Comparsion of cross sections of initial state polarisations and unpolarized sample for VBF di-Higgs production

Dear Madgraph experts,

I have simulated a vector boson fusion process where the two vector bosons that fuse are polarised and give a Higgs pair. I looked at the three polarisation combinations both bosons are longitudinal polarised (LL), one boson is longitudinal and one transversal polarised (LT) and both bosons are transversal polarised (TT) using the following process definitions:

LL:
generate w+{0} w-{0} > h h QED=2 QCD=0
add process z{0} z{0} > h h QED=2 QCD=0

LT:
generate w+{T} w-{0} > h h QED=2 QCD=0
add process w+{0} w-{T} > h h QED=2 QCD=0
add process z{0} z{T} > h h QED=2 QCD=0
add process z{T} z{0} > h h QED=2 QCD=0

TT:
generate w+{T} w-{T} > h h QED=2 QCD=0
add process z{T} z{T} > h h QED=2 QCD=0

Additionally I produced the unpolarised process (full) as a comparison point:

Full:
generate w+ w- > h h QED=2 QCD=0
add process z z > h h QED=2 QCD=0

As a test I compared the cross section between the full process and the sum of the different polarisations. As far as I understand in order to compare the cross section I need to take into account that Madgraph averages over the initial state polarisations. Since there are 3 different polarisations for the vector bosons (longitudinal, left-handed and right-handed) I need to divide the cross section of each of the polarisation combinations by 9 before summing over them. I noticed that by doing this I don’t get a good closure in some cases (up to a difference of ~60%). I noticed that using LL/9, LT/2.25 and TT/2.25 instead leads to a good closure. My suspicion is that Madgraph already averages over the left-and right-handed polarisation that are included in the transversal polarisation meaning I need to account for that by dividing the factor 9 by 4 for the two combinations that include the transversal polarisation.

My question is if that is the correct explanation and the 1/9 for LL and 1/2.25 for LT and TT are the right factors or what else could lead to the non-closure?
I also simulated samples where I explicitly used the left- and right-handed polarisation instead of the transversal one and in this case I get to a good closure by dividing every combination by 9.