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Math Help - 3 of a kind

  1. #1
    Member oldguynewstudent's Avatar
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    3 of a kind

    Counting the three-of-a-kind hands requires some care. One approach encodes each hand as a 5-list (A,B,C,D,E) where
    • A is the denomination of the three-of-a-kind; 13 ways to choose A
    • B is the 3-set of suits for the three-of-a-kind; \left({4\atop 3}\right) ways to choose B
    • C is a 2-set of denominations for the other two cards; \left({12\atop 2}\right) ways to choose C
    • D is the suit of the smaller denomination in C; 4 ways to choose D
    • E is the suit of the larger denomination in C; 4 ways to choose E
    By the product principle there are 13 * \left({4\atop 3}\right) * \left({12\atop 2}\right) * 4^2 = 54,912 hands.

    Question 24: Explain what is wrong with the following reasoning: There are 52 ways to select the first card that is a part of the three-of-a-kind. Then there are \left({3\atop 2}\right) ways to pick the other two cards to make up the three-of-a-kind. Finally there are \left({48\atop 2}\right) ways to pick the other two cards.

    The answer would be that 3 choose 2 ways to pick the other two cards to make up the three-of-a-kind is not independent of the card that is first picked, therefore the product principle is not used correctly. In other words, for every denomination picked in the first step, there are 3 choices that lead to the same 3-list. Is this the correct reason why the above will not work?
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  2. #2
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    Hello, oldguynewstudent!

    Counting the 3-of-a-kind hands requires some care.

    Question 24: Explain what is wrong with the following reasoning:

    There are 52 ways to select the first card that is a part of the 3-of-a-kind. . True!
    Then there are {3\choose 2} ways to pick the other two cards to make up the 3-of-a-kind. . Yes!
    Finally, there are {48\choose2} ways to pick the other two cards. . ← Here!

    Picking any two of the other 48 cards, we could get a Pair, forming a Full House.

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