I was going over some questions in basic counting and I got stuck on this question :
What is the probability that if I draw $65$ cards uniformly at random, with replacement, from the deck, every card is drawn at least once?
I know the answer has something to do with Stirling's number but the formula my lecturer gave me doesn't make sense to me. Any help would be appreciated.
The short answer is $$\frac{{65\brace52}52!}{52^{65}}.$$
Explanation: Suppose we have a deck of $k$ cards, say $C=\{c_1,c_2,\ldots, c_k\}$, and we draw, with replacement, $n\geq k$ of them. Let $(d_1,\ldots, d_n)$ be the sequence of draws (so each $d_j\in C$). I'll call this a drawing.
Obviously there are $k^n$ possible drawings.
In how many drawings is each card chosen at least once? I'll call these good drawings. Well, let $(d_1,\ldots, d_n)$ be a drawing and suppose we have $k$ buckets labelled $\{c_1,c_2,\ldots, c_k\}$. Place $d_j$ into bucket $c_i$ when $d_j=c_i$. Then we know that the drawing is good if and only if each bucket is non-empty.
Now, suppose we have $k$ buckets, and we label them with labels $c_1,\ldots, c_k$, each bucket getting a different label. There are $k!$ ways of doing this.
Now place the symbols $\{D_1,\ldots, D_n\}$ into the buckets, each bucket getting at least one symbol. There are ${n\brace k}$ ways of doing this, where ${n\brace k}$ is the Stirling number of the second kind. Then we can define a good drawing $(d_1,\ldots, d_n)$ by setting $d_j=c_i$ if and only if $D_j$ was placed in bucket labelled $c_i$.
Thus there are exactly ${n\brace k}k!$ good drawings and so the probability that a drawing is good is $$\frac{{n\brace k} k!}{k^{n}}.$$
