Chance a cube ends in the digits 11

Pick $x$ uniformly from $\{1,2,\dots,10^{100}\}$. What is the probability that the last two digits of $x^3$ are $11$?

$0.01$. Writing $x=a+10b$ with $a$ the units digit, $x^3\equiv a^3+30a^2b\pmod{100}$. The units digit forces $a=1$, and then $30b\equiv10\pmod{100}$ forces $b$ to end in $7$. So exactly the $x$ ending in $71$ work: one residue in $100$.