If we have some random variable X, and it's distributed geometrically.

$\displaystyle X = G(p)$

and

$\displaystyle p = \frac{1}{6}$

To get expected value it's just:

$\displaystyle E(X) = \frac{1}{\frac{1}{6}} = 6 $

But how to get this probability:

$\displaystyle P( E(X)-2 < E(X) < E(X)+2 )$

thank you for your help.