I'll post my work, but I'm not sure how to calculate variance. The question asks for the expected sum of 3 dice rolls and the variance. I think I got the expected sum.

Any help would be awesome :) thanks!

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  • 1
    $\begingroup$ If the random variables are independent (which is the case here), the variance of their sum will be equal to the sum of their variances. $\endgroup$
    – Hoda
    Apr 14, 2014 at 15:29

2 Answers 2


The variance calculation is incorrect. Let random variables $X_1,X_2,X_3$ denote the results on the first roll, the second, and the third. The $X_i$ are independent. The variance of a sum of independent random variables is the sum of the variances. Since the variance of each roll is the same, and there are three die rolls, our desired variance is $3\operatorname{Var}(X_1)$.

To calculate the variance of $X_1$, we calculate $E(X_1^2)-(E(X_1))^2$. And $$E(X_1^2)=\frac{1}{6}\left(1^2+2^2+\cdots+6^2\right).$$

  • $\begingroup$ I kinda get that? So I'd need to do variance for $n_1,n_2,n_3$? I'm still decently confused $\endgroup$
    – Cozen
    Apr 14, 2014 at 15:43
  • $\begingroup$ I called them $X_1$, $X_2$, $X_3$. You need the variance of each, then add. The variances are all the same, so we find the variance of $X_1$ and multiply by $3$. I gave detailed instructions for finding the the variance of $X_1$. If you follow them and simplify, you will find that the variance of $X_1$ is $\frac{35}{12}$, so the variance of the sum is $\frac{35}{4}$. (Recall that you know that $E(X_1)=\frac{7}{2}$.) $\endgroup$ Apr 14, 2014 at 15:48
  • $\begingroup$ This helped me understand this explanation a bit: answers.yahoo.com/question/index?qid=20070804123222AAXyGez $\endgroup$
    – Alan Wolfe
    Jul 22, 2017 at 4:20

If your dice are "independant" then the variance of the sum is the sum of the variance

  • $\begingroup$ So the variance would also be 10.5? $\endgroup$
    – Cozen
    Apr 14, 2014 at 15:32

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