Probability of picking a ball that has already previously been picked Given $n$ unique balls, you pick out a fraction $f$ of them and put them back. You then repeat. 
What is the probability $p$ as a function of the number of iterations $t$ of picking a ball that has already previously been picked?
For instance, if $n=3$ and a single ball is picked each iteration ($f=\frac{1}{3}$), $p(t)=0,\frac{1}{3},\frac{5}{9},\frac{19}{27},...$, and then my brute-force-by-hand-method kinda breaks down. 
Any help will be much appreciated, thanks!
 A: Let $A_t$ be the number of balls which have been seen after $t$ iterations.  Then starting with $P(A_0=0)=1$, you have $$\displaystyle P(A_{t+1}=a_{t+1}|A_{t}=a_{t}) = \dfrac{{a_{t}\choose fn-a_{t+1}+a_{t}}{n-a_{t}\choose a_{t+1}-a_{t}}}{{n\choose fn}} $$ and thus $$\displaystyle P(A_{t+1}=a_{t+1})= \sum_{a_{t}} \dfrac{{a_{t}\choose fn-a_{t+1}+a_{t}}{n-a_{t}\choose a_{t+1}-a_{t}}}{{n\choose fn}} P(A_{t}=a_{t}) $$ with the probability you are looking for being  $$p(t+1) = 1- P(A_{t+1} = A_{t}+fn) = 1-\sum_{a_t} \dfrac{{n-a_{t}\choose fn}}{{n\choose fn}} P(A_{t}=a_{t}).$$
I have no idea whether this can be simplified, though $E[A_n]$ certainly can be.
In your example of $n=3$ and $fn=1$, this will give 


*

*$P(A_{t+1}=1|A_{t}=0) =1$

*$P(A_{t+1}=1|A_{t}=1) =\frac13$ 

*$P(A_{t+1}=2|A_{t}=1) =\frac23$ 

*$P(A_{t+1}=2|A_{t}=2) =\frac23$ 

*$P(A_{t+1}=3|A_{t}=2) =\frac13$ 

*$P(A_{t+1}=3|A_{t}=3) =1$ 

*$P(A_{t+1}=1)= \dfrac{1}{3^t}$ 

*$P(A_{t+1}=2)= \dfrac{2^{t+1}-2}{3^t}$ 

*$P(A_{t+1}=3)= \dfrac{3^t-2^{t+1}-1}{3^t}$ 

*$p(t+1) = 1- P(A_{t+1} = A_{t}+1)= 1-\dfrac{2^n}{3^n}$ 


so your sequence continues $0, \frac{1}{3},  \frac{5}{9},  \frac{19}{27},  \frac{65}{81},  \frac{211}{243},  \frac{665}{729},  \frac{2059}{2187},  \frac{6305}{6561},  \frac{19171}{19683},  \frac{58025}{59049},  \frac{175099}{177147}, 
 \ldots$
