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I was playing a card game where I thought of the problem and could not solve. The rule is that you have to put down the bigger cards than the previous player did. The order is: 3,4,5,6,7,8,9,10,J,Q,K,A,2. The player is dealt 13 cards randomly. What is the possibility that 6 of my cards are double cards of continuity? For example: 2 Jacks, 2 Queens, and 2 Kings. They can be any cards within the deck dealt as the player can rearrange the cards to their liking later. Note: all 4 2s can't be part of the sequence as they're the biggest cards already.
Thanks!

Guava
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  • Clarification requested: "What is the possibility that 6 of my cards are double cards of continuity? For example: 2 Jacks, 2 Queens, and 2 Kings." I am having trouble interpreting your question. It seems as if it is assumed that you are dealt 13 cards at random from a standard 52 card deck. Under this assumption, it seems as if you are asking for the precise probability of the 13 cards constituting exactly 10 different ranks out of the 13, with 3 of those ranks having exactly 2 cards each. ...see next comment – user2661923 Oct 05 '21 at 09:28
  • However, now I wonder if there is the added constraint that the 3 ranks with pairs must be consecutive ranks. I am also wondering whether (for example) having 2 Kings, 2 Aces, and 2 deuces would count. Further, I am wondering why you provided the background of a specific game. You could have asked the same question by simply specifying that Aces are counted low only, and that the three pairs must be consecutive ranks. Please edit your question to clarify what you are asking. Clarifications belong in the edited question, not in the comments. – user2661923 Oct 05 '21 at 09:31
  • Also I am having trouble interpreting "Note: all 4 2s can't be part of the sequence as they're the biggest cards already." I understand that if the ranks must be consecutive, that this precludes the ranks being AA,22,33, since that would constitute rounding the corner. However, it still leaves open whether you can have KK,AA,22. Also, it leaves open whether you could have (for example) 33,44,55,6,7,8,9,10,J,2. – user2661923 Oct 05 '21 at 09:39

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