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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 $k$.

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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 $X_i$

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  • $\begingroup$ I think what confused me was that $p$ was variable in this case, but $p_i$ is specific to an $X_i$ and is dependent only on the position of $X_i$ in the sequence of $X$'s, which has nothing to do with an event in the experiment ({X_1=a_1,...X_{i-1}=a_{i-1}}) for which $X_i$ is conditioned on? $\endgroup$ – Yandle May 20 at 17:52
  • $\begingroup$ $p$ is not variable. It is a function of index $i$. Once you know how many different cards you got - "$i-1$", you can calculate $p_i$. $\endgroup$ – Markoff Chainz May 20 at 18:21

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