on CAP and a 1% of no one home at all".

 

Unfortunately, neither of these conditional setups are options in the scenario editor.

 

On the other hand, if you want a 1% chance of nothing being home using independent group appearance probabilities, you would need

 

(1 - pA) x (1 - pB) x (1 - pC) = .01, and, assuming pA = pB = pC,

(1 - pA) = .215, so

pA = .785

 

But if you need to assign such a high probability to the appearance of the individual groups, then the average result will be that there will be a lot of airplanes at the base. In this case, the average result will be that 2.355 groups will be present.

It's been a little longer than that for me Akula, but the way you put it certainly makes sense to me. Similar to computing the odds when throwing dice.

 

Then agai, I never was the math prodigy that some of the other guys involved in this thread seem to be.

 

Buddha

So if I set a 33% chance of MiG-21s, a 33% of MiG-25s and a 33% chance of MiG-29s AND each element is assigned a CAP Formation Mission; I would have a 99% chance of one of the three types of a/c being present AND being on CAP and a 1% of no one home at all?

 

Here is the complete calculation for this situation.

 

I am going to modify the situation slightly for ease of calculation.

 

Premise:

 

I have a base at which 3 different units, A, B, C, may or may not appear. Each unit has an appearance probability of 1/3. So the probability of any unit NOT appearing is 2/3. Each appearance probability is independent of all the others.

 

Outcomes:

 

All 3 units appear: 1/3 x 1/3 x 1/3 = 1/27

Two units appear: 1/3 x 1/3 x 2/3 = 2/27 (can happen 3 different ways)

One unit appears: 1/3 x 2/3 x 2/3 = 4/27 (can happen 3 different ways)

No units appear: 2/3 x 2/3 x 2/3 = 8/27

 

Total probability: 1/27 + (3 x 2/27) + (3 x 4/27) + 8/27 = 27/27 = 1

(i.e. it is guaranteed that one of the possible outcomes actually happens. If the total probability does not equal one, there is a mistake in the calculation.)

 

Average number of units appearing:

 

(3 x p3) + (2 x p2) + (1 x p1) + (0 x p0) =

(3 x 1/27) + (2 x 3 x 2/27) + (1 x 3 x 4/27) + (0 x p0) =

(3 + 12 + 12)/27 = 27/27 = 1

 

One of the units appears on average, and each one is equally likely.

 

The average result can be obtained by just looking at the statement of the premise, but the individual outcomes need to be calculated in detail.

 

Ok, now that I can comprehend. And now I can even reproduce the results myself if I had to. Thanks Victor, you actually taught an old dog a new trick here.

It's been a little longer than that for me Akula, but the way you put it certainly makes sense to me. Similar to computing the odds when throwing dice.

 

Then agai, I never was the math prodigy that some of the other guys involved in this thread seem to be.

 

Buddha

 

Don't feel bad, I pretty much need a calculator for anything more advanced than 2 + 2.

Ok, now that I can comprehend. And now I can even reproduce the results myself if I had to. Thanks Victor, you actually taught an old dog a new trick here.

Happy to be of service. I was, after all, teaching physics for a living, before I was downsized.