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

Apr 30, 2018

$7$

#### Explanation:

Intuitively we would expect the sum of a single die to be the average of the possible outcomes, ie:

$S = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5$

And so we would predict the sum of a two die to be twice that of one die, ie we would predict the expected value to be $7$

If we consider the possible outcomes from the throw of two dice:

And so if we define $X$ as a random variable denoting the sum of the two dices, then we get the following distribution:

So then we compute the expected value, using

$E \left(X\right) = \sum x \left(P \left(X\right) = x\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 2 \cdot \frac{1}{36} + 3 \cdot \frac{2}{36} + 4 \cdot \frac{3}{36} + 5 \cdot \frac{4}{36}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus + 6 \cdot \frac{5}{36} + 7 \cdot \frac{6}{36} + 8 \cdot \frac{5}{36} + 9 \cdot \frac{4}{36}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus + 10 \cdot \frac{3}{36} + 11 \cdot \frac{2}{36} + 12 \cdot \frac{1}{36}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12}{36}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{252}{36}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 7$