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I played a card game with my fam where we are in 2 teams of 2 (4 of us total). Each player gets a hand of 7 cards from the 52 deck and between myself and my partner we needed a king OR an ace. In a total of 3 hands each (so 6 hands total between me and my partner) neither of us got a king OR an ace. This seems very unlikely to me. I can't do the maths on it so I need some help and an explanation of how to do it (dumbed down please) would be appreciated!

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  • $\begingroup$ How many ways can you draw seven non-Aces and non-Kings? What does it mean mathematically if you do this three times? How many ways can you draw any seven cards? $\endgroup$ Commented Oct 24, 2020 at 11:51
  • $\begingroup$ This sound interesting, but we need a bit more detail. How did the redraws happen? Did everyone redraw or only you and your partner redraw? Are all cards reshuffled or are seen cards removed from the game? (e.g. If the other two players held all kings and aces, you would never draw any kings or aces if they don't redraw as well) $\endgroup$
    – player3236
    Commented Oct 24, 2020 at 11:51

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What is the probability that a $14$-card hand (you and your partner's hands combined) contains no kings or aces ($8$ such cards in a $52$-card deck)?

We need to choose $14$ cards from the $52-8=44$ non-king-non-aces to make such an "unlucky" hand. All possible hands may be generated by taking $14$ out of $52$. Thus the probability in the blockquote above is $$\binom{44}{14}/\binom{52}{14}=0.0649847\dots$$ To get the probability that the above happens for three rounds, cube the one-round result. You get $$0.000274431\dots$$ or $0.027\%$.

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