Coupon collector's problem worst case time? The expected time is $n Hn$. So for 9 coupons we get ~26 trials. But what is the probability that all coupons have been collected after 26 trials? How do we know the number of trials required to collect all with arbitrary certainty?

(* Wolfram Mathematica: *)

(*Exact, but slow for n > 200*)
trialsToP2[c_,n_,k_] := (n!/((n-c)! n^k)) StirlingS2[k,n]

(*Approximate*)
trialsToP[n_,t_] := Block[{u,p},
  u = (t-n Log[n])/n;
  p = Exp[-Exp[-u]];
  p
  ]

trialsFromP[n_,p_] := Block[{u,t},
  u = Log[1/Log[1/p]];
  t = n * Log[n] + n *u;
  t
  ]


The results seem pretty close even for small values of N:
Input          | Results
N   P          | trialsFromP trialsToP2 trialsToP
9   EulerGamma | 25.16       0.5904     0.5772
    0.99       | 61.1764     0.9932     0.99
               |
100 EulerGamma | 520.4       0.5790     0.5772
    0.99       | 920.532     0.9905     0.99


a = test[9, EulerGamma] 
b = test[9, 0.99]
c = test[100, EulerGamma]
d = test[100, 0.99]

test[n_, p_] := Block[{t,p1,p2},
  t  = trialsFromP[n, p];
  p1 = trialsToP2[n, n, Round[t]];
  p2 = trialsToP[n, t];

  {N[t, 4], N[p1, 4], N[p2, 4]}
]

 A: If there are $n$ types of coupons, and you sample $k$ times with replacement, the probability you have $C=c$ different coupons has the recursion
$$\Pr(C=c|n,k)=\frac{n-c+1}{n}\Pr(C=c-1|n,k-1) + \frac{c}{n}\Pr(C=c|n,k-1)$$
starting at $\Pr(C=0|n,0)=1$ and $\Pr(C=c|n,0)=0$ for $c \ne 0$.
You can do the calculation for example on a spreadsheet to find $\Pr(C=9|n=9,k=26) \approx 0.62912$. 
More precisely, this is $\dfrac{9!}{9^{26}}S_2(26,9)$ where  $S_2(r,n)$ is a Stirling number of the second kind and in general   $$\Pr(C=c|n,k) =  \dfrac{n!}{(n-c)!n^{k}} S_2(k,c).$$
To have $99\%$ certainty with $9$ types of coupons, you need to sample up to $58$ times;  $99.9\%$ certainty requires $78$.
A: Let $T_n$ denote the time needed to complete all $n$ coupons. Then $T_n/(n\log n)\to1$ almost surely when $n\to\infty$, in particular, $P[T_n\gt (1+x)n\log n]\to0$ and $P[T_n\lt(1-x)n\log n]\to0$ when $n\to\infty$, for every positive $x$. 
To refine this, note that $P[T_n\lt n\log n+nu]\to\mathrm e^{-\mathrm e^{-u}}$ for every real number $u$. 
In other words, $T_n=n\log n+nU_n$ where $U_n\to U$ in distribution and the density $f_U$ of $U$ is such that $f_U(u)=\mathrm e^{-u-\mathrm e^{-u}}$ for every real number $u$.
If $n=9$, then $n\log n+nu=26$ for $u=.692$ and $\exp(-\mathrm e^{-.692})\approx.606$. This suggests that for 9 coupons, all coupons are collected at or before the 26th trial with probability roughly 60%. Assuming 9 is large enough for the asymptotics to be relevant...
