What is the expected number of throws of a die (with 6 faces numbered 1 through 6), until we see four ‘sixes’ turn up regardless of any order? We throw a die until four 'sixes' turn up regardless of any order. If 'X' represents the number of throws of the die, then what is the probability mass function for 'X'? What is the value of E(X) in this case?
 A: $\ X=X_1+X_2+X_3+X_4\ $, where $\ X_1\ $ is the number of the throw on which the first six turns up, and, for $\ 2\le n\le 4\ $, and $\ \sum_\limits{i=1}^n X_i\ $ is the number of the throw on which the $\ n^\text{th}\ $ six turns up.  The random variables $\ X_i\ $ are independent with identical PMF
$$
\text{Prob}\left(X_i=t\right) = \frac{1}{6}\left(\frac{5}{6}\right)^{t-1}\ ,
$$
so the PMF of $\ X\ $ is the fourfold convolution of this one:
\begin{align}
\text{Prob}\left(X_1+X_2=t\,\right)&=\text{Prob}\left(X_3+X_4=t\,\right)\\
&=\frac{1}{36}\sum_{i=1}^{t-1}\left(\frac{5}{6}\right)^{i-1}\left(\frac{5}{6}\right)^{t-i-1}\\
&=\frac{t-1}{36}\left(\frac{5}{6}\right)^{t-2}\ \text{for }\ t\ge 2\\
 \therefore \text{Prob}\left(X=t\,\right)&=\frac{1}{36^2}\sum_{i=2}^{t-2}(i-1)(t-i-1)\left(\frac{5}{6}\right)^{i-2}\left(\frac{5}{6}\right)^{t-i-2}\\
&=\frac{t-1\choose 3}{36^2}\left(\frac{5}{6}\right)^{t-4}\ \text{for }\ t\ge 4\ .
\end{align}
As Don Thousand has already noted, $\ E(X)=4E\left(X_i\right)\ $, where
\begin{align}
E\left(X_i\right)&=\frac{1}{6}\sum_\limits{t=1}^\infty t\left(\frac{5}{6}\right)^{t-1}\\
&=6\ .
\end{align}
Generalisation: The form of the expressions for these PMFs suggests that if $\ T_n\ $ is the number of the throw on which the $\ n^\text{th}\ $ six turns up for any $\ n\ $, then
$$
\text{Prob}\left(T_n=t\right)=\frac{t-1\choose n-1}{6^n}\left(\frac{5}{6}\right)^{t-n}\ .\,
$$
and, in fact, this identity isn't difficult to prove by induction.
