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Math Help - Dice Pair Matching

  1. #1
    Junior Member utopiaNow's Avatar
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    Dice Pair Matching

    4 players roll a die. They score 1 point if there is a matching pair. What are the probabilities of getting the possible scores.

    For example: for the case where 2 people get ”1”, and the other two roll a ”3” and a ”4” respectively, then the team’s score would be 1. Similarly, for the case that 3 people throw a ”5” and the other one does not, the score would be 3, since there are 3 different pairs of players that have the same number.

    Proposed solution:
    Imagine players as slots to fill with possibilities of rolling dice.

    Score 0:
    All 4 players roll different numbers. So 6*5*4*3 = 360 ways of getting score 0.

    Score 1:
    2 Players roll the same, other 2 are different from each other and the matching pair. So 6*1*5*4 = 120 ways of getting score 0.

    Score 2:
    2 pairs of players match. So 6*1*5*1 = 30 ways. The ones are the corresponding matches to the preceding rolls.

    Score 3:
    3 matches and 1 different so 6*1*1*5 = 30 ways again. The 4th person can have any one of the 5 choices left over.

    Score 4:
    All roll the same number so. 6*1*1*1 = 6 ways of getting score 4.

    My problem is these all add up to 546, whereas I would imagine they'd have to add up to 1296 = 6^4 possibilities.

    Any suggestions?
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  2. #2
    MHF Contributor

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    Let’s look at the case score 2. That could happen if the players roll \boxed{2}\boxed{2}\;\boxed{6}\boxed{6}.
    But that can happen is \frac{4!}{2^2} ways.
    Moreover, there are {6 \choose 2} ways to have two different pairs.
    It seems to me that have under-counted this case.

    If I have miss-read the problem please say why.
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  3. #3
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    Hello, utopiaNow!

    You're not considering WHO gets the pairs, etc.


    Four players roll a die.
    They score 1 point if there is a matching pair.
    What are the probabilities of getting the possible scores?

    For example:
    2 people get ”1”, and the other two roll a ”3” and a ”4” resp., then the score would be 1.
    Similarly, that 3 people throw a ”5” and the other one does not, the score would be 3.

    Score 0
    Four different numbers: abcd
    Number of ways: . 6\cdot5\cdot4\cdot3 \:=\:{\color{blue}360}


    Score 1
    A pair and two singles: aabc
    There are: . {4\choose2} = 6 distributions . . . \{aabc, abac, abca, baac, baca, bcaa\}
    Number of ways: . 6\cdot(6\cdot5\cdot4) \:=\:{\color{blue}720}


    Score 2
    Two pairs: aabb
    There are 3 distributions.
    . . If the four players are A,B,C,D, they can be paired in 3 ways:
    . . . . \{AB,\,CD\},\;\{AC,\,BD\},\:\{AD,\,BC\}
    Number of ways: . 3\cdot(6\cdot5) \;=\;{\color{blue}90}


    Score 3
    A triple and a single: aaab
    There are 4 choices of who gets the triple.
    Number of ways: . 4\cdot(6\cdot5) \;=\;{\color{blue}120}


    Score 4
    A quadruple: aaaa
    The only choice is the value of the quadruple.
    Number of ways: . {\color{blue}6}


    Check: . 360 + 720 + 90 + 120 + 6 \;=\;{\bf1296}

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  4. #4
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, utopiaNow!

    You're not considering WHO gets the pairs, etc.



    Score 0
    Four different numbers: abcd
    Number of ways: . 6\cdot5\cdot4\cdot3 \:=\:{\color{blue}360}


    Score 1
    A pair and two singles: aabc
    There are: . {4\choose2} = 6 distributions . . . \{aabc, abac, abca, baac, baca, bcaa\}
    Number of ways: . 6\cdot(6\cdot5\cdot4) \:=\:{\color{blue}720}


    Score 2
    Two pairs: aabb
    There are 3 distributions.
    . . If the four players are A,B,C,D, they can be paired in 3 ways:
    . . . . \{AB,\,CD\},\;\{AC,\,BD\},\:\{AD,\,BC\}
    Number of ways: . 3\cdot(6\cdot5) \;=\;{\color{blue}90}


    Score 3
    A triple and a single: aaab
    There are 4 choices of who gets the triple.
    Number of ways: . 4\cdot(6\cdot5) \;=\;{\color{blue}120}


    Score 4
    A quadruple: aaaa
    The only choice is the value of the quadruple.
    Number of ways: . {\color{blue}6}


    Check: . 360 + 720 + 90 + 120 + 6 \;=\;{\bf1296}


    Ah, yes that's it. Thanks Soroban for your help and explanation.
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