Does that mean you want a greater weight for school 4 or you’re saying you’d get one with my approach?
round(c(.5/2,.5/2,.5/3,.5/3,.5/3),2)
[1] 0.25 0.25 0.17 0.17 0.17
The weights for school 2 sum to .5 and so do the weights for school 4.
I think what is confusing here is the combination of odd number of slots for even membership, or vise versa. Odd-odd or even-even is straightforward:
for [1,1,2,2], c(1,1,1,1) gives you equal weights , c(1,1,2,2) gives school 2 twice the influence. same for [1,2,3], c(1,1,2) gives the third school twice the influence of the other two. But for [1,1,2], I think you need to split the weight for school one to respect the dummy coding: c(.5,.5,1) for equal weights or c(.5,.5,2) for school 2 having twice the influence of 1.
EDIT: I am wrong!
Have a look at this