Solution:

Let $a_n$ be the expected number of rolls starting with an $n$-sided die. We see immediately that $a_1 = 0$, and $a_n = 1 + \frac{1}{n} \sum_{i=1}^{n} a_i$ for $n > 1$. Thus $a_2 = 2$, and for $n \geq 3$, $a_n = 1 + \frac{1}{n} a_n + \frac{n-1}{n}(a_{n-1} - 1)$, or $a_n = a_{n-1} + \frac{1}{n-1}$. Thus $a_n = 1 + \sum_{i=1}^{n-1} \frac{1}{i}$ for $n \geq 2$, so $a_6 = 1 + \frac{60 + 30 + 20 + 15 + 12}{60} = \frac{197}{60}$.