<s>[INST] How many bags would need to be bought to have all the red noses? There are 9 types of Red Noses, for comic relief this year:

Each one is sold in an opaque packet, so it is lucky dip which one you will get.
Assuming there is the same amount of each type (${1\over9}$th). On average, how many bags would need to be bought for someone to get all 9 of them?
 [/INST] 
I'm going assume that there are, effectively, infinite bags of noses—if we only had 9 bags... well, you get it. ;)
Say we have $n-1$ of the $9$ unique noses. The probability of getting a new style of nose on the next bag is $p_n = 1 - \frac{n-1}{9}$. Then, on average, it takes $\frac{1}{p_n}$ bags to get the next unique nose. (i.e. - Say we have $0$ unique noses. Then, $n = 1$ and  $p_n = 1 - \frac{0}{1} = 1$ and it will take only one bag to get a new nose. Similarly, if we have $4$ nose-styles, $n = 5$ and it will take $1 - \frac{4}{9} = \frac{9}{5}$ bags to get a $5^{\text{th}}$ nose.)
The probabilities here are independant. That is, the chance of getting the $5^{\text{th}}$ new nose on the next bag doesn't depend on the chance of getting the $4^{\text{th}}$ one (once you have the $4^{\text{th}}$ nose, the probability of getting the $5^{\text{th}}$ will always be $\frac{1}{p_5}$). Thus, we can just sum the average of each $p_n$ until $n = 9$.
So, we have: 
$$\frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} + \frac{1}{p_4} + \frac{1}{p_5} + \frac{1}{p_6} + \frac{1}{p_7} + \frac{1}{p_8} + \frac{1}{p_9} + = \\ 
1 + \frac{9}{8} + \frac{9}{7} + \frac{9}{6} + \frac{9}{5} + \frac{9}{4} + \frac{9}{3} + \frac{9}{2} + \frac{9}{1} = \\
9 \sum_{n=1}^{9} \frac{1}{n} = \frac{9 \times 7129}{2520} = 25.46$$
That is, on average, it'd take about 25 bags to get all nine noses.
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