Suppose you played two rounds of a game of cards. Each round you are independently dealt an $n$-card hand from a shuffled deck (e.g., n=5 $n=13$ for Bridge, $n=7$ for Hold'em, etc) .
Q1 (Motivating): In a game involving $n$-card hands, what is the probability that your first hand is disjoint from your second hand? That is, what is the probability your successive hands share no cards in common?
Where I'm getting stuck is generalizing this question to an $m$-card deck, $n$-card hand, and, in particular, a $k$-round game:
Q2 (General): After $k$ independent rounds of a game of $n$-card hands from a shuffled $m$-card deck, what's the probability all hands are disjoint (i.e., that you don't see the any card more than once)?
Clearly $P(disjoint\; hands)=1$ after just $k=1$ round, and by the pigeonhole principle $P(disjoint\; hands)=0$ after $k=\lceil\frac{n}{m}\rceil$ rounds. But I'm stuck reasoning through a closed-form equation.
Ultimately I'm interested in the $P=0.5$ scenario, something akin to the birthday paradox, namely:
Q3 (Goal): How many rounds of a (generalized) card game do you have to play before you would expect to see one (or more) cards repeated across hands?
Thanks for the help!
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