First of all, I must say that this is not a problem that I have been asked to solve, but a personal thinking that has come to mind, and for which I can't find a satisfactory answer.
Here it is:
A company markets toys (there are $N$ different types) that it conceals in chocolate eggs and it is assumed that each chocolate egg purchased contains one toy.
Now suppose there are $N-1$ types of toys that are evenly distributed in the eggs (there are as many marketed toys of one kind as there are toys of another kind). On the other hand, the $N^{th}$ type of toy is considered “rare”: it exists in eggs in limited numbers.
I assume I can model this by assuming that $T$ eggs are offered for sale and that:
- each toy of type $i$ (for $1 \leq{i} \leq{N-1}$) is present $x$ times: taking an egg at random, the probability of finding a toy of type $i$, ($1 \leq{i} \leq{N-1}$) is: $\frac{x}{T}$
- the type $N$ toy is present $y$ times (with $y<x$ by assumption): taking an egg at random, the probability of finding a toy of this type is: $\frac{y}{T}$
- we have: $(N-1)x+y=T$
The problem is to estimate how many eggs I have to buy to hope to complete the collection of $N$ toys, knowing that successive purchases will almost certainly lead to accumulating toys that I already have...
Since I didn't know how to start, I thought that, to better understand the problem, I should first solve the following question: I buy an egg; as each egg contains a toy, I already have $1$ element of my collection (of any type...); I buy a second egg; what is the probability that it is the same type as the one I already have?
So, I wrote:
Event “second toy identical to first”: “$2=1$”
Event “toy $1$ of type $i \,(1<=i<=N-1)$”: “$1=i$”
Event “toy $2$ of type $i \,(1<=i<=N-1)$”: “$2=i$”
Event “toy $1$ of type $N$”: “$1=N$”
Event “toy $2$ of type $N$”: “$2=N$”
Then:
$$P(2=1)=P(1=i \land 2=i)+ P(1=N \land 2=N)=P(1=i) \times P(2=i | 1=i) + P(1=N) \times P(2=N | 1=N) = \frac{x}{T} \frac{x-1}{T-1}+\frac{y}{T} \frac{y-1}{T-1}$$
- And:
$$P(2\neq1)=1-P(2=1)$$ If now I designate by $A_1$ the random variable giving the number of purchases that I must make to find a different toy from the one I already have, then it is possible that $A_1=1$ (if from the new purchase, I find right away a different toy…), or that $A_1=n$ (if during the first $n-1$ purchases, I find a toy identical to the one I already have, and if I only find a different toy during the $n^{th}$ purchase …).
So it seems to me that $A_1$ follows a geometric law with parameter $P(2\neq1)$: $$P(A_1=n)= {P(2=1)}^{n-1} \times P(2\neq1)$$ With: $$E(A_1)= \frac{1}{ P(2\neq1)}$$ To be honest, this way of doing things seems very complicated to me and I don't see how to continue... Did I make mistakes in my writings? Do I complicate things too much? Is there an easier way to solve this collector problem? Any directions appreciated. Thank you.