Zinc finger nuclease specificity
The human diploid genome is ~6.5 billion bp, so you need to find the length of a sequence that has a probability of occurring only once (assuming the sequence is completely random):

$4^n=6.5\cdot10^9$ 

$n\cdot log(4)=log(6.5)+9$

$n=\frac{log(6.5)+9}{log(4)}$

$n\approx16.30$

According to this article, each finger of a ZFN recognizes 3 bp, which means you'd need $\frac{18}{3}=6$ fingers to probabilistically recognize a unique site. Interestingly, the article also says:


  The requirement for dimerization is a great advantage for this reason: because a monomer is not active, cleavage does not occur at single binding sites. The cleavage reagent is assembled only at the target if the fingers have adequate specificity, and the combined requirement for binding two proteins brings the overall specificity into a very useful range; e.g., two three-finger proteins specify the location of 18 bp, which is sufficient, in principle, to pick out a single target, even in a complex genome.