Last digit of a random power equal to one

Choose $m$ uniformly from $\{111,133,155,177,199\}$ and, independently, $n$ uniformly from the twenty integers $2004,2005,\dots,2023$. Find the probability that the last digit of $m^n$ is $1$.

$\tfrac25$. Only last digits matter. Powers of $1$ always end in $1$ (20 values of $n$); $3^n$ and $7^n$ end in $1$ when $4\mid n$ (5 values each); $9^n$ does when $n$ is even (10 values); $5^n$ never does. That is $40$ of the $100$ equally likely pairs.