College Algebra, Chapter 10, Review Exercises, Section Review Exercises, Problem 42

Two dice are rolled. John gets $\$ 5$ if they show the same number, or he pays $\$ 1$ if they show different numbers. What is the expected value of this game?

There are six out of 36 possible outcomes to show the numbers $(1,1), (2,2), (3,3), (4,4), (5,5)$ and $(6,6)$. The probability that two dice show the same number is $\displaystyle \frac{6}{36} = \frac{1}{6}$. Thus, the probability that the two dice show two different numbers is $\displaystyle 1 - \frac{1}{6} = \frac{5}{6}$. Thus, the expected value of this game is

$\displaystyle 5 \left( \frac{1}{6} \right) - 1 \left( \frac{5}{6} \right) = 0$

In other words, if you play this game many times, you would accept to earn nothing on every dollar you bet on average.