Assume that, every time you buy a box of Wheaties, you receive a picture of one of the $n$ baseball player. Let $X_k$ be the number of additional boxes you have to buy, after you have obtained $k-1$ different pictures, in order to obtain the next new picture. Thus $X_1 = 1$, $X_2$ is the number of boxes bought after this to obtain a picture different from the first pictured obtained, and so forth.
If I want to find the variance for the number of boxes before getting half of the players' pictures (assume there are $2n$ players), my book states the following:
$$p_{X_k} = \frac{2n-k+1}{2n}$$
Since this is a geometric distribution, $V(X)=\frac{1-p}{p^2}$, so
$$V(X_k) = \frac{2n(k-1)}{(2n-k+1)^2}$$
So variance for the total number of boxes before getting the first half of the players' pictures is:
$$\sum_{k=1}^{13}\frac{26(k-1)}{(26-k+1)^2}$$
For expected value I get that regardless of whether $X_1,X_2..,X_k$ are independent $E(X_1+X_2+..+X_k)=E(X_1)+E(X_2)...+E(X_k)$ but I don't think this is true for variance. Based on the above solution it seems like that the number of boxes to get the kth players' picture (random variables $X_1,X_2..X_k$) mutually independent? If so, is there a way to show this? This isn't intuitively clear to me, since $p_k$ is variable and dependent on $k$.
Variables $X_1,X_2,...X_k$ are indeed mutually independent.
Of course $X_1$ = $1$, lets say you received Lebrons picture.
Imagine that $X_2$ = $1000$. So it means that you bought $999$ cards and for all of them you received Lebrons picture. Only on try $1000$ you received a new picture.
Does it somehow affect the $X_3$? the fact that you had to buy $1001$ boxes to obtain two different pictures doesnt change the probability of obtaining new card. All it matters for $X_3$ is that you have only two different pictures. So is for any $X_i$