I can see why you might be confused, so I'll try to explain how it works.

Let's examine this case: 100 armies attacking 60 defending armies with 19% SR luck.

In order to calculate the probability that these armies will take the territory, we do these steps:

1) First of all, we calculate binomial probabilities (also called 100% luck in WZ).

[1] 1.606938e-40 2.410407e-38 1.789727e-36 8.769664e-35 3.189965e-33 9.187099e-32 2.181936e-30 4.395043e-29 7.663856e-28 1.175125e-26 1.604045e-25
[12] 1.968601e-24 2.190068e-23 2.223762e-22 2.072864e-21 1.782663e-20 1.420559e-19 1.052885e-18 7.282455e-18 4.714432e-17 2.864017e-16 1.636581e-15
[23] 8.815222e-15 4.484265e-14 2.158053e-13 9.840720e-13 4.258004e-12 1.750513e-11 6.845755e-11 2.549454e-10 9.050560e-10 3.065512e-09 9.915016e-09
[34] 3.064641e-08 9.058719e-08 2.562323e-07 6.939626e-07 1.800552e-06 4.477688e-06 1.067756e-05 2.442492e-05 5.361569e-05 1.129759e-04 2.285792e-04
[45] 4.441709e-04 8.291190e-04 1.487007e-03 2.562714e-03 4.244495e-03 6.756543e-03 1.033751e-02 1.520222e-02 2.148776e-02 2.919091e-02 3.811036e-02
[56] 4.781118e-02 5.762955e-02 6.672895e-02 7.420719e-02 7.923819e-02 8.121914e-02 7.988768e-02 7.537790e-02 6.819905e-02 5.914136e-02 4.913282e-02
[67] 3.908293e-02 2.974969e-02 2.165603e-02 1.506506e-02 1.000750e-02 6.342785e-03 3.832099e-03 2.204769e-03 1.206664e-03 6.274654e-04 3.096047e-04
[78] 1.447502e-04 6.402414e-05 2.674426e-05 1.053055e-05 3.900205e-06 1.355559e-06 4.409650e-07 1.338644e-07 3.779700e-08 9.888749e-09 2.386939e-09
[89] 5.289241e-10 1.069734e-10 1.961179e-11 3.232713e-12 4.743655e-13 6.120845e-14 6.837114e-15 6.477266e-16 5.060364e-17 3.130122e-18 1.437301e-19
[100] 4.355457e-21 6.533186e-23

2) Calculate how many armies are going to be killed by interpolating 0% and 100% results based on the luck modifier (19%), and round up these numbers:

[1] 49 49 49 49 49 50 50 50 50 50 50 51 51 51 51 51 52 52 52 52 52 53 53 53 53 53 54 54 54 54 54 54 55 55 55 55 55 56 56 56 56 56 57 57 57 57 57 58
[49] 58 58 58 58 58 59 59 59 59 59 60 60 60 60 60 61 61 61 61 61 62 62 62 62 62 62 63 63 63 63 63 64 64 64 64 64 65 65 65 65 65 66 66 66 66 66 66 67
[97] 67 67 67 67 68

3) Now sum up all these binomial probabilities which gives at least 60 kills and you'll get a solution:

0.6967399

===========

I skipped some details, because they were already mentioned by Fizzer, but I'll give you my code written in "R"

A <- 100
D <- 60
L <- 0.19
binomial <- c()
kills <- c()
output <- 0
for (i in 0:A) {
binomial <- c(binomial,factorial(A)/factorial(i)/factorial(A-i)*0.6^i*0.4^(A-i))
kills <- c(kills, round((i-0.6*A)*L+0.6*A))
}
output <- sum(binomial[match(D, kills):A])
output