Step 1: Sort the numbers by remainder when divided by 3.
Each number from $1$ to $50$ leaves remainder $0$, $1$, or $2$ on division by 3. The sum $m + n$ is a multiple of 3 only in these cases: both remainders are $0$, or one is $1$ and the other is $2$.
Step 2: Count how many numbers fall in each remainder group.
Remainder $0$ (the multiples of 3: $3, 6, \dots, 48$): there are $16$ of them. Remainder $1$ ($1, 4, \dots, 49$): $17$ of them. Remainder $2$ ($2, 5, \dots, 50$): $17$ of them. Total $16 + 17 + 17 = 50$.
Step 3: Turn counts into probabilities.
$P(\text{rem } 0) = \frac{16}{50}$, $P(\text{rem } 1) = \frac{17}{50}$, $P(\text{rem } 2) = \frac{17}{50}$.
Step 4: Probability both have remainder 0.
Since the picks are independent, this is $\frac{16}{50} \cdot \frac{16}{50} = \frac{256}{2500}$.
Step 5: Probability remainders are 1 and 2 in either order.
This is $2 \cdot \frac{17}{50} \cdot \frac{17}{50} = \frac{578}{2500}$.
Step 6: Add and simplify.
Total favourable probability $= \frac{256 + 578}{2500} = \frac{834}{2500} = 0.3336$.
\[ \boxed{0.3336} \]