Berkeley Math Circle: Monthly Contest 1

Solution:

The number of rolls is always at least $2$ and at most $7$. For there to be at least $k$ rolls, the first $k-1$ rolls have to be distinct and in decreasing order. The number of ways this can happen is $\binom{6}{k-1}$, and the total number of ways to have $k-1$ rolls is $6^{k-1}$. Thus, the expected value is

$$\begin{aligned} E(\# \text{ of rolls }) & = P(\geq 1 \text{ roll }) + P(\geq 2 \text{ rolls }) + \cdots + P(\geq 7 \text{ rolls }) \\ & = 1 + \frac{\binom{6}{1}}{6} + \frac{\binom{6}{2}}{6^2} + \frac{\binom{6}{3}}{6^3} + \frac{\binom{6}{4}}{6^4} + \frac{\binom{6}{5}}{6^5} + \frac{\binom{6}{6}}{6^6} . \end{aligned}$$

By the binomial theorem, this is $\left(1+\frac{1}{6}\right)^6 \approx 2.52$.