The question is from Introduction to Probability by Sheldon Ross.
Suppose that there are N distinct types of coupons and that each time one obtains a coupon, it is, independently of previous selections, equally likely to be any one of the N types. One random variable of interest is T, the number of coupons that needs to be collected until one obtains a complete set of at least one of each type. Rather than derive $P(T = n)$ directly, let us start by considering the probability that T is greater than n.
I was trying to understand why deriving $P(T=n)$ is hard and if possible, to find an expression directly. Here is my attempt.
We have n spaces to fill $(n>N)$. First, I choose one element from each of the N different types of coupons and then put them in any N spaces out of n spaces. For first element, we have N choices each with probability $\frac{1}{N}$. For the second element, we have $(N-1)$ choices each with probability $\frac{1}{N}$. Continuing like this, we get
$$\binom{n}{N}\frac{N}{N}\frac{N-1}{N}…\frac{N-(N-1)}{N}$$
Next, we need to fill the remaining $(n-N)$ spaces with any type of the coupon. Suppose, we fill the remaining spaces with just one type of coupon. Then, we need to divide by $(n-N+1)!$ because it will surely match any one of the different types of coupons already placed. We can choose 1 coupon of a particular type in $\binom{N}{1}$ ways. However, this is just one case. There will be cases where the remaining spaces are filled with $(n-N-1)$ coupons of same type and 1 coupon of some other type. We can choose such combinations in $\binom{N}{2}$ ways. We also need to divide by $((n-N-1)+1)!(1+1)!$. Here, I encounter the problem of how to partition $(n-N)$ as sum of various natural numbers. I have heard of partition function, but it doesn't have a closed form (source: Wikipedia). I was thinking of some expression like
$$\binom{n}{N}\frac{N}{N}\frac{N-1}{N}…\frac{N-(N-1)}{N}\cdot(\binom{N}{1}\frac{1}{(n-N+1)!}+\binom{N}{2}\frac{1}{(n-N)!2!}+…)$$
The general term inside the second bracket would be something like
$$\binom{n}{N}\frac{N}{N}\frac{N-1}{N}…\frac{N-(N-1)}{N}\cdot(\binom{N}{1}\frac{1}{(n-N+1)!}+\binom{N}{2}\frac{1}{(n-N)!2!}+…+\binom{N}{a}\frac{1}{(x+1)!(y+1)!(z+1)!…})$$
where $$x+y+z…=n-N$$ and $a$ is the number of natural numbers into which $(n-N)$ has been partitioned
I understand why the approach is difficult and seems to lead nowhere, but I would appreciate if someone could verify whether my line of reasoning is correct and if possible, to provide a form for directly calculating $P(T=n)$. I understand the other way approaching it, as evident in the last line of the quote, but I wanted to do it in a different way.
I am taking an Introductory course on Probability currently. This is my first question on stackexchange. I have tried my best to explain my line of reasoning but if you feel that the question is vague, please let me know.
Best Answer
We'll use the inclusion-exclusion principle to find $P(T\neq n)$. Label the cards $C_1,\ldots, C_N$ and let $E_i(n)$ be the event that $C_i$ has not been picked after $n$ selections. Hence, we have \begin{equation}\label{eq} P(T\neq n)=P\left(\bigcup_{i=1}^NE_i(n)\right)\end{equation} Now for distinct $i,j,k\ldots$ we have \begin{align*} P\left(E_{i}(n)\right)&=\frac{(N-1)^n}{N^n}\\ P\left(E_{i}(n)\cap E_j(n)\right)&=\frac{(N-2)^n}{N^n}\\ P\left(E_{i}(n)\cap E_j(n)\cap E_k(n)\right)&=\frac{(N-3)^n}{N^n}\\ &\ldots \end{align*} Hence, the inclusion-exclusion principle yields \begin{align*} P(T\neq n)&=P\left(\bigcup_{i=1}^NE_i(n)\right)\\ &=\sum_{1\leq i<N}P(E_i(n))-\sum_{1\leq i<j\leq N}P(E_i(n)\cap E_j(n))+\ldots\\ &=\sum_{r=1}^N(-1)^{r-1}\binom{N}{r}\frac{(N-r)^n}{N^n}\\ &=\frac{(-1)^{N-1}}{N^n}\sum_{r=0}^{N-1}(-1)^{r}\binom{N}{r}r^n\\ &=1+\frac{(-1)^{N-1}}{N^n}\sum_{r=0}^{N}(-1)^{r}\binom{N}{r}r^n \end{align*} where the last step uses the symmetry of binomial coefficients. I don't think this expression has a closed form (although I don't know for sure). Finally, we obtain $$P(T=n)=1-P(T\neq n)=\frac{(-1)^{N}}{N^n}\sum_{r=0}^{N}(-1)^{r}\binom{N}{r}r^n$$ It is well known that the above summation equals $0$ for $0\leq n<N$, and here we see that as a straightforward corollary of the above equation.