Inspired by Pi Han

$0.222 \dot {2} _4 = \frac {2}{4^1} + \frac {2}{4^2} + \frac {2}{4^3} + ... = 2( \frac {1}{4^1} + \frac {1}{4^2} + \frac {1}{4^3} + ... ) =$

$= 2 ( \frac { \frac {1}{4} }{ 1 - \frac {1}{4} } ) = 2 ( \frac { \frac {1}{4} }{ \frac {3}{4} } ) = 2 × \frac {1}{3} = \frac {2}{3} _ {10}$

$0.333 \dot {3} _5 = \frac {3}{5^1} + \frac {3}{5^2} + \frac {3}{5^3} + ... = 3( \frac {1}{5^1} + \frac {1}{5^2} + \frac {1}{5^3} + ... ) =$

$= 3 ( \frac { \frac {1}{5} }{ 1 - \frac {1}{5} } ) = 3 ( \frac { \frac {1}{5} }{ \frac {4}{5} } ) = 3 × \frac {1}{4} = \frac {3}{4}$

$\text {Since } \frac {3}{4} > \frac {2}{3} \iff \frac {9}{12} > \frac {8}{12} \text { , therefore our answer is:}$

$\boxed {0.33333..._5}$