Formation of halohydrin vs alkyl dihalide For the electrophilic addition of $\ce{X2}$ to alkene in $\ce{CCl4}$ solvent, a dihaloalkane is formed. This is a halogen addition reaction, with the intermediate being a halonium ion.



However, if instead of $\ce{CCl4}$, $\ce{H2O}$ is used instead, a halohydrin would be formed. 



However, in the second case (halohydrin formation reaction), the $\ce{X-}$ ion is still present in the reaction. Furthermore, $\ce{X-}$ is a stronger nucleophile than $\ce{H2O}$. Why is it then that it is $\ce{H2O}$ that reacts with the halonium rather than $\ce{X-}$, the stronger nucleophile?
The halohydrin is usually formed by means of a solvolysis reaction where water is used as the solvent. So, it is present in such high amounts that it overpowers the larger nucleophilic strength of the $\ce{X-}$ ion since the latter is only present in relatively minute quantities. Furthermore, water is a polar protic solvent and thus it has the effect of weakening the nucleophilic strength of $\ce{X-}$ in solution by "encasing" it in a large hydration shell.