Lets assume you are talking rolling a fair die. The prior history of rolls does not effect the future outcomes, so:

\(\displaystyle E(X|X>2)=E(X)+2\)

and:

\(\displaystyle \displaystyle E(X)=\sum_{k=1}^{\infty} k \left(\frac{5}{6}\right)^{k-1}\frac{1}{6}=\frac{1}{6} \sum_{k=1}^{\infty} k \left(\frac{5}{6}\right)^{k-1}\)