Probability that each number obtained by throwing a die is no smaller than the preceding number 
A fair die is thrown 4 times. Find the probability that the each number obtained is no smaller than the preceding number.



*

*If all numbers obtained are same, number of such outcomes $$=\dbinom{6}{1}=6$$

*If 3 numbers obtained are same, number of ways is equal  to the number of ways of choosing any two numbers $=\dbinom{6}{2}$. But these two number can be arranged as $x,y,y,y$ or $x,x,x,y$ where $x<y$. So $$2\times\dbinom{6}{2}$$

*If 2 numbers are same, the cases are $a_1<a_2<a_3=a_4$, $a_1<a_2=a_3<a_4$, $a_1=a_2<a_3<a_4$, $a_1=a_2<a_3=a_4$. Number of ways $$=3\times\dbinom{6}{3}+\dbinom{6}{2}$$

*if all 4 numbers are different, $$\dbinom{6}{4}$$


Total number of possibilities $=6^4$.
Even though I got the correct answer, is there any shorter method?
 A: 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 $Pr = \dfrac{\binom95}{6^4}$
A: 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.
$$\begin{array}{|l:l|} \hline
\rm a{=}a{=}a{=}a & \dbinom 3 0 \dbinom 6 1
\\\hdashline\rm a{=}a{=}a{>}b , a{=}a{>}b{=}b, a{>}b{=}b{=}b & \dbinom 3 1\dbinom 6 2
\\\hdashline\rm a{=}a{>}b{>}c, a{>}b{=}b{>}c, a{>}b{>}c{=}c & \dbinom 3 2 \dbinom 6 3
\\\hdashline\rm a{>}b{>}c{>}d & \dbinom 3 3 \dbinom 6 4
\\ \hline\end{array}$$
$$\dfrac{\sum_{n=1}^4 \dbinom{3}{n-1}\dbinom{6}{n}}{6^4} = \dfrac{\dbinom 6 1 + 3\dbinom 6 2 + 3 \dbinom 6 3 + \dbinom 6 4}{6^4}$$
A: 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 $$|A|=|B|={9 \choose 4}.$$
A: If a sequence of outcomes is non-decreasing, then it is completely characterized by the number of times each outcome occurs.  Let $x_k$ be the number of times the outcome $k$ appears in the four throws, where $1 \leq k \leq 6$.  Then
$$x_1 + x_2 + x_3 + x_4 + x_4 + x_6 = 4$$
which is an equation in the non-negative integers.  A particular solution in the non-negative integers corresponds to placing five addition signs in a row of four ones.  For instance,
$$+ + 1 + 1 1 + + 1$$
corresponds to the sequence of throws $(3, 4, 4, 6)$, while 
$$1 + 1 + + 1 + 1 +$$
corresponds to the sequence of throws $(1, 2, 4, 5)$.  The number of such solutions is the number of ways five addition signs can be inserted into a row of four ones, which is 
$$\binom{4 + 5}{5} = \binom{9}{5}$$ 
since we must select which five of the nine symbols (five addition signs and four ones) will be addition signs.  
Since the total number of possible sequences is $6^4$, the probability that the sequence of outcomes is non-decreasing when a fair die is thrown four times is $$\frac{\binom{9}{5}}{6^4}$$   
A: Let us count the number of ways in which this is possible.
Suppose between each dice throw, the number that appears on the die increases by a certain amount. Note that since we are only counting the number of favourable outcomes, this is a valid assumption. There are $3$ places between the dice throws where this increase can occur. But the initial throw can also produce a number greater than one, and the last can also produce a number less than six. So, we add $2$ more places for increase: before the first throw and after the last throw. We get the following diagram:
(Start at $1$)
Increase $\#1$
Die roll $\#1$
Increase $\#2$
Die roll $\#2$
Increase $\#3$
Die roll $\#3$
Increase $\#4$
Die roll $\#4$
Increase $\#5$
(End at $6$)
There are $5$ spots for increase, and $5$ units of increase from $1$ to $6$.
So, the answer is equal to the no. of ways of distributing $5$ similar objects into $5$ distinct boxes.
For this problem, the formula is $\binom{5+5-1}{5-1} = \binom94 = 126$
So no. of favourable outcomes is $126$.
Total possible outcomes are $6^4 = 1296$
$$\frac{126}{1296}=\frac7{72}$$
Credit: Keyur Joshi (on Quora)
