2007 AMC 10B Problems/Problem 22

Problem 22

A player chooses one of the numbers $1$ through $4$ . After the choice has been made, two regular four-sided (tetrahedral) dice are rolled, with the sides of the dice numbered $1$ through $4.$ If the number chosen appears on the bottom of exactly one die after it has been rolled, then the player wins $1$ dollar. If the number chosen appears on the bottom of both of the dice, then the player wins $2$ dollars. If the number chosen does not appear on the bottom of either of the dice, the player loses $1$ dollar. What is the expected return to the player, in dollars, for one roll of the dice?

$\textbf{(A) } -\frac{1}{8} \qquad\textbf{(B) } -\frac{1}{16} \qquad\textbf{(C) } 0 \qquad\textbf{(D) } \frac{1}{16} \qquad\textbf{(E) } \frac{1}{8}$

Solution

There are $2 \cdot 3 \cdot 1 = 6$ ways for your number to show up once, $1 \cdot 1 = 1$ way for your number to show up twice, and $3 \cdot 3 = 9$ ways for your number to not show up at all. Think of this as a set of $16$ numbers with six $1s,$ one $2,$ and nine $-1s.$ The average of this set is the expected return to the player. $\frac{6(1)+1(2)-9(1)}{16} = \frac{6+2-9}{16} = \boxed{\mathrm{(B) \ } -\frac{1}{16}}$

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2007 AMC 10B Problems/Problem 22

Problem 22

Solution

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