AMC 12 Mock Exam 4 Problem 25

Let $N$ be the number of rolls. For $k\ge0$, $N>k$ exactly when the first $k$ rolls form a nondecreasing sequence (for $k=0,1$ this is automatic). A nondecreasing sequence of length $k$ from $\{1,\dots,6\}$ is determined by how many times each face appears, so there are $\binom{k+5}{5}$ of them (stars and bars), and $P(N>k)=\binom{k+5}{5}6^{-k}$.

Since $N$ is a positive integer, $N=\sum_{k\ge0}[N>k]$, so $E[N]=\sum_{k\ge0}P(N>k)=\sum_{k\ge0}\binom{k+5}{5}\left(\tfrac16\right)^k$.

To evaluate this sum, let $x=\frac16$ and multiply six copies of the geometric series $1+x+x^2+\cdots=\frac{1}{1-x}=\frac65$: $$\left(\tfrac65\right)^6=\prod_{j=1}^{6}\left(\sum_{a_j\ge0}x^{a_j}\right)=\sum_{a_1,\dots,a_6\ge0}x^{a_1+\cdots+a_6}.$$ (All terms are positive, so expanding and regrouping is legitimate.) Grouping the terms by $k=a_1+\cdots+a_6$, the number of terms with a given $k$ is the number of ways to write $k$ as an ordered sum of six nonnegative integers, which is $\binom{k+5}{5}$ by stars and bars. Hence $\sum_{k\ge0}\binom{k+5}5x^k=\left(\frac65\right)^6$, and $$E[N]=\boxed{\left(\tfrac65\right)^6}=\frac{46656}{15625}\approx2.99.$$