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Suppose you have a 4-sided die, a 6-sided die, and a 12-sided die. You roll the three dice and add up the numbers that show up. What is the expected value of the sum of the rolls?

My attempt solution is using indicators. Here is the outline:

Let $I_A=\{4-sided\}$, $I_B=\{6-sided\}$, $I_C=\{12-sided\}$ and $X=I_A+I_B+I_C$. So, $$E[X]=E[I_A]+E[I_B]+E[I_C]=P(A)+P(B)+P(C)=2.5+3.5+6.5=12.5$$ where $P(A)$ is the expected sum of rolling a 4-sided die, and similar for the other two.

Is my approach correct? If we want the expected value of the three rolls shouldn't we add the expected sum of each of die?

I have doubts because continuing with this I get a negative variance which shouldn't happen. Unless I miscalculated the variance.

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2 Answers 2

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The variables aren't indicator functions. Let $A\sim \text{Unif}\{1,2,3,4\}$, $B\sim \text{Unif}\{1,2,3,4, 5, 6\}$ and $C\sim \text{Unif}\{1,2,\dotsc, 12\}$ be uniform random variables on their respective sets. They represent the value of the individual dice rolls. Then put $X=A+B+C$ to be the sum of the dice rolls $$ EX=EA+EB+EC=2.5+3.5+6.5 $$

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  • $\begingroup$ Oh ok got it. Thanks $\endgroup$
    – user480875
    Apr 22, 2019 at 21:49
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Thanks to @Foobaz John, the variance is,

$$Var(X)=Var(A)+Var(B)+Var(C)=\frac{(4-1)^2}{12}+\frac{(6-1)^2}{12}+\frac{(12-1)^2}{12}\approx12.92$$

by independence of variables.

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