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Help with Expectation of Rolled Die 3 times

  1. Jun 21, 2005 #1
    I am new here and grateful to have found this site! I have a problem:

    The question is:

    Roll a fair die 3 times. Find the math expectation of the numerical sum of the outcomes of the rolls.

    I have no clue. I have looked over all notes, and I just don't "get it". Can someone help explain this to me in "layman's terms"?

    Thanks for any help

  2. jcsd
  3. Jun 21, 2005 #2
    The outcome of each roll is a(n independent) random variable. What you're looking for in your notes is the expectation of a function of a random variable, in this case the function happens to be a sum.
    Although, you needn't go into your notes to logic this problem out. :smile:
    Last edited: Jun 21, 2005
  4. Jun 21, 2005 #3
    Thanks, but I see these "formulas" in my notes:

    E(x) = 1/p and I see P(x=n) = (1-p)^n-1 p

    I soooo don't understand this. Haven't had to do math in almost 20 years. So, I just add up the 3 rolled numbers (that I choose?) and get the average for them, correct? Do I ignore the above formulas for this equation? Sorry to sound so stupid. :eek:)
  5. Jun 21, 2005 #4
    The E(x) you have is for a more involved problem where the probability of one success is p (thus the probability of failure is 1-p). The P(x=n) you have is the probability that you get one success (and n-1 failures) on the simple success/failure "experiment". This probability mass function corresponds to the geometric distribution. You would use these two equations if your variable were the number of experiments conducted before a "success" occurs.
    Your question is asking about the value on the die. If the die is fair, then the probability of each number facing up after one roll is the same, 1/6. So you have P(n) = 1/6 for n=1,...,6. The expectation of a discrete random variable as you have here is just [tex]E(X) = \sum_x x p(x)[/tex]
    It is a mean, but in the event of an unfair die or different random variables, it accounts for the weighted mean by multiplying each outcome by its probability. From calculating the expectation for one die, it is easy to see the expected value of throwing three independent dies without further calculation.
    Last edited: Jun 21, 2005
  6. Jun 22, 2005 #5
    Oh, thank you so much! I really appreciate the help. I used to love math, I guess it's so fast paced now and the instructors forget that there are some who haven't seen an algebra problem, let alone calculus, etc. in many years. :eek:) They just assume you know what "nth" means. Which by the way, I don't. LOL Again, thank you for your time and help. I'm sure I will be putting another one out here. :biggrin:
  7. Jun 22, 2005 #6
    Ok, don't laugh...this is what I came up with:

    P(n) = 1/6
    = (1)*(1/6) + (2)*(1/6) + (3)*(1/6)
    = 1 + 2 + 3 * (1/6)
    = 6 * 1/6
    = 6/6 = 1

    Am I anywhere close?

  8. Jun 22, 2005 #7
    I'm assuming that's the expectation of X where X is the sum of a single die throw. It doesn't make much sense though, as it seems your die has only 3 sides and the total probability of it having any result at all is 3/6. :wink: The x in the expectation formula is the value of each possible X, and expectation is a weighted average over all possible outcomes. The way to use the expectation for a single die throw is as follows:
    E(X) = 1*(1/6) + 2*(1/6) + ... + 6*(1/6) = 3.5
    That's the expectation (weighted average) of the outcome of a single die throw given the probability of each possible outcome. The reason we are using the numbers on the die is because the random variable X deals with the sum of the outcome, though in the above case we're only summing one die. Note that it need not be a number actually on the die as it is the experiment's average outcome that should approach the expectation as the number of experiments (throws of a single die) increases without bound. Now generalize to summing three independent dice being thrown.
    Last edited: Jun 22, 2005
  9. Jun 22, 2005 #8
    Thanks to you again! Ok... hey, at least I was close. I was thinking I only had to do it x 3 since I was only rolling it 3 times. LOL I see what you are saying. Do I do the same with one that tells me to roll a fair die twice. X1 and X2 are the outcomes of the roll one and two and find a distribution function? Same kinda way? Yes, I have to be the dumbest person ever to have to do stats/prob. Apparently my brain isn't wired correctly for it. :uhh:

  10. Jun 22, 2005 #9
    Pretty much. Since X1 and X2 are independent throws, and use the same probability distribution, do you see that E(X1 + X2) is the same as 2*E(X1), in essence the expected mean in a simpler experiment that just multiplies the outcome of one die throw by 2 ?
  11. Jun 24, 2005 #10

    I really don't see that, but I will give it a shot. Right now I am correcting one I got wrong on a previous assignment. do faint! LOL N straight lines on a plane - gotta find min and max number of intersections. Yeah, right. :confused: I'll show you what I come up with on the above one soon as I work on this other one. Thank you for all of your help!
  12. Jun 25, 2005 #11
    haven't forgotten....just still working on the other ones .... pray for me :smile:
  13. Jun 26, 2005 #12

    Ok....here is what I came up with:

    P(n) 1/6 E(X + X )
    = (1)*(1/6) + (2)*(1/6) + (3)*(1/6) + (4)*(1/6) + (5)*(1/6) + (6)*(1/6)
    = 21/6
    = 3.5
    = E(X + X )
    = 3.5 + 3.5
    = 7

    Close? Oh, please say yes! LOL :rofl:

    ps I do have the "1" and "2" on the big Xs, E(X + X) but it didn't show up when I copied and pasted it.
  14. Jul 10, 2005 #13
    mircat: Close? Oh, please say yes!

    Sure, the expectation on any throw is 3.5, however, you originally postulated three throws.
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