Last coupon collected in the coupon collectors problem Consider the classical coupon collectors problem. Given a particular coupon $i$ we can ask for the probability that $i$ is the last coupon collected. We asked this question on cstheory and got a wonderful answer from James Martin introducing us to the idea of Poissonization in coupon collecting:
$Pr(L=i) = \int_{0}^{\infty} n_i e^{n_i t} \prod_{j\neq i} (1-e^{n_j t}) dt$
where $L$ is the last coupon collected and $n_i$ is the number of coupons of type $i$, with $\sum_i n_i = k$. 
For example:
$\int_0^\infty a e^{-a t} \left(1-e^{-b t}\right) \left(1-e^{-c t}\right) \left(1-e^{-d t}\right) \left(1-e^{-e t}\right) \left(1-e^{-f t}\right) \, dt$,
gives the following in Mathematica:
$a \left[ \frac{1}{a}-\frac{1}{a+b}-\frac{1}{a+c}+\frac{1}{a+b+c}-\frac{1}{a+d}+\frac{1}{a+b+d}+\frac{1}{a+c+d}-\frac{1}{a+b+c+d}- \right.$
$\left. \frac{1}{a+e}+\frac{1}{a+b+e}+\frac{1}{a+c+e}-\frac{1}{a+b+c+e}+\frac{1}{a+d+e}-\frac{1}{a+b+d+e}-\frac{1}{a+c+d+e}+ \right.$
$\left. \frac{1}{a+b+c+d+e}-\frac{1}{a+f}+\frac{1}{a+b+f}+\frac{1}{a+c+f}-\frac{1}{a+b+c+f}+\frac{1}{a+d+f}-\frac{1}{a+b+d+f}- \right.$
$\left. \frac{1}{a+c+d+f}+\frac{1}{a+b+c+d+f}+\frac{1}{a+e+f}-\frac{1}{a+b+e+f}-\frac{1}{a+c+e+f}+\frac{1}{a+b+c+e+f}- \right.$
$\left. \frac{1}{a+d+e+f}+\frac{1}{a+b+d+e+f}+\frac{1}{a+c+d+e+f}-\frac{1}{a+b+c+d+e+f} \right]$.
The trouble is that this expression takes exponential time in the number of coupons to evaluate and has an exponential number of terms. We are looking for a way to bound this asymptotically in $k$, but any further references or information would be greatly appreciated.

I have no idea if this bound will be of any use to you or not (or even how it compares to the inclusion-exclusion bound already posted), but you can get a quick(er) upper bound as follows.  Without loss of generality let's assume the coupon we care about is coupon $k$ and that $n_1 \leq n_2 \leq \dots \leq n_{k-1}$.    We can view the event that coupon $k$ is last as the intersection of $k-1$ different events.  
$E_1$: The first coupon selected is not of type $k$.
$E_2$: The first coupon selected which does not match the first type of coupon drawn is not of type $k$.
...
$E_{n-1}$ the first coupon drawn does not match any of the first $n-2$ types of coupons drawn is not of type $k$.  
The probability of $E_1$ is just $(1-\frac{n_k}{n})$.  $P(E_2)$ is more complicated, but we know an expression for the probability of $E_2$ given that the first coupon drawn is of type $j$: $(1-\frac{n_k}{n-n_j})$, which is largest for $j=1$.  It follows by averaging over all $j$ that 
$$P(E_2 | E_1) \leq 1-\frac{n_k}{n-n_1}$$
In general, we have 
$$P(E_j| E_1, E_2, \dots, E_{j-1}) \leq 1-\frac{n_k}{n-n_1-n_2-\dots-n_{j-1}}$$
And you can multiply this bound over all $1 \leq j \leq n-1$ to get a bound on the probability that $k$ is last:
$$P(k \textrm{ last }) \leq \prod_{i=1}^{j-1} \left(1-\frac{n_k}{n-S_{k-1}}\right)$$
where $S_k=n_1+\dots+n_k$.  
This bound is exact when all of the $n_j$ are equal, but in general it gets worse the further apart the $n_j$ are.  
