6. Given the function
$$
f(x)=\frac{1}{3} x^{3}-(a-1) x^{2}+b^{2} x,
$$

where, $a \in\{1,2,3,4\}, b \in\{1,2,3\}$.
Then the probability that the function $f(x)$ is increasing on $\mathbf{R}$ is
6. $\frac{3}{4}$.

Since $f(x)$ is an increasing function on $\mathbf{R}$, we have
$$
f^{\prime}(x)=x^{2}-2(a-1) x+b^{2} \geqslant 0
$$

always holds, which means
$$
\Delta=4(a-1)^{2}-4 b^{2} \leqslant 0 .
$$

Solving this, we get $|a-1| \leqslant|b|$.
$$
\begin{array}{l}
\text { Therefore, }(a, b) \\
=(1,1),(2,1),(1,2),(2,2),(3,2), \\
\quad(1,3),(2,3),(3,3),(4,3),
\end{array}
$$

There are 9 such pairs.
Also, the total number of possible pairs $(a, b)$ is $4 \times 3=12$, so the probability that $f(x)$ is an increasing function on $\mathbf{R}$ is $\frac{9}{12}=\frac{3}{4}$.