$$X \ge 6$$: The probability for

$${{{5^5}} \over {{6^6}}} + {{{5^6}} \over {{6^7}}}\, + \,...\, + \,\infty = {{{5^5}} \over {{6^6}}}\left( {{1 \over {1 - 5/6}}} \right) = {\left( {{5 \over 6}} \right)^5}$$

For $$X>3$$, we have

$${{{5^3}} \over {{6^4}}} + {{{5^4}} \over {{6^5}}}\, + {{{5^5}} \over {{6^6}}}\, + ...\, + \,\infty = {\left( {{5 \over 6}} \right)^3}$$

Hence, the conditional probability is

$${{{{(5/6)}^6}} \over {{{(5/6)}^3}}} = {{25} \over {36}}$$