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$\newcommand{\V}{\operatorname{V}}\newcommand{\Cov}{\operatorname{Cov}}$Here's my work so far:

$X_i$ is the number of rolls with $i$ dots. $Y$ is the sum of the odd numbers, so $Y=X_1+3X_3+5X_5$.

$$\V[Y]=V[X_1]+V[3X_3]+V[5X_5]+2(\Cov(X_1,3X_3)+\Cov(X_1,5X_5)+\Cov(3X_3,5X_5))$$

$$\V[X_1]=100\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)=13.89, \V[3X_3]=9*13.89=125, \V[5X_5]=347.25$$

Now I'm stuck trying to find the covariances... Help please?

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  • $\begingroup$ why do you think that there are non-zero covariances? $\endgroup$ – Math-fun Dec 3 '14 at 20:47
  • $\begingroup$ If I roll a die 100 times, then the number of times I roll a 1 is affected by the number of times I roll a 3... If I roll a 3 90 times then the number of times I roll a 1 is forced to be 10 or less. Therefore they are not independent, so if the covariance is zero then I must show it. $\endgroup$ – tim Dec 3 '14 at 20:49
  • $\begingroup$ Shouldn't $Y$ be $X_1+3X_3+5X_5$? It's "the sum of the odd numbers", so rather than counting the number of times you get a $3$, thus adding $1$ every time a $3$ appears, it should add $3$ every time a $3$ appears. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 3 '14 at 20:52
  • $\begingroup$ Oh shoot, you're right... $\endgroup$ – tim Dec 3 '14 at 20:53
  • $\begingroup$ Edited to reflect what $Y$ should be... $\endgroup$ – tim Dec 3 '14 at 20:56
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Let $X_i=0$ if the $i$-th toss is even, and let $X_i$ be the number obtained on the $i$-th toss if that number is odd. Then if $Y=X_1+\dots +X_n$, where $n=100$, then $Y$ is the sum of the odd numbers tossed.

The $X_i$ are independent, so the variance of $Y$ is the sum of the variances of the $X_i$. This is $n\text{Var}(X_1)$.

I will leave finding the variance of $X_1$ to you. It is straightforward to calculate, since $X_1=0$ with probability $\frac{1}{2}$, and $X_1=\frac{1}{6}$ for each of $1$, $3$, and $5$.

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