Throw Until Matched

We can define $E_n$ as being the expected number of rolls to finish after having rolled $n$ distinct outcomes. We know that $E_6 = 1$, as it will only take one more roll to get a match (whatever it is!):

• $E_6$ means you finished six rolls. The fact that you reached the seventh roll means that the first six rolls were all distinct numbers. Since a six-sided die only has six distinct outcomes, it means the seventh throw will always finish this game.

Working backwards, for $E_5$, there's a 1/6 chance of needing another roll, and a 5/6 chance that this roll finishes the game. We already know $E_6 = 1$. Let's work this out: \begin{equation} E_5 = \frac{1}{6} (1 + E_6) + \frac{5}{6} 1 \end{equation} \begin{equation} E_5 = \frac{7}{6}  \end{equation} \begin{equation} E_4 = \frac{2}{6} (1 + E_5) + \frac{4}{6} 1 \end{equation} \begin{equation} E_4 = \frac{25}{18}  \end{equation} \begin{equation} E_3 = \frac{3}{6} (1 + E_4) + \frac{3}{6} 1 \end{equation} \begin{equation} E_3 = \frac{61}{36}  \end{equation} \begin{equation} E_2 = \frac{4}{6} (1 + E_3) + \frac{2}{6} 1 \end{equation} \begin{equation} E_2 = \frac{115}{54}  \end{equation} \begin{equation} E_1 = \frac{5}{6} (1 + E_2) + \frac{1}{6} 1 \end{equation} \begin{equation} E_1 = \frac{899}{324}  \end{equation} \begin{equation} E_0 = 1 + E_1 \approx 3.775 \end{equation}