Probability & Statistics: Random variables I have a problem similar to the well-known "Coupon Collector Problem." 
A box of a certain brand of cereal comes with a special toy. There are 10 different toys in all.  How many packs you will need to buy until you find the first toy already acquired in previous purchases?
 A: IMHO, this is not possible.
Suppose you get the toy of type $T_1$ in the $1^{st}$ box.
Then, 


*

*The probability of getting the toy of type $T_1$ again in $2^{nd}$ box is $\frac{9}{10}\frac{1}{10}$

*The probability of getting the toy of type $T_1$ again in $3^{rd}$ box (and failing in previous attempts) is $\frac{9}{10}\frac{9}{10}\frac{1}{10} = {(\frac{9}{10}})^2\frac{1}{10}$  

*The probability of getting the toy of type $T_1$ again in $4^{th}$ box  (and failing in previous attempts)  is ${(\frac{9}{10}})^3\frac{1}{10}$

*The probability of getting the toy of type $T_1$ again in $n^{th}$ box  (and failing in previous attempts) is ${(\frac{9}{10}})^{n-1}\frac{1}{10}$
If you succeed after picking $n$ boxes, then :
${(\frac{9}{10}})^{n-1}\frac{1}{10} = 1$
Just take $\log$ of both the sides and convince yourself that you get a non-realistic value of $n$.
A: The probability of not drawing a duplicate before and drawing a duplicate in the $n^\text{th}$ box would be
$$
\underbrace{\left(1-\frac0{10}\right)\dots\left(1-\frac{n-2}{10}\right)}_{n-1\text{ factors}}\frac{n-1}{10}
=\frac{10!(n-1)}{10^n(11-n)!}
$$
Thus, the expected box would be
$$
\sum_{n=1}^{11}n\frac{10!(n-1)}{10^n(11-n)!}=\frac{7281587}{1562500}=4.66021568
$$

A Note on the Distribution
$$
\frac{10!(n-1)}{10^n(11-n)!}
=\frac{10!}{10^{10}}\left(\frac{10^{11-n}}{(11-n)!}-\frac{10^{10-n}}{(10-n)!}\right)
$$
which shows that
$$
\begin{align}
&\sum_{n=1}^{11}\frac{10!(n-1)}{10^n(11-n)!}\\
&=\sum_{n=1}^{11}\frac{10!}{10^{10}}\left(\frac{10^{11-n}}{(11-n)!}-\frac{10^{10-n}}{(10-n)!}\right)\\
&=\frac{10!}{10^{10}}\left(\frac{10^{10}}{10!}-\frac{10^{-1}}{(-1)!}\right)\\[9pt]
&=1
\end{align}
$$
Which verifies that the expected value of $1$ is $1$ (the distribution is a valid probability distribution).
A: I see this a simple geometric distribution (Special case negative binomial distribution)
In the first cereal box, you will find a toy, let's call this $T_{1}$. $T_1$ is one random toy out of the 10 possible toy. Once the first toy is selected, the probability of getting the same toy again is $P(T_{1})=\dfrac{1}{10}
 $.
Let's call getting the event of getting the same toy again success. The probability of success in 0.1. Let X be the no. of trials/attempts before success.
The E(X), which is the expected value is the average number of trials before success is achieved. The formula for the expected value of a geometric distribution is $E(X)=\dfrac{(1-p)}{p}$ where $p$ is the probability of success.
In this case, $p=0.1$,
$$E(X)=\dfrac{0.9}{0.1}=9$$
So, on average it will take 9 attempts before success is achieved. Further distribution can be estimated using the Tschebycheff's inequality. 
Answer open to criticism.
