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

  1. If all numbers obtained are same, number of such outcomes $$=\dbinom{6}{1}=6$$
  2. 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 $$2\times\dbinom{6}{2}$$
  3. 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 $$=3\times\dbinom{6}{3}+\dbinom{6}{2}$$
  4. 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?

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up vote 5 down vote accepted

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

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Great solution, Anil. Wish I could give more than +1. – Shailesh Jan 22 at 6:08
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@Shailesh: Thanks for your kind comments. – true blue anil Jan 22 at 6:12
    
@trueblueanil can you explain how 200101 became 1146? – Aditya Dev Jan 22 at 7:19
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@AdityaDev These are boxes numbered 1...6, so 2 in box 1 represents choosing 1 twice, i.e. 1,1 and 1 in box 4 represents choosing 4 and 1 in box 6 represents choosing the number 6. Its truly a very nice solution. – Shailesh Jan 22 at 7:22
    
@AdityaDev: I have added a line in the answer that should also help. – true blue anil Jan 22 at 8:33

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.

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

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

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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 $$|A|=|B|={9 \choose 4}.$$

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@Shailesh This method actually does give the correct answer ($126/6^4$); If we change the variables in set $B$ to $b_1,b_2\ldots$ (to make things clearer), we have the condition $b_1<b_2<b_3<b_4$. Then, since we have $b_1=a_1,b_2=a_2+1\ldots$, that condition becomes $a_1<a_2+1<a_3+2<a_4+3$, which, since the $a_i$ are integers, is equivalent to the original $a_1\le a_2\le a_3\le a_4$. – 2012rcampion Jan 22 at 7:18
    
@2012rcampion I withdraw my comment – Shailesh Jan 22 at 7:20

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 $$\frac{\binom{9}{5}}{6^4}$$

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