A conditional probability problem on coupon collection Suppose that there are $n$ types of coupons, and that the type of each new coupon obtained is independent of past selections and is equally likely to be any of $n$ types. Suppose one continues collecting until a complete set of at least one of each type is obtained. 
(a) Find the probability that there is exactly one type $i$ coupon in the final collection. 
Hint : Condition on T, the number of types that are collected before the first type $i$ appears. 
Now, because of each new coupon obtained is independent of past selections and is equally likely to be any of $n$ types, I feel that $T$ is uniformly distributed between $0$ and $n-1$. But, I am not able to prove it.
 A: What is the probability that $T=0$? The first coupon must be $i$, hence it's $1/n$. What about $T=1$? Then we can get any positive number of any single other type, of which there are $n-1$, followed by coupon $i$. $(n-1)/n\times (1+1/n+1/n^2+\cdots)\times 1/n=(n-1)/n\times 1/(1-1/n)\times 1/n=1/n$. That's a good sign...
So can we come up with a simple argument along the lines you suggest? Well, let $X_j$ be the first time a $j$ is collected. Then all the $X_j$ will end up in some particular order (there is 0 probability they are infinity, so we don't bother with that). We want to know where $X_i$ falls. But note that the probability of any one specific order like $4,8,2,1,5,6,3,7$ is equal to that of any other order since we can just relabel all the coupons and nothing can change; there is nothing special about any one labelling. Therefore, the order is a uniform random permutation of $1,\ldots,n$. Then it follows that the location of $X_i$ is uniformly distributed in the order as you require.
I'll leave you to think about how to use this for the rest of the question.
A: I'll write out a more complete answer for future reference:
By symmetry, type $i$ is equally likely to occur as the $k$-th last type for any $k$ up to $n$. There will then be a single instance of it in the end if it is not drawn again before all of the remaining $k-1$ types have been drawn. Again by symmetry, the probability for type $i$ to be drawn last of the $k$ types is $\frac 1k$. Thus the total probability to end up with one instance of type $i$ is
$$
\frac1n\sum_{k=1}^n\frac1k=\frac{H_n}n\sim\frac{\log n}n\;,
$$
where $H_n$ is the $n$-th harmonic number.
