I think it might be helpful to work backwards...

Let C(n,m) be your expected number of hits with n attacks to come if you have m images remaining. Then (assuming for simplicity that we pre-cast MI) we are interested in C(n,3) as compared to C(n,0).

Let T(m) be the chance that an image is targeted when you have m of them, and let x and y be as you defined them in the OP. We have T(0) = 0, T(1) = 10/20, T(2) = 13/20, T(3) = 15/20.

Then

C(n,0) = nx

C(1,m) = (1 - T(m)) * x

and for m > 0 and n > 1:

C(n,m) = C(1,m) + (1 - T(m) + T(m)*(1-y)) * C(n-1,m) + T(m) * y * C(n-1,m-1)