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Suppose I have a deck of $n$ cards. I draw $k$ cards from the deck and note down which cards I have seen. I then replace the cards that I drew and shuffle the deck. I repeat the process of drawing $k$ cards from the deck for a total of $r$ times.

What is the probability after $r$ rounds that I have seen every card in the deck of $n$ cards.

I was able to simulate this for $n=52$, $k=34$, and $r=9$ using a Monte Carlo simulation in Python and got $\approx .995$. I am hoping that this is generalizable and that there is a way to directly calculate the probability. I know that the denominator would be $(_nC_k)^r$ but the numerator is proving difficult to calculate. Thanks!

RobPratt
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namremir
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The probability is $$ \sum_{i=0}^{n}(-1)^i \binom ni \left(\frac{\binom{n-i}{k}}{\binom nk}\right)^r $$ You can prove this with the principle of inclusion exclusion. Start with $1$, and then for each card, subtract out the probability that the card was not chosen. There are $n$ cards, so we subtract $n\cdot \left[\binom{n-1}k/\binom nk\right]^r$. However, events where two cards were not chosen have now been subtracted twice, so these need to be added back in, meaning we add in $\binom n2 \left[\binom{n-2}k/\binom nk\right]^r$. Continuing this pattern of corrections leads to the stated formula.

In this answer, joriki used this calculation as an intermediate step in determining the expected number of drawings it takes to see all cards, when you draw $k$ each time, giving a generalization of the coupon collector's problem.

Mike Earnest
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