NMR of a bridging Methyl complex
Seems to me there is an error in the last line of this list:


  
  No $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.66 \times 0.66$
  One $\ce{^{195}Pt}$ resulting in a doublet: intensity $2 \times 0.66 \times 0.33$
  Two $\ce{^{195}Pt}$ resulting in a singlet: intensity $0.33 \times 0.33$
  This results in a final ratio of 1:8:18:8:1
  


Simultaneous J-couplings to two $\ce{^{195}Pt}$ should result in a triplet, not a singlet, given that the J-couplings are identical in magnitude.

Therefore the expected relative intensities in the quintet are as follows:

$$\begin{align} &amp;0.34^2\times \frac14 &amp;= 0.029      \tag{1}
\\&amp;0.66\cdot0.34\times1 &amp;=0.224    \tag{2} \\&amp; 0.34^2\times \frac24 + 0.66^2 \times 1 &amp;=0.493   \tag{3} \\&amp; 0.66\times0.34\times1 &amp;=0.224   \tag{4}  \\&amp; 0.34^2\times \frac14 &amp;=0.029\ \tag{5} \end{align}$$

Note there is an implicit assumption that molecules with two $\ce{^{195}Pt}$ result in a 1:2:1 triplet contributing to peaks (1) (3) and (5) above, and that molecules with one $\ce{^{195}Pt}$ account entirely for peaks (2) and (4).

Normalized so that the central peak (3) has an integral of 18 the expected integrals become

$$\begin{align} &amp; 1.0      \\&amp;8.2     \\&amp; 18.0    \\&amp; 8.2     \\&amp;1.0\end{align}$$

Separately I attempted digitizing and integrating the provided image of the spectrum. I obtained the following integrals: 

$$\begin{align} &amp; 0.25      \\&amp;7.0     \\&amp; 18.0    \\&amp; 6.2     \\&amp;0.7\end{align}$$

Given the resolution perhaps the disagreement is acceptable.