Percent Chance of Unit being present

[Akmatov]

The idea of beefing up the usefulness of recon aircraft got me thinking. I've putzed around with Harpoon over the years and am coming back to it again and really liking HCE. I usually find myself overwhelmed with the huge scenarios with platforms all over the landscape doing all sorts of things. I prefer smaller scenarios where the problems are more limited - like escorting a convoy across the Atlantic in the face of Badgers, Bears and Victors or conducting a detailed carrier strike against a target.

 

Currently I'm setting up a 'test range' with a base in Greenland from which I'll be running flights down to Keflivik to get a feel for what the various aircraft can do. So far, I've noticed that my EW a/c can provide intel on aircraft present via ESM; which in reality would be hugely helpful, ie is the base protected by MiG-21s or MiG-29s.

 

Given that I'm writing my own scenarios, I kinda know what the Red side has. BUT when assigning a unit to a base, one is asked the percent chance this unit will be present. Does this work? If so, it would be a great way to increase the uncertainty factor and justify the use of pre-strike recon.

[CV32]

BUT when assigning a unit to a base, one is asked the percent chance this unit will be present. Does this work? If so, it would be a great way to increase the uncertainty factor and justify the use of pre-strike recon.

 

Yes, it works. You have to be careful, though, that when using it, you don't also remove the possibility of a particular Red patrol or strike mission actually happening.

[Akmatov]

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?

 

This would be very nice

[CV32]

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? This would be very nice

 

Akmatov, your question screams intervention by VitP, but yeah, generally it would work the way you intend.

[Akmatov]

screams intervention by VitP

 

Huh?

[CV32]

screams intervention by VitP

 

Huh?

 

A joke, or a facetious reference on my part to HG member Victor in the Pacific, and his obvious penchant for math. Never mind.

[VictorInThePacific]

screams intervention by VitP

 

Huh?

 

A joke, or a facetious reference on my part to HG member Victor in the Pacific, and his obvious penchant for math. Never mind.

 

Ask, and ye shall receive.

 

Akmatov, the numbers you have stated (33% chance of each of 3 units A,B,C being present) would appear to mean that 1% of the time you would have no units. However, the specific possibility of getting zero units in a particular test would require the simultaneous failure of all three units to appear. The associated probability is obtained by multiplying 67% by 67% by 67%, approximately 2/3 cubed (= 8/27 = 30% done in my head; you can plug it into a calculator). So if your intention is to have some unit there 99% of the time, this would not be the way to go. In this case, I would suggest that, unless you want to spend a lot of time calculating these possibilities, you allocate a group with a 99% appearance probability, and then the other groups as appropriate.

[CV32]

Ask, and ye shall receive. Akmatov, the numbers you have stated (33% chance of each of 3 units A,B,C being present) would appear to mean that 1% of the time you would have no units. However, the specific possibility of getting zero units in a particular test would require the simultaneous failure of all three units to appear. The associated probability is obtained by multiplying 67% by 67% by 67%, approximately 2/3 cubed (= 8/27 = 30% done in my head; you can plug it into a calculator). So if your intention is to have some unit there 99% of the time, this would not be the way to go. In this case, I would suggest that, unless you want to spend a lot of time calculating these possibilities, you allocate a group with a 99% appearance probability, and then the other groups as appropriate.

 

Thanks, VitP, I knew you'd come through.

 

Akmatov, the point is that you can use the "chance of appearing" mechanism to ensure that there is always a CAP in place.

[Akula]

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?

 

This would be very nice

 

Not exactly, you still have only a 33% chance of any of the aircraft showing up. Each group is working with its own variable, so the end result is that you have a 67% chance that none of those aircraft will make it in to the scenario.

[TonyE]

VitP so far has this most appropriate answer. Math police signing off...

[CV32]

Results oriented approach --> the "chance of appearing" mechanism does work

 

Mathematical analysis --> see VitP's comments

 

Okay?

[Akmatov]

So if your intention is to have some unit there 99% of the time, this would not be the way to go. In this case, I would suggest that, unless you want to spend a lot of time calculating these possibilities, you allocate a group with a 99% appearance probability, and then the other groups as appropriate.

 

OK, I am now thoroughly confused and mathematically humbled.

 

My goal is to create a situation where there is an element of uncertainty regarding which units will be present in a self-created scenario. A 1% 'no show' chance is not important, it just came about as a result of my choosing 33% rather than 33.33% in my example.

 

If 33.3% x 3 groups or 25% x 4 groups doesn't arrive at that result, what percentage value should I be using?

[Herman]

If 33.3% x 3 groups or 25% x 4 groups doesn't arrive at that result, what percentage value should I be using?

I thought that the math would have been, .33 x .33 x .33 = .035937 chance of all three showing up a the same time.

[Akula]

OK, since you have 3 groups each with a 1 in 3 chance of appearing, then wouldn't you have a 1 in 9 chance of all 3 showing up in any given circumstance?

 

After all, you have 3 groups with 3 possible outcomes each. In order for all three to appear in the scenario, all 3 of the groups have to have the 1 result that allows them to be present. So you have 3 groups * 3 outcomes with only 1 possible outcome of the 9 total having all groups present, thus a 1 in 9 chance of having all 3 groups present in any given scenario.

 

And before you get technical I suck at math but this is how I remember it from high school....16 years ago.

[VictorInThePacific]

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.