Nonsubstituted Cycloalkene HBr Reaction
Cyclohexene is very symmetric. At first approximation, you can draw two planes of symmetry: one containing the double bond and one orthogonal, bisecting the double bond.

These two planes of symmetry mean, that both sides of the double bond are equal as are both carbon atoms. Therefore, the hydrogen atom protonating the double bond in $\ce{HBr}$ addition cannot sense any difference between either of the four possible attack pathways and chooses one at random. The bromide then has to choose the other carbon, but can again choose the plane from which it attacks (‘above’ or ‘below’ — in inverted commas because they are equivalent!) at random.

Even after the reaction, the product bromocyclohexane is highly symmetric. You can again draw a plane of symmetry through it; this time, the plane must include the $\ce{C-Br}$ bond and also bisect the carbon atom on the other side of the ring. The presence of a plane of symmetry means that the molecule cannot exist in enantiomeric forms, and since you cannot say on which side the hydrogen was previously added (they, too, are all identical), you only get one possible product.



We might consider a more complicated case and decide to add $\ce{DBr}$ instead of $\ce{HBr}$. In this case, we would indeed observe more than one type of product since a deuterium atom cannot be mapped onto a hydrogen atom by symmetry. However, we would obtain all symmetry-equivalent products at equal ratios because of the high symmetry of the starting compound.



In general, if you can find planes of symmetry then both sides are equally likely to react and you cannot predict a preference.