Three fair coins are tossed. If all three coins show heads, then the player wins $15. If all three coins show tails, then the player wins $10. If it costs $5 to play the game, what is the player's expected net gain or loss at the end of two games?

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Three fair coins are tossed. If all three coins show heads, then the player wins $15. If all three coins show tails, then the player wins $10. If it costs $5 to play the game, what is the player's expected net gain or loss at the end of two games?

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Answer 1 · 2017-04-26T03:22:36

$- $ 3.75$ $-$3.75$

Explanation:

expectation is defined as $n \sum i P (x) x$ $sum_i^n P(x)x$ . 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 $P (x = x_{1}) = \frac{1}{8}$ $P(x=x_1)=1/8$ because there is only 1 way to accomplish this.
$P (x = x_{2}) = \frac{1}{8}$ $P(x=x_2)=1/8$ because this is the only way for this one too.
this means that $P (x = x_{3}) = \frac{6}{8}$ $P(x=x_3)=6/8$

for two games we could have
$P (x = x_{1}, x = x_{1}) = \frac{1}{64}, 30$ $P(x=x_1, x= x_1)=1/64, 30$
$P (x = x_{1}, x = x_{2}) = \frac{1}{64}, 25$ $P(x=x_1, x= x_2)=1/64, 25$
$P (x = x_{1}, x = x_{3}) = \frac{6}{64}, 15$ $P(x=x_1, x= x_3)=6/64, 15$
$P (x = x_{2}, x = x_{1}) = \frac{1}{64}, 25$ $P(x=x_2, x= x_1)=1/64, 25$
$P (x = x_{2}, x = x_{2}) = \frac{1}{64}, 20$ $P(x=x_2, x= x_2)=1/64, 20$
$P (x = x_{2}, x = x_{3}) = \frac{6}{64}, 10$ $P(x=x_2, x= x_3)=6/64, 10$
$P (x = x_{3}, x = x_{1}) = \frac{6}{64}, 15$ $P(x=x_3, x= x_1)=6/64, 15$
$P (x = x_{3}, x = x_{2}) = \frac{6}{64}, 10$ $P(x=x_3, x= x_2)=6/64, 10$
$P (x = x_{3}, x = x_{3}) = \frac{36}{64}, 0$ $P(x=x_3, x= x_3)=36/64, 0$

Leading to
$\frac{1}{64} \cdot 30 + \frac{1}{64} \cdot 25 + \frac{6}{64} \cdot 15 + \frac{1}{64} \cdot 25 + \frac{1}{64} \cdot 20 + \frac{6}{64} \cdot 10 + \frac{6}{64} \cdot 25 + \frac{6}{64} \cdot 20$ $1/64*30 + 1/64*25 + 6/64*15 + 1/64* 25 + 1/64*20 + 6/64* 10 + 6/64* 25 + 6/64* 20$

simplifying

$\frac{1}{64} (30 + 25 + 25 + 20) + \frac{6}{64} (15 + 10 + 15 + 10)$ $1/64(30 + 25 + 25 +20) + 6/64(15 + 10 + 15 + 10)$
$\frac{1}{64} (100) + \frac{6}{64} (50) = \frac{400}{64} = 6.25$ $1/64(100) + 6/64(50) = 400/64 = 6.25$

Another way to think about this is knowing that the expectation for a two games can be stated as $E (g a m e 1 + g a m e 2) = E (g a m e 1) + E (g a m e 2)$ $E(game1+game2) = E(game1) +E(game2)$ thus $15 \cdot \frac{1}{8} + 10 \cdot \frac{1}{8} + 0 \cdot \frac{6}{8} = \frac{25}{8}$ $15*1/8 + 10 *1/8 +0*6/8 = 25/8$ for one game.
for two games it should be $\frac{50}{8} = 6.25$ $50/8 = 6.25$ which it is. Now we can subtract the amount to play per game leaving

$6.25 - 10 = 3.75$ $6.25 - 10 = 3.75$ . Playing the two games should result in a loss of $$ 3.75$ $$3.75$

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Three fair coins are tossed. If all three coins show heads, then the player wins $15. If all three coins show tails, then the player wins $10. If it costs $5 to play the game, what is the player's expected net gain or loss at the end of two games?

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