Probability that each number obtained by throwing a dice is no smaller than the preceeding number

Since you are wanting a shortcut method,
count the number of ways $4$ balls can be placed in $6$ bins marked $1-6$, using stars and bars

Note that each of the $\binom{4+6-1}{6-1}$ results thus obtained can yield only one non-decreasing sequence.

A result of $\;\;\fbox{2}\fbox{0}\fbox{0}\fbox{1}\fbox{0}\fbox{1}\;$, e.g. means obtaining $1-1-4-6$ in sequence.

Thus $Pr = \dfrac{\binom95}{6^4}$

Nope, that seems likely to be the most concise way to do it.

Count ways to pick $n\in\{1,2,3,4\}$ unique numbers, and to arrange them with $n-1$ "$>$" signs, to meet the criteria.

$$\begin{array}{|l:l|} \hline \rm a{=}a{=}a{=}a & \dbinom 3 0 \dbinom 6 1 \\\hdashline\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \dbinom 3 1\dbinom 6 2 \\\hdashline\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \dbinom 3 2 \dbinom 6 3 \\\hdashline\rm a{>}b{>}c{>}d & \dbinom 3 3 \dbinom 6 4 \\ \hline\end{array}$$

$$\dfrac{\sum_{n=1}^4 \dbinom{3}{n-1}\dbinom{6}{n}}{6^4} = \dfrac{\dbinom 6 1 + 3\dbinom 6 2 + 3 \dbinom 6 3 + \dbinom 6 4}{6^4}$$

The problem reduces to finding the cardinality of the following set $$A=\{(a_1,a_2,a_3,a_4)\in\{1,\ldots,6\}^4: \ a_1\leq a_2\leq a_3 \leq a_4\}.$$ Define $$B=\{(a_1,a_2,a_3,a_4)\}\in\{1,\ldots 9\}^4: \ a_1 < a_2 < a_3 < a_4\}$$ and $$f:A\to B, \quad f(a_1,a_2,a_3,a_4) = (a_1,a_2+1,a_3+2,a_4+3).$$ Since $f$ is a bijection we get that $$|A|=|B|={9 \choose 4}.$$