What is the highest relative abundance for diatomic bromine? I was introduced to this thing today, and saw a graph showing the detection of ions with different relative abundances.

Just as an example here, $\ce{Br2}$ is shown a detection chart. And at $160~\mathrm{m/z}$ the ion has the highest relative abundance, but why? Since the $\mathrm{m/z}$ ratio is just the mass, for $\mathrm{z}$ is always $+1$, the one which is supposed to have the highest peak should be at $162$?
In $\ce{Br2}$, the possible combinations of isotopes are:


$\ce{^{79}Br-^{79}Br} : \mathrm{m/z}~158$
$\ce{^{79}Br-^{81}Br} : \mathrm{m/z}~160$
$\ce{^{81}Br-^{79}Br} : \mathrm{m/z}~160$
$\ce{^{81}Br-^{81}Br} : \mathrm{m/z}~162$


Bromine's isotopic distribution is essentially a $50:50$ ratio of mass $79$ and mass $81$. That means that there's a $25\%$ chance of $\mathrm{m/z}~158$ $(0.5 \cdot 0.5 = 0.25)$ arising from $\ce{^{79}Br-^{79}Br}$. There's a $50\%$ chance of $\mathrm{m/z}~160$ since there are two combinations ($\ce{^{79}Br-^{81}Br}$ and $\ce{^{81}Br-^{79}Br}$) that give that mass $(0.5\cdot0.5 + 0.5\cdot0.5 = 0.25)$. There's a $25\%$ chance of $\mathrm{m/z}~162$  $(0.5\cdot 0.5 = 0.25)$ arising from $\ce{^{81}Br-^{81}Br}$.