College Algebra, Chapter 10, 10.5, Section 10.5, Problem 20

A bag contains two silver dollars and six slugs. A game consists of reaching into the bag and drawing a coin, which you get to keep. Determine the "fair price" of playing this game, that is, the price at which the player can be expected to break even if he or she plays the game many times (in other words, the price at which the player's expectation is zero.)

There are total of eight outcomes in this game. If we let $x$ be the number of silver dollars to break even, then the expected value is represented by

$P(2) + P(1) + P(0) - x = 0$

where

$P(2)$ is the probability that the two coins are picked

$P(1)$ is the probability that only one coin is picked.

$P(0)$ is the probability that only slugs are picked.


$
\begin{equation}
\begin{aligned}

P(2) =& \frac{2}{8} \times \frac{1}{7} = \frac{2}{56} = \frac{1}{28}
\\
\\
P(1) =& \frac{2}{8} \times \frac{6}{7} + \frac{6}{8} \times \frac{2}{7} = \frac{12}{56} + \frac{12}{56} = \frac{3}{7}
\\
\\
P(0) =& \frac{6}{8} \times \frac{5}{7} = \frac{30}{56} = \frac{15}{28}

\end{aligned}
\end{equation}
$



Thus, we have


$
\begin{equation}
\begin{aligned}

\frac{1}{28} + \frac{3}{7} + \frac{15}{28} - x =& 0
\\
x =& 1

\end{aligned}
\end{equation}
$



It shows that one would need to pay $\$ 1.00$ to break even, on the average.