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What's the expected value of this problem (random variable)?

  1. Jul 22, 2013 #1
    1. The problem statement, all variables and given/known data

    What's the expected value of this problem (random variable)?

    X: represent the result of dice number 1 - result of dice number 2

    example dice 1 first roll = 2; second roll = 3
    dice 2 first roll = 1; second roll = 2

    X = 2+3 -(1+2) = 2

    what's the expected value of X

    2. Relevant equations



    3. The attempt at a solution

    I think it's 0, because the Expected value of dice 1 is the same as that of dice 2
     
  2. jcsd
  3. Jul 22, 2013 #2
    I assume you know how to create a table for discrete random variable, showing the outcome of ##X## and it's probability ##P(X)##. Do that first before you try to calculate the ##E(x)##
     
  4. Jul 22, 2013 #3
    I don't even know that. Our teacher didn't show us any of that. He just skipped over a lot of material.

    It's not a homework problem. I just tried to think of an example that may eventually help me in doing my assignment.

    Also, if you happen to have a document with a lot of examples that shows you how to compute E(X) that is easy to understand, don't hesitate. Thank you.
     
  5. Jul 22, 2013 #4
    I'm referring to this table:
    Expectations06.gif

    Ok why don't you look at this page here: (You can no doubt search more in google also)
    http://gwydir.demon.co.uk/jo/probability/calcdice.htm

    Or this video from Khan's Academy
    https://www.khanacademy.org/math/probability/random-variables-topic/random_variables_prob_dist/v/expected-value--e-x [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Jul 23, 2013 #5

    haruspex

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    I don't understand the relationship between that table and the OP. That table appears to be more to do with tossing a coin 5 times than with rolling a die twice.
    More usefully, if you have two random variables, X and Y, then E(X+Y) = E(X)+E(Y). You can use this both to get the expected value of the sum of two rolls of a die, and to get the expected difference from the other sum.
     
  7. Jul 23, 2013 #6
    I'm just giving an example of discrete distribution table from the internet. Maybe he can recall if it was previously taught.
     
  8. Jul 23, 2013 #7

    HallsofIvy

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    The smallest possible value of a single roll of a single die is 1. The largest possible value is 6. The smallest possible value of "the result of die #1- the result of die #2" is 1- 6= -5 and that is possible only with "1" and "5". What is the probability of that?

    You can get -4 in two different ways: 1- 5 and 2- 6. What are the probabilities of "1 and 5" and "2 and 6"? So what is the probability of "1 and 5 or 2 and 6"?

    Continue that until you get to +5, given by 6- 1. You should quickly see a pattern that will simplify the calculations.
     
  9. Jul 23, 2013 #8
    Didn't see that. I don't even know what that means.
     
  10. Jul 23, 2013 #9
    What did you do? Did you use the Discrete Binomial Distribution and why do you have 10/32 in the middle?
     
  11. Jul 23, 2013 #10
    The pattern I see leads me to believe it's 0.
     
  12. Jul 23, 2013 #11

    HallsofIvy

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    I was hoping for a pattern for the probabilities of "-5", "-4", ..., "4", "5": probability of -5 is 1/36, -4 is 2/36= 1/18, -3 is 3/36= 1/12, etc. But if you mean that the expected value is 0, yes, that is correct because of the symmetry: a- b and b- a are equally likely.
     
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