Packs of gum probability On each gum pack there's a prize. There's $6$ different prizes and each gum pack has the same probability to have each prize. Johnny buys a gum pack each day to collect all the different prizes and only then he will stop. 
I need to calculate the standard deviation of the number of days that Johnny buys packs of gum.
SO I see it's Geometric distribution and I think I need to calculate the variance first and from that to take a square root to calculate the std. So, to calculate the variance I need the probability to get a prize. So I need to calculate the variance for each prize? First, $1/6$, then $1/5$ etc, and then to sum the variances up?
 A: You have 6 independent Geomtric rvs: $X_1 \sim G(1), X_2 \sim G(\frac{5}{6}), \ldots X_6 \sim G(\frac{1}{6})$, so you can get the variance of time in Coupon Collector's problem directly:
$$
Var(T) = \sum_{k=1}^{6}Var(X_k)
$$
EDIT: you are counting the number of failures until the first success, and the probability of success is known and defined as the number of 'untouched' types of gums/total types. The events of sampling gums are independent, so $Cov(X,Y)=0$. Now you can use the definition of variance of Geometric rv, sum them and take the square root to get standard deviation
A: Let $\mathbf{p}_n=(p_1,p_2,p_3,p_4,p_5,p_6)^T$ and $p_i$ be the probability of having $i$ distinct prizes after opening $n$ gum packs. $\mathbf{p}_1=(1,0,0,0,0,0)^T$ and $\mathbf{p}_n=A\mathbf{p}_{n-1}$ where
$$A=\begin{pmatrix}
\frac{1}{6} && \frac{5}{6} && 0 && 0 && 0 && 0\\
0 && \frac{2}{6} && \frac{4}{6} && 0 && 0 && 0\\
0 && 0 && \frac{3}{6} && \frac{3}{6} && 0 && 0\\
0 && 0 && 0 && \frac{4}{6} && \frac{2}{6} && 0\\
0 && 0 && 0 && 0 && \frac{5}{6} && \frac{1}{6}\\
0 && 0 && 0 && 0 && 0 && 1
\end{pmatrix}^T$$
The probability of getting the full of $6$ distinct prizes after opening exactly $n$ gum packs is $\frac16(0,0,0,0,1,0)\mathbf{p}_{n-1}$ as Johnny have to have exactly $5$ distinct prizes after opening $n-1$ gum packs and the $6$th will be different with the probability of $\frac16$.
By computing diagonalization of $A=SDS^{-1}$ we will be able to find $p_5(n-1)$ explicitly, as $\mathbf{p}_{n-1}=A^{n-2}\mathbf{p}_{1}$ and $A^{n-2}=SD^{n-2}S^{-1}$
Diagonalization of $A^T$:
$$S=\begin{pmatrix}
1 && 5 && 10 && 10 && 5 && 1\\
0 && 1 && 4 && 6 && 4 && 1\\
0 && 0 && 1 && 3 && 3 && 1\\
0 && 0 && 0 && 1 && 2 && 1\\
0 && 0 && 0 && 0 && 1 && 1\\
0 && 0 && 0 && 0 && 0 && 1
\end{pmatrix}
$$
$$D=\operatorname{diag}\left(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6},1\right)$$
So $p_5(n-1)=15\cdot2^{3 - n} - 20\cdot3^{2 - n} - 5\cdot2^n\cdot3^{2 - n} + 5\cdot6^{2 - n} + 5^{-1 + n}\cdot6^{2 - n}$
Performing the explicit summations $M[X]=\frac16\sum\limits_{n=2}^\infty n\cdot p_5(n-1)$$=\frac{147}{10}$ and 
$M[X^2]=\frac16\sum\limits_{n=2}^\infty n^2\cdot p_5(n-1)$$=\frac{6377}{25}$ and thus the variance is $\sigma^2=D[X]=M[X^2]-\left(M[X]\right)^2=\frac{3899}{100}$
