What is the electron count in this nickel complex? Our class has started learning about electron counting using the ionic method. I was having a little difficulty, especially when there are two metals in one complex, so I looked at Wikipedia for help.

The Wikipedia page for organonickel complexes says:


  In $\ce{(allyl)2Ni2Br2}$ and $\ce{(allyl)Ni(C5H5)}$, nickel is
  assigned to oxidation number +2, and the electron counts are 16 and
  18, respectively.


I understand that in $\ce{(allyl)Ni(C5H5)}$, the allyl and cyclopentadienyl ligands both have a $-1$ "charge", so the nickel has a +2 "charge". Thus, the electron count is: $4 + 8 + 6 = 18$.

However, I am struggling to calculate the electron count in $\ce{(allyl)2Ni2Br2}$. Using the same logic, $\ce{Ni}$ is in the +2 oxidation state. Since the structure (see below) does not have a metal-metal bond, we do not have to add $1$ when counting electrons. So, I think the electron count should be $\frac{4*2 + 8*2 + 2*2}{2} = 14$, not $16$. How does Wikipedia get that $\ce{Ni}$ has an electron count of $16$ in $\ce{(allyl)2Ni2Br2}$?
The bromide ions contribute an electron pair on each side of the square they form with the nickel centers (thus the bromine has a positive formal charge like a bridged bromonium ion), so in your fraction you need $4×2$ in the numerator where you have $2×2$.  Then it will come out to $16$.