Conquer Club

If you throw 3 dice, and they throw 1, call the probability of winning "a". For 3v2, call a win "b", a loss "c" and a draw "d". c = 1- b - d, so I'll make this switch wherever I can. Call the expected loss against n armies x_n.
x_1 = a*0 + (1-a)*(1 + x_1), because if you win, then your loss is 0, and if not, then it's one more than it was before, and you have to try again.
x_1 = 1/a -1 = (1296-855)/855 = 49/95 = 0.516

x_2 = c(2 + x_2) + d(1 + x_1) = (d*x_1 + 2c + d)/(b + d) since b + d = 1 - c
x_2 = 808234/522595 = 1.547

x_3 leads us to the general case, since it is the first value that cannot be taken in one throw. I need the value for initial recurrence conditions, but I'll just write the general equation.
x_n = c*(2+ x_n) + d*(1 + x_(n-1)) + b*x_(n-2)
Since this is an inhomogenous recurrence (it contains 2c + d on the right side, with no x values attached), we convert it to a homogenous recurrence by subtracting x_(n-1) from both sides.
x_n - x_(n-1) = c*(2+ x_n) + d*(1 + x_(n-1)) + b*x_(n-2) - c*(2+ x_(n-1)) - d*(1 + x_(n-2)) - b*x_(n-3) or more simply
x_n = [(b + 2d)*(x_(n-1)) + (b - d)*(x_(n-2)) - b*x_(n-3)]/(b + d) and canceling all the fractions, we get
x_n = [8112*(x_(n-1)) + 279*(x_(n-2)) - 2890*x_(n-3)]/(5501)

We solve this in the usual way, by solving x^n = [8112*(x^(n-1)) + 279*(x^(n-2)) - 2890*x^(n-3)]/(5501), and find that x=1 (twice) or x = -2890/5501. So we look for a general solution of the form x_n = A + Bn + C(-2890/5501)^n

A little bit of leg work (and using x_0 = 0) gives us A = -(2a - 2b - d)(b + d)/[a(2b+d)^2], C = (2a - 2b - d)(b + d)/[a(2b+d)^2], and B = (2 - 2b -d)/(2b + d), or A = -0.2213412, C = 0.2213412, and B = 0.8534144