Pratt is given a fair die. He repeatedly throw the die until he get at least each number (1 to 6).
Define the random variable $X$ to be the total number of trials that pratt throws the die. or How many times he has to throw a die so that each side of the die appears at least once. Determine the expected value $E(X)$.
This is a special case of the Coupon Collector's Problem.
Throw the die once. Of course we get a "new" number. Now throw the die again, and again, until we have obtained each of $1$ to $6$ at least once.
Let random variable $X_2$ be the "waiting time" (number of additional throws) until we get a number different from the result on the first throw. Let $X_3$ be the waiting time (number of additional throws) between the time we get the second new number and the time we get the third. Let $X_4$ be the waiting time between the third new number and the fourth. Define $X_5$ and $X_6$ analogously. Then $X=1+X_2+X_3+X_4+X_5+X_6$. By the linearity of expectation we have $E(X)=1+E(X_2)+\cdots+E(X_6)$.
The various $X_i$ have geometric distribution. Consider for example $X_2$. After the first throw, the probability that a result is "new" is $\frac{5}{6}$, so $E(X_2)=\frac{6}{5}$. Similarly, $E(X_3)=\frac{6}{4}$, and so on. Now put the pieces together.
$X=1+Y_{1}+Y_{2}+Y_{3}+Y_{4}+Y_{5}$ where $Y_{i}$ stands for the number of rolls needed to arrive at a new face of the die when $i$ faces have allready shown up.
Here $Y_{i}$ has a geometric distribution with parameter $p_{i}=\frac{6-i}{6}$. Then $\mathbb{E}Y_{i}=\frac{1}{p_{i}}=\frac{6}{6-i}$ leading to $\mathbb{E}X=1+\mathbb{E}Y_{1}+\mathbb{E}Y_{2}+\mathbb{E}Y_{3}+\mathbb{E}Y_{4}+\mathbb{E}Y_{5}=\sum_{i=0}^{5}\frac{6}{6-i}=6\sum_{k=1}^{6}k^{-1}$
