Originally Posted by

**Soroban** Hello, solace!

Rolling a **pair** of dice, there are 36 possible outcomes.

The totals and their respective probabilities are:

. . $\displaystyle \begin{array}{cc}2 & \frac{1}{36} \\ 3 & \frac{2}{36} \\ 4 & \frac{3}{36} \\ 5 & \frac{4}{36} \\ 6 & \frac{5}{36} \\ 7 & \frac{6}{36}\end{array}$

. .$\displaystyle \begin{array}{cc}8 & \frac{5}{36} \\ 9 & \frac{4}{36} \\ 10 & \frac{3}{36} \\ 11 & \frac{2}{36} \\ 12 & \frac{1}{36}$$\displaystyle

\end{array}$

$\displaystyle P(\text{5 or lower: win \$3}) \;=\;$$\displaystyle

\frac{1}{36} + \frac{2}{36} + \frac{3}{36} + \frac{4}{36} \;=\;\frac{10}{36}$

$\displaystyle P(\text{11 or 12: win \$10}) \;=\;\frac{2}{36} + \frac{1}{36}\;=\;\frac{3}{36}$

$\displaystyle P(\text{lose \$2}) \;=\;\frac{5}{36} + \frac{6}{36} + \frac{5}{36} + \frac{4}{36} + \frac{3}{36} \;=\;\frac{23}{36}$

Hence: .$\displaystyle E \;=\;\frac{10}{36}(3) + \frac{3}{36}(10) + \frac{23}{36}(-2) \;=\;\frac{14}{36} \;=\;0.30555... $

Therefore, you can expect to win an average of about 30½¢ per game.