Question

What is the expected value of the sum of two rolls of a six sided die?

Answer 1

Hint: To find the sum of two rolls of a six sided die, we have to list down the sample space and find the sum of each of the outcomes. Then, we have to find the probability of occurrence of each of the sums. Then, we have to use the formula $E\left(X\right)=\sum xP(X=x)$ to find the expectation.

Complete step-by-step answer:
We have to find the sum of two rolls of a six sided die. Let us consider the sample space of rolling two die.
$S=\left\{\begin{array}{rl}& (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ & (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ & (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ & (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ & (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ & (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}$
Now, we have to sum each of the outcomes, that is $(1,1)=(1+1)=2$ . The result of such addition will be
$S=\left\{\begin{array}{rl}& 2,3,4,5,6,7\\ & 3,4,5,6,7,8\\ & 4,5,6,7,8,9\\ & 5,6,7,8,9,10\\ & 6,7,8,9,10,11\\ & 7,8,9,10,11,12\end{array}\right\}$
We have to define X as a random variable denoting the sum of the two dices.
Now, we have to find the probability of each sum. We know that probability of an event is the ratio of the number of favourable outcomes to the total number of outcomes. Here, we can say that the total number of outcomes $=36$ . Now, we have to find the probability of the sum to be 2.
$P(X=2)={\displaystyle \frac{1}{36}}$
Similarly, we have to find other sums.
$\begin{array}{rl}& P(X=3)={\displaystyle \frac{2}{36}}\\ & P(X=4)={\displaystyle \frac{3}{36}}\\ & P(X=5)={\displaystyle \frac{4}{36}}\\ & P(X=6)={\displaystyle \frac{5}{36}}\\ & P(X=7)={\displaystyle \frac{6}{36}}\\ & P(X=8)={\displaystyle \frac{5}{36}}\\ & P(X=9)={\displaystyle \frac{4}{36}}\\ & P(X=10)={\displaystyle \frac{3}{36}}\\ & P(X=11)={\displaystyle \frac{2}{36}}\\ & P(X=12)={\displaystyle \frac{1}{36}}\end{array}$
Now, let us tabulate the distribution.

X	2	3	4	5	6	7	8	9	10	11	12
$P(X=x)$	${\displaystyle \frac{1}{36}}$	${\displaystyle \frac{2}{36}}$	${\displaystyle \frac{3}{36}}$	${\displaystyle \frac{4}{36}}$	${\displaystyle \frac{5}{36}}$	${\displaystyle \frac{6}{36}}$	${\displaystyle \frac{5}{36}}$	${\displaystyle \frac{4}{36}}$	${\displaystyle \frac{3}{36}}$	${\displaystyle \frac{2}{36}}$	${\displaystyle \frac{1}{36}}$

We know that expected value of a geometric random variable is given by
$E\left(X\right)=\sum xP(X=x)$
Therefore, the expected value of the sum of two rolls of a six sided die can be found as follows.
$\Rightarrow E\left(X\right)=2\times {\displaystyle \frac{1}{36}}+3\times {\displaystyle \frac{2}{36}}+4\times {\displaystyle \frac{3}{36}}+5\times {\displaystyle \frac{4}{36}}+6\times {\displaystyle \frac{5}{36}}+7\times {\displaystyle \frac{6}{36}}+8\times {\displaystyle \frac{5}{36}}+9\times {\displaystyle \frac{4}{36}}+10\times {\displaystyle \frac{3}{36}}+11\times {\displaystyle \frac{2}{36}}+12\times {\displaystyle \frac{1}{36}}$
Let us simplify the above result.
$\Rightarrow E\left(X\right)={\displaystyle \frac{2}{36}}+{\displaystyle \frac{6}{36}}+{\displaystyle \frac{12}{36}}+{\displaystyle \frac{20}{36}}+{\displaystyle \frac{30}{36}}+{\displaystyle \frac{42}{36}}+{\displaystyle \frac{40}{36}}+{\displaystyle \frac{36}{36}}+{\displaystyle \frac{30}{36}}+{\displaystyle \frac{22}{36}}+{\displaystyle \frac{12}{36}}$
Let us add the terms.
$\begin{array}{rl}& \Rightarrow E\left(X\right)={\displaystyle \frac{2+6+12+20+30+42+40+36+30+22+12}{36}}\\ & \Rightarrow E\left(X\right)={\displaystyle \frac{252}{36}}\\ & \Rightarrow E\left(X\right)=7\end{array}$
Hence, the expected value of the sum of two rolls of a six sided die is 7.

Note: Students must be very thorough with the formula of expected value. We can also call the expected value of mean. The expected value of the sum of two rolls of a six sided die will be twice that of a single dice. Let us see the expected value of a single die which is the sum of all the probabilities or the average of all the possible outcomes.. Let us denote this expectation as $E\left(Y\right)$ . There will be only 6 outcomes each of probability ${\displaystyle \frac{1}{6}}$ .
$$\begin{array}{rl}& \Rightarrow E\left(Y\right)=1\times {\displaystyle \frac{1}{6}}+2\times {\displaystyle \frac{1}{6}}+3\times {\displaystyle \frac{1}{6}}+4\times {\displaystyle \frac{1}{6}}+5\times {\displaystyle \frac{1}{6}}+6\times {\displaystyle \frac{1}{6}}\\ & \Rightarrow E\left(Y\right)={\displaystyle \frac{1+2+3+4+5+6}{6}}\\ & \Rightarrow E\left(Y\right)={\displaystyle \frac{21}{6}}=3.5\end{array}$$
We can see that $E\left(X\right)=2E\left(Y\right)$