(1.512 + 1.672 + 1.832 + 1.992) / 4 = 1.752

I assume you used the above equation to calculate the percentage of killing 1 or 2 armies. However, what this equation fails to take into account is the fact that R is more likely to be some numbers than others. In this case, since it's a relatively simple problem, we can use a probability tree to illustrate this.

Note: P is for pass, F is for Fail. The numbers in brackets represent number of armies passed over the total number rolled so far.

Start: _[0]_
P / \ F
(60%) (40%)
/ \
Army #1: _[1/1] [0/1]_
P / \ F P / \ F
(60%) (40%) (60%) (40%)
/ \ / \
Army #2: [2/2] [1/2] [0/2]
P / \ F P / \ F P / \ F
(60%) (40%)(60%) (40%)(60%) (40%)
/ \ / \ / \
Army #3: [3/3] [2/3] [1/3] [0/3]

As we can see here, getting 0 kills here is the least likely, while 2 is the most likely. To be exact, we can calculate this by finding the probability of taking each path.

0 armies: 40%*40%*40% = 6.4%

1 army: (40%*40%*60%)*3 = 28.8%

2 armies: (40%*60%*60%)*3 - 43.2%

3 armies: (60%*60%*60%) = 21.6%

6.4% + 28.8% + 43.2% + 21.6% = 100%.

So, we would then weigh the numbers using the percentages above...

1.512*6.4% + 1.672*28.8% + 1.832*43.2% + 1.992*21.6% = 1.8 armies.

As expected, we end up with 1.8 armies, which is the number you put for E... and you will always end up with E. Thus, there isn't any point in using this formula to find out the average number of armies killed. This formula still holds up even on 100% luck - try plugging in 0, 1, 2, and 3 instead of those decimal numbers and you will still end up with 1.8 as the answer.

To find the average amount of armies killed, multiply the number of armies killed by the kill rate. (Yes, this is the same as finding E in my last post.)

What the formula I've posted above could be used for, however, is finding the minimum and maximum values for armies killed. For the minimum, simply replace R with zero and either round down (Weighted Random) or round normally (Straight Round). For the maximum, replace R with the number of armies you attack with, and round up (Weighted Random) or round normally (Straight Round).

As an example, here's why strategic 1v1 uses 16% luck: (Note: This is for 4v2)

On 16% luck: K=0*16% + 2.4*84% = 2.016 armies. This guarantees that 4 will kill 2 armies.

On 17% luck: K=0*17% + 2.4*83% = 1.992 armies. This means that 4 can fail to take 2 armies, although the chance is minuscule. (Usually, this will show up as 100% on the analyze graph, even though it isn't 100% - this is due to the fact that Warlight's analyzer simply simulates the situation hundreds of times, and shows the results.)