Anti-Markovnikov addition to a #pi# bond requires the addition of the non-hydrogen group to the less substituted carbon.
When a carbocation intermediate forms, it usually seeks to stabilize itself through rearrangements: which are accomplished through methyl or hydride shifts.
Hence, it will generally become more substituted, and Markovnikov addition will take place, as a result.
When we have a radical initiator, like #HOOH#, we can ensure that the radical intermediate (that has had the halogen added to the #pi# bond, already, picture below) becomes the most stable which will undergo hydrogen abstraction with #HBr#.
If you want a detailed mechanism, ask!