Resampling Probability I've had a student working on some resampling problems. Here's a question I've posed to her:
M consecutive draws of size K are taken from an urn with N balls. Balls are replaced between draws. Obviously N>K and M>1. 
What's the probability that at least one ball is drawn twice (i.e. resampled)?
 A: I will start you on an solution. First I will restate the problem, such I understood and interpreted it.
On eurn have $N$ identifiable balls (we can think of them as numbered from $1,2,\dots,N$).
We draw, without replacement, a sample of $K$ balls, then we replace them, and draw again a sample of $K$ balls, and so on, for a total of $M$ draws. Later, when I say draw I mean such a simultaneous draw of $K$ balls. $N > K, M > 1$. Define the event
$$
   A = \{\text{no ball drawn twice}\}
$$
We ask for the $P(A^c)$. Obviously, if $MK>N$ we have $P(A)=0$, so we must assume that 
$MK \le N$ to get an interesting solution.
No let the number of draws vary, $m=1, \dots, M$ and let $P_m(A)$ be the probability we seek when we do only $m$ draws. Then
$$
   P_1(A)=1
$$ by definition. Now we can compute for increasing $m$.
$$
P_2(A)=P(\text{balls in second draw distinct from balls in first draw}) \\
   =  \frac{\binom{K}{0}\binom{N-K}{K}}{\binom{N}{K}}
$$
which is a hypergeometric probability. 
Then you can continue in this way, inductively, using
$$
P_m(A)=P(\text{no hits in mth draw}|\text{no hits up to (m-1)th draw})\cdot P_{m-1}(A)
$$
leaving the rest for you!
