Solution verification: Calculating expected value of sum of three unusual dice

Question

Problem: We have three fair dice. Their numbering is unusual, the $6$ numbers on the respective dice are: $$\text{Die}\,\,\#1:\,1,3,5,7,9,10,\quad\text{Die}\,\,\#2:\,1,2,2,3,3,3,\quad\text{Die}\,\,\#3:\,2,2,4,4,4,4.$$ We roll all three of these dice, and denote by $X$ the sum of the three numbers that are showing. Find the expected value $X$.

My Attempt: I have a hunch that the indicator approach might somehow work here but I do not see how to execute it. Therefore, I will attempt to solve the problem the hard way, by finding the probability mass function of $X$. Note that $X$ takes values in the set $A=\{4,5,6,7,8,9,10,11,12,13,14,15,16,17\}.$ Now we have the following calculations using the independence of the die rolls and some basic counting rules $$P(X=4)=\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{2}{6}=\frac{2}{6^3},\, \\P(X=5)=\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{2}{6}=\frac{4}{6^3},\, \\P(X=6)=\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{2}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{2}{6}=\frac{12}{6^3},$$ $$P(X=7)=\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{2}{6}=\frac{12}{6^3},\, \\P(X=8)=\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{2}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{2}{6}=\frac{24}{6^3},$$ $$P(X=9)=\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{2}{6}=\frac{12}{6^3},\, \\P(X=10)=\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{2}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{2}{6}=\frac{24}{6^3},$$ $$P(X=11)=\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{2}{6}=\frac{12}{6^3},\, \\P(X=12)=\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{2}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{2}{6}=\frac{24}{6^3},$$ $$P(X=13)=\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{2}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{2}{6}=\frac{14}{6^3},$$ $$P(X=14)=\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{2}{6}+\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{2}{6}=\frac{26}{6^3},$$ $$P(X=15)=\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{2}{6}=\frac{18}{6^3},$$ $$P(X=16)=\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{4}{6}+\frac{1}{6}\cdot\frac{2}{6}\cdot\frac{4}{6}=\frac{20}{6^3},\, \\P(X=17)=\frac{1}{6}\cdot\frac{3}{6}\cdot\frac{4}{6}=\frac{12}{6^3}.$$ One can see that $\sum_{k=4}^{17}P(X=k)=1$, so we have a legitimate probability mass function. Now let us calculate the expectation of $X$. Using the definition we have $$E[X]=\sum_{k=4}^{17}k P(X=k)=\frac{1}{6^3}[4\cdot2+5\cdot4+6\cdot12+7\cdot12+8\cdot24+9\cdot12+10\cdot24+11\cdot12+12\cdot24+13\cdot14+14\cdot26+15\cdot18+16\cdot20+17\cdot12]=11.5$$

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Could anybody please give me some feedback about my approach? If my exposition is unclear, please let me know and I will do my best to improve it. In addition, if anyone has a simpler way of tackling the problem, any hints on how to go about such a method would be much appreciated.
Thank you very much for your time and feedback.

Answer 1

To elaborate JMoravitz's comment, let $X$, $Y$ and $Z$ denote the outcome of three dice respectively (renameing the sum as $W$). Then what you want is $E(X+Y+Z)$, which by linearity of the expectation: $$ E(X+Y+Z) = E(X)+E(Y)+E(Z)\;, $$ which can be found by $$ E(X) = \frac16(1+3+5+7+9+10) = \frac{35}{6}, $$ $$ E(Y) = \frac13\cdot 2+\frac12\cdot 3+\frac16\cdot 1=\frac{14}{6}, $$ and $$ E(Z) = \frac13\cdot 2+\frac23\cdot 4=\frac{10}{3}\;. $$

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[Added:]

Remark. This is actually connected to your attempt as follows (without using linearity). $$ \begin{align} E(W) &= \sum_{w\in A}wP(X+Y+Z=w) \end{align} $$ But for each $w\in A$, the event $\{X+Y+Z=w\}$ can be written as a disjoint union as $$ \{X+Y+Z=w\} = \bigcup_{w_1+w_2+w_3=w\\w_k\in D_k, k=1,2,3}(X = w_1,Y=w_2,Z =w_3) $$ Thus, $$ \begin{align} E(W) &= \sum_{w_k\in D_k}(w_1+w_2+w_3)P(X = w_1)P(Y = w_2)P(Z=w_3)\\ &=\sum_{w_k\in D_k}w_1P(X = w_1)P(Y = w_2)P(Z=w_3)\\ &+\sum_{w_k\in D_k}w_2P(X = w_1)P(Y = w_2)P(Z=w_3)\\ &+\sum_{w_k\in D_k}w_3P(X = w_1)P(Y = w_2)P(Z=w_3)\\ &= \sum_{w_1\in D_1}w_1P(X=w_1) +\sum_{w_2\in D_2}w_2P(Y=w_2) +\sum_{w_3\in D_3}w_3P(Z=w_3) \\ &= E(X)+E(Y)+E(Z)\;. \end{align} $$

Answer 2

$$E[X]=\frac{1+3+5+7+9+10}{6}+\frac{1+2+2+3+3+3}{6}+\frac{2+2+4+4+4+4}{6}=\frac{69}{6}=11.5$$

Answer 3

Let me illustrate the suggestion made by JMoravitz in the comments. Define $X_1$, $X_2$, and $X_3$ as the random variables whose outcomes are the roll of the first, second, and third dice, respectively. Then $$X = X_1 + X_2 + X_3,$$ so $$E(X) = E(X_1 + X_2 + X_3) = E(X_1) + E(X_2) + E(X_3)$$ by linearity of expectation. We have: $$E(X_1) = \sum_x x P(X_1 = x) = \frac{1}{6} \left(1 + 3 + 5 + 7 + 9 + 10 \right) = \frac{35}{6}.$$ $$E(X_2) = \sum_x x P(X_2 = x) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{2}{6} + 3 \cdot \frac{3}{6} = \frac{14}{6}.$$ $$E(X_3) = \sum_x x P(X_3 = x) = 2 \cdot \frac{2}{6} + 4 \cdot \frac{4}{6} = \frac{20}{6}.$$ Hence: $$E(X) = \frac{35 + 14 + 20}{6} = \frac{69}{6} = 11.5.$$