Free-radical addition of bromine to alkene

If light and radical initiators are present, radical addition of $\ce{HBr}$ to alkenes takes place and outcompetes electrophilic addition of $\ce{HBr}$ to alkenes. Why isn't a similar thing same happening if we use $\ce{Br2}$ instead? Why does the addition of $\ce{Br2}$ to alkenes only take place through electrophilic addition? What is wrong with the following radical mechanism?


$$
\begin{align}
&\text{Initiation:} &\quad \ce{Br2 &->[$h\nu$] 2 Br^.} \tag{R1}\\
&\text{Propagation:} &\quad \ce{CH2=CH2 + Br^. &-> ^.CH2-CH2Br} \tag{R2}\\
& &\quad \ce{^.CH2-CH2Br + Br2 &-> CH2Br-CH2Br + Br^.} \tag{R3}\\
&\text{Termination:} &\quad &\ldots
\end{align}
$$




The addition of bromine on a double bond occurs without light. So the molecule $\ce{Br2}$ is not broken into Bromine atoms. But it reacts with the alkene $\ce{R1R2C=CR3R4}$, producing a triangular cation $\ce{C2R1R2R3R4Br^+}$. And this cation reacts later with the remaining Bromide ion to produce the saturated compound $\ce{C2R1R2R3R4Br2}$.


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