so you form a bromonium ion first, (three membered ring with cationic bromine). well, although it's drawn with a positive charge on the bromine, the positive charge is distributed slightly, especially onto the carbons in the bromonium ring. you can see this if you draw two resonance contributors by kicking the electrons from the C-Br bond onto the Br to make a carbocation and a neutral Br.
anyway, since the two carbons in the bromonium ring have partial positive charges, the trick is to see which one will have the greater partial positive charge. this will be the more substituted carbon for the same reason a more substituted carbocation is more stable than a less substituted carbocation.
so the short answer is, the more substituted carbon has a greater partial positive charge (delta plus) so it's more susceptible to attack by the nucleophile (water). thus, the bromine will end up on the less substituted carbon and the hydroxyl will end up on the more substituted carbon.
