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Throw three dice

  1. Jun 8, 2013 #1
    we throw three ordinary dice and X,Y,Z their results and W=X+Y+Z how many differents values have the random values E(W/X) and E(W/X+Y)?

    can anyone explain me how to beggin because i am confused.. i will use E(X+Y+Z/X)=E(X/X)+E(Y/X)+E(Z/X)=? for the first one?
     
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  3. Jun 8, 2013 #2

    HallsofIvy

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    I think you need to go back and look at the problem again. I don't know what you mean by "how many differents values have the random values E(W/X) and E(W/X+Y)?"
    How many different values" of what? E(W/X) and E(W/(X+ Y) are specific numbers. Do you mean how many (X, Y, Z) combinations give W that are equal to those expectations. Actually, I would be surprized if those were integer values. Or do you mean simply "find E(W/X) and E(W/(X+Y))"?

    It's not all that difficult to determine the [itex]6^3= 216[/itex] combinations of three dice. W ranges in value from 3 to 18.
     
  4. Jun 9, 2013 #3

    haruspex

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    Assuming you mean E((X+Y+Z)/X), that's a good start. E(X/X) is obvious. Since X and Y are independent, can you expand E(Y/X) into separate functions of X and Y?
     
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