Product of four random digits ending in 1, 3, 7 or 9

Four integers are chosen independently and uniformly from $1,2,\dots,9$ (repeats are allowed). What is the probability that their product ends in the digit $1$, $3$, $7$ or $9$?

$\tfrac{256}{6561}$. The product's last digit is odd and not $5$ exactly when no factor is even and no factor is $5$, i.e. every choice lies in $\{1,3,7,9\}$. Each pick does so with probability $\tfrac49$, giving $(\tfrac49)^4=\tfrac{256}{6561}$.