TITLE: Limit probability
QUESTION [5 upvotes]: You start with a bag of N recognizable balls. You pick them one by one and replace them until they have all been picked up at least once. So when you stop the ball you pick has not been picked before but all the others have been picked once or more.
Let $P_N$ be the probability that all the others were actually picked twice or more.
Questions: does $P_N$ have a limit as  $N\to \infty$, is this limit 0 or can you compute it?
Note: This is a question that was asked at the Yahoo answers forum, by gianlino. Since no one has found an answer in that forum, I forwarded it here. Here is the original link. http://answers.yahoo.com/question/index;_ylt=ApZ_Q6sS137DnEcNoUBqn.vty6IX;_ylv=3?qid=20100704050924AAfLYYJ

REPLY [2 votes]: Also, here's a heuristic argument to show that the probability behaves like $1/N$ as $N$ becomes large.
The last ball to be chosen is chosen for the first time after roughly $N\log N$ steps. (It's the maximum of $N$ random variables which are geometric with mean $N$). By this stage, the number of times that any particular other ball has been selected is Poisson($\log N$). So the probability a particular other ball has been selected only once is roughly $\log N e^{-\log N}$ which is $\log N/N$. Hence the probability that no other ball has been selected only once is roughly
$(1-\log N/N)^{N-1}$, which is $\exp[-\log N + O(\log N/N)] \sim 1/N$.
Of course the previous paragraph ignores various dependencies, but as $N$ becomes large these will be negligible and no doubt one could formalise the argument if desired.