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Math Help - Conditional distribution for two dice values

  1. #1
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    Conditional distribution for two dice values

    Hi!

    I have two dice values X, Y and should calculate the conditional distributions
    X+Y given Y = 3 or 5
    Y given X>Y

    I don't know how to start with that. Can anyone give me a hint?
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  2. #2
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    Quote Originally Posted by mork View Post
    I have two dice values X, Y and should calculate the conditional distributions
    X+Y given
    1) Y = 3 or 5
    2) Y given X>Y
    For #1, your given subspace is B=\{(1,3),(2,3),(3,3),(4,3),(5,3)(6,3),(1,5),(2,5)  ,(3,5),(4,5),(5,5)(6,5)\}.
    Now P(X+Y=5|B)=\frac{1}{12}, WHY?

    But P(X+Y=7|B)=\frac{2}{12}, WHY?.

    Now you finish.
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  3. #3
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    Quote Originally Posted by Plato View Post
    For #1, your given subspace is B=\{(1,3),(2,3),(3,3),(4,3),(5,3)(6,3),(1,5),(2,5)  ,(3,5),(4,5),(5,5)(6,5)\}.
    Now P(X+Y=5|B)=\frac{1}{12}, WHY?

    But P(X+Y=7|B)=\frac{2}{12}, WHY?.

    Now you finish.
    OK, I have your B already and I know how to calculate P(X+Y=n|B) for every n, but how should the conditional distribution look like? Is it formula which can be derived from this or do I have to "see" what it should look like?

    The only thing I can see is that P is 2\cdot\frac{1}{12}, 4\cdot\frac{2}{12},2\cdot\frac{1}{12} between 4 and 11. I cannot come up with a formula that returns this behaviour.
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  4. #4
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    Quote Originally Posted by mork View Post
    I cannot come up with a formula that returns this behaviour.
    There is no formula.
    It is just a matter of counting. There is only one way X+Y=4.
    There are two ways X+Y=7.
    Count all the other cases.
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  5. #5
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    Quote Originally Posted by Plato View Post
    There is no formula.
    It is just a matter of counting. There is only one way X+Y=4.
    There are two ways X+Y=7.
    Count all the other cases.
    So I was looking for something which isn't there.
    How do I write the result?
    P(X+Y|Y \in \{3,5\}) = \{\frac{1}{12},\frac{1}{12},\frac{2}{12},\frac{2}{  12},\frac{2}{12},\frac{2}{12},\frac{1}{12},\frac{1  }{12}\}

    By the way: thank you so far for your patience with me
    Last edited by mork; March 12th 2011 at 11:36 AM.
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  6. #6
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    Quote Originally Posted by mork View Post
    So I was looking for something which isn't there.
    How do I write the result?
    P(X+Y|Y \in \{3,5\}) = \{\frac{1}{12},\frac{1}{12},\frac{2}{12},\frac{2}{  12},\frac{2}{12},\frac{2}{12},\frac{1}{12},\frac{1  }{12}\}

    By the way: thank you so far for your patience with me
    This makes no sense because

    1. probability is a number. Your right hand side is not a number.

    2. What is the condition X + Y has to satisfy?

    And what is Y \in \{3,5\} meant to mean? Does it mean that Y = 3 or Y = 5?

    Plato's replies say exactly what to do. I'm not sure what the trouble can stiil be ....? The result (that is, I assume, the answer) is given in your very first post!
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