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Suppose you roll a fair 6-sided die four times.

Let C be the event that the value of each roll is at least as large as the value of the previous roll.

What is the probability of C?

I know that $$\omega = 6^4 = 1296$$

I also know that to get P(C), I need to divide C by $\omega$

But what is the fastest/simpler way to get C?

I could write all the sets that would qualify but that would

A: be time consuming.

B: error prone.

Is there a formula to find C in this case?

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2 Answers

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A non-decreasing sequence is completely determined by the number of times each outcome occurs. Hence, the number of non-decreasing sequences of four rolls is the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 4 \tag{1}$$ in the non-negative integers. A particular solution of equation corresponds to the placement of five addition signs in a row of four ones. For instance, $$+ + 1 1 + + 1 + 1$$ corresponds to $x_1 = x_2 = 0$, $x_3 = 2$, $x_4 = 0$, $x_5 = x_6 = 1$ (that is, to the sequence of rolls $3, 3, 5, 6$). The number of such solutions is the number of ways five addition signs can be placed in a row of four ones, which is $$\binom{4 + 5}{5} = \binom{9}{5}$$ since we must choose which five of the nine symbols (four ones and five addition signs) will be addition signs.

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Finding $|C|$ is equivalent to the problem that finding the number of functions from $X=\{1,2,3,4\}$ to $Y=\{1,2,3,4,5,6\}$ and $f(1)\le f(2)\le f(3)\le f(4)$. Then, just select elements of $Y$ with repetition. Since $f$ monotonely increases, $f(1),f(2),f(3),f(4)$ are determined automatically. Thus $$ |C|=\left(\!\binom{6}{4}\!\right) = \binom{6+4-1}{4}=\binom{9}{4}. $$

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