#### Explanation:

expectation is defined as ${\sum}_{i}^{n} P \left(x\right) x$. In this case we know that the possible outcomes for a round is

• x_1 =3 heads, $15 • x_2 = 3 tails,$10
• x_3 =anything else, 0$so $P \left(x = {x}_{1}\right) = \frac{1}{8}$because there is only 1 way to accomplish this. $P \left(x = {x}_{2}\right) = \frac{1}{8}$because this is the only way for this one too. this means that $P \left(x = {x}_{3}\right) = \frac{6}{8}$for two games we could have $P \left(x = {x}_{1} , x = {x}_{1}\right) = \frac{1}{64} , 30$$P \left(x = {x}_{1} , x = {x}_{2}\right) = \frac{1}{64} , 25$$P \left(x = {x}_{1} , x = {x}_{3}\right) = \frac{6}{64} , 15$$P \left(x = {x}_{2} , x = {x}_{1}\right) = \frac{1}{64} , 25$$P \left(x = {x}_{2} , x = {x}_{2}\right) = \frac{1}{64} , 20$$P \left(x = {x}_{2} , x = {x}_{3}\right) = \frac{6}{64} , 10$$P \left(x = {x}_{3} , x = {x}_{1}\right) = \frac{6}{64} , 15$$P \left(x = {x}_{3} , x = {x}_{2}\right) = \frac{6}{64} , 10$$P \left(x = {x}_{3} , x = {x}_{3}\right) = \frac{36}{64} , 0$Leading to 1/64*30 + 1/64*25 + 6/64*15 + 1/64* 25 + 1/64*20 + 6/64* 10 + 6/64* 25 + 6/64* 20 simplifying $\frac{1}{64} \left(30 + 25 + 25 + 20\right) + \frac{6}{64} \left(15 + 10 + 15 + 10\right)$$\frac{1}{64} \left(100\right) + \frac{6}{64} \left(50\right) = \frac{400}{64} = 6.25$Another way to think about this is knowing that the expectation for a two games can be stated as $E \left(g a m e 1 + g a m e 2\right) = E \left(g a m e 1\right) + E \left(g a m e 2\right)$thus $15 \cdot \frac{1}{8} + 10 \cdot \frac{1}{8} + 0 \cdot \frac{6}{8} = \frac{25}{8}$for one game. for two games it should be $\frac{50}{8} = 6.25$which it is. Now we can subtract the amount to play per game leaving $6.25 - 10 = 3.75$. Playing the two games should result in a loss of $3.75