smc

Let ${\log }_{a}b=z$, so $a^z=b$. Define $a=2^x$, $b=2^y$; then ${\left(2^x\right)}^z=2^{xz}=2^y$, so $x|y$. Here we can just make a table and count the number of values of $y$ per value of $x$. The largest possible value of $x$ is 12, and we get ${\sum }_{x=1}^{12}\lfloor \frac{25}{x}-1\rfloor =24+11+7+5+4+3+2+2+1+1+1+1=62$. The total number of ways to pick two distinct numbers is $\frac{25!}{(25-2)!}=25\cdot 24=600$, so we get a probability of $\frac{62}{600}=\boxed{\frac{31}{300}}$.