Solution:
$\frac{23}{144}$
To read "HMMT," there are $\binom{8}{4}$ ways to place the letters, and $\frac{4!}{2}$ ways to place the numbers. Similarly, there are $\binom{8}{4} \frac{4!}{2}$ arrangements where one can read "2005." The number of arrangements in which one can read both is just $\binom{8}{4}$. The total number of arrangements is $\frac{8!}{4}$, thus the answer is
$$\frac{\binom{8}{4} \frac{4!}{2} + \binom{8}{4} \frac{4!}{2} - \binom{8}{4}}{\frac{8!}{4}} = \binom{8}{4} \frac{4}{8!} \cdot 23 = \frac{23}{144}.$$