Singletons in coupon collecting problem There are $n$ types of coupons. All types are equally likely to turn up and each "draw" of a coupon is independent of others. If someone collects coupons until they have a complete set of all the $n$ types, what is the expected value of the number of coupons that only appear once in this complete set?

In the book, they give this solution:

Let $X$ be the number of singletons. Let $T_i$ be the $i$th type of coupon collected and $A_i$ the event that there is only one $T_i$ coupon in the set. Then $$\mathbb{E} [X]=\sum_{i=1}^n \mathbb{P}(A_i)$$
This much I understand. What I do not understand is the following:

Now, at the moment when the first type $T_i$ coupon is collected, there remain $n − i$
types that need to be collected to have a complete set. Because, starting at this moment,
each of these $n − i + 1$ types (the $n − i$ not yet collected and type $T_i$) is equally
likely to be the last of these types to be collected, it follows that the type $T_i$ will be the
last of these types (and so will be a singleton) with probability $\frac {1}{
n−i+1}$.

I do not understand this derivation of the probability. If $T_i$ just got collected, how is it that it can be collected last of the $n-i$ not yet collected types? 

They are looking for the chance that you get another $T_i$.  You can collect it again until you have found all the other types you are looking for.  After you get the first $T_i$, they say make a list of the first occurrence after now of $T_i$ and all the coupons you have not found yet.  If $T_i$ is the last, you will only have one $T_i$ when you complete your set.  If it is not the last, you will have a duplicate $T_i$ when you complete your set.  
As a specific example, say you have all the coupons except $a,b,c$.  Now you draw your first $a$ (this is $T_i$ in the above).  They say you should look at the order of the next draw of $a,b,c$.  There are $3!$ possible orders, of which $2$ have $a$ after the others, so you now have $\frac 23$ chance of getting a second $a$ before you finish the set and a $\frac 13$ chance you get your complete set while you have only one $a$.
