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    Question about well-suited poker hand




    Hi, I have the following question:
    A peculiar deck of playing cards is like a regular deck, but has five suits: spades, hearts, diamonds, clubs, and clovers. There are 13 cards in each suit: Ace, 2, 3, , 10, Jack, Queen, King. There are thus 513 = 65 cards in this deck. Consider dealing a 6-card hand from this deck.

    A well-suited hand is one that contains at least one card of every suit, no more than one card of any kind (no pairs, etc.), and not all cards of consecutive kinds. An example of a well-suited hand is {2 of spades, 4 of hearts, 5 of diamonds, 8 of clubs, 10 of clubs, Jack of clovers}.

    How many different 6-card hands can be dealt from this deck that are well-suited?
    ----------
    I started off by using 5c1 to choose the one suit that gets a repeat, then multiply by 13c2 for the possible combinations of two cards to choose from that suit after that. I thought at this point that I could do 13^4 for the remaining cards, but then I realized that I have to avoid repeating anything (to avoid pairs). How can I do that? And how do I avoid getting all consecutive cards?

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    Re: Question about well-suited poker hand

    I started off by using 5c1 to choose the one suit that gets a repeat
    Good starting point. Next, you can multiply by

    \binom {13} {6} \times 6!

    which is the number of permutations of 6 distinct kinds (some write 13P6). Then divide by

    2!

    to remove the redundant permutation of two kinds in the same suit.

    Finally, to adjust the 6 consecutive kinds, you need to subtract the number of consecutive kinds from

    \binom {13} {6}

    before you do the multiplication.

  3. #3
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    Re: Question about well-suited poker hand


    Okay, so to be clear on this the 13c6*6! is basically the number of ways to choose 6 unique cards (2 vs. 3 vs. J, etc.) times the number of ways to order them?

    How does the 2! eliminate the possibility of two kinds in the same suit? I'm not certain how we can guarantee that all 5 suits are present.

    And for number of consecutive hands, I did this by adding up all the possible permutations where that occurs (ex: 2, 3, 4, 5, 6, 7) which was 13 and subtracted that.

    Thank you for the help!

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