AMC 10 · 2007 · #22

Grade 7 arithmetic

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

expected-valueprobability-basiccasework systematic-enumeration ↑ Prerequisites: probability-basic

📏 Medium solution 💡 2 insights

Problem

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?

Pick an answer.

(A)

$-\frac{1}{8}$

(B)

$-\frac{1}{16}$

(C)

(D)

$\frac{1}{16}$

(E)

$\frac{1}{8}$

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: You pick one number from $1$ to $4$. Two fair four-sided dice (faces $1$–$4$) are rolled, and each die's result is the number on its bottom face. You win $\$1$ if your number is on the bottom of exactly one die, win $\$2$ if it is on the bottom of both, and lose $\$1$ if it is on neither. Find the expected return for one roll of the two dice.

Givens: You choose a fixed number from $\{1,2,3,4\}$ before rolling; Two fair tetrahedral dice are rolled; each bottom face is equally likely to be $1,2,3,$ or $4$; Payoffs: $+\$1$ if your number is on exactly one bottom, $+\$2$ if on both bottoms, $-\$1$ if on neither; Answer choices: (A) $-\frac{1}{8}$, (B) $-\frac{1}{16}$, (C) $0$, (D) $\frac{1}{16}$, (E) $\frac{1}{8}$

Unknowns: The expected (average) dollar return for one roll of the two dice

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #7 Identify Subproblems, #8 Analyze the Units

Tool #2 (Make a Systematic List) — each die's bottom is one of $4$ numbers, so the two dice make $4\times4=16$ equally likely bottom-pairs. Listing them lets us just count, instead of juggling probability formulas. Tool #7 (Identify Subproblems) — the three payoffs correspond to three separate cases (your number on exactly one bottom, on both, on neither); count each case's outcomes on its own. Tool #8 (Analyze the Units) — the answer is in dollars, and 'expected return' means the average payoff: multiply each case's dollar value by its chance and add, so the units stay dollars throughout.

Execute — Answer: B

#2 Make a Systematic List 7.SP.C.7 Step 1

Set up 16 equal outcomes

Each die's bottom face is equally likely to be $1,2,3,$ or $4$, and the dice are independent.
So an outcome is an ordered pair (first die's bottom, second die's bottom), and there are $4\times4=16$ equally likely pairs.
Fix your chosen number — call it your number — since by symmetry the answer is the same for any choice.
Now every question becomes: in how many of the $16$ pairs does your number appear?

$$4\times4=16\text{ equally likely outcomes}$$

💡 Two independent 4-way choices make a clean grid of $16$ equally likely outcomes to count over.

#7 Identify Subproblems 7.SP.C.8 Step 2

Count 'exactly one' outcomes

Your number is on exactly one bottom in two disjoint ways: it is on the first die but not the second, or on the second but not the first.
Each die shows your number in $1$ way and shows something else in $3$ ways.
So 'first yes, second no' is $1\times3=3$ pairs, and 'first no, second yes' is $3\times1=3$ pairs, giving $3+3=6$ pairs.
That is a $\frac{6}{16}$ chance, worth $+\$1$.

$$1\cdot3+3\cdot1=6\;\Rightarrow\;P=\tfrac{6}{16}$$

💡 Exactly one hit means one die matches and the other misses, and there are two ways to assign which die does which.

#7 Identify Subproblems 7.SP.C.8 Step 3

Count 'both' and 'neither'

Both bottoms show your number in only $1\times1=1$ way, a $\frac{1}{16}$ chance worth $+\$2$. Neither bottom shows your number when each die shows one of the other $3$ numbers, which is $3\times3=9$ ways, a $\frac{9}{16}$ chance worth $-\$1$.
Check the cases cover everything: $6+1+9=16$, all outcomes accounted for.

$$1\cdot1=1,\quad 3\cdot3=9,\quad 6+1+9=16$$

💡 Both-hit is the single all-match pair; no-hit is both dice avoiding your number, and the three counts must fill all $16$.

#8 Analyze the Units 7.NS.A.3 Step 4

Average the dollar payoff

Expected return is each payoff times its chance, all added up.
Put every case over $16$: $\frac{6}{16}$ pays $+1$, $\frac{1}{16}$ pays $+2$, and $\frac{9}{16}$ pays $-1$.
So the expected return is $\frac{6(1)+1(2)+9(-1)}{16}=\frac{6+2-9}{16}=\frac{-1}{16}$.
The player expects to lose $\frac{1}{16}$ of a dollar per roll, which is choice (B).

$$\dfrac{6(1)+1(2)+9(-1)}{16}=\dfrac{-1}{16}\;\Rightarrow\;\textbf{(B)}$$

💡 Weighting each dollar amount by how often it happens gives the true long-run average per roll.

[1] #2 7.SP.C.7 Each die's bottom face is equally likely to be $1,2,3,$ or $4$, and the dice are

[2] #7 7.SP.C.8 Your number is on exactly one bottom in two disjoint ways: it is on the first di

[3] #7 7.SP.C.8 Both bottoms show your number in only $1\times1=1$ way, a $\frac{1}{16}$ chance

[4] #8 7.NS.A.3 Expected return is each payoff times its chance, all added up. Put every case ov

Review

Reasonableness: The result $-\frac{1}{16}$ is a small negative number, which fits: the big $\frac{9}{16}$ chance of losing $\$1$ nearly cancels the combined $\frac{7}{16}$ chance of winning, so the game is only slightly unfair to the player. The numerator $6+2-9=-1$ is a tiny negative, ruling out $0$ (C) and the positive choices (D),(E); and it is not as negative as $-\frac{2}{16}=-\frac{1}{8}$ (A), so (B) is the only fit. As a sanity check, the total probability $\frac{6+1+9}{16}=1$, so no outcome was missed or double-counted.

Alternative: Use expectation per die (Tool #7). For a single die, the chosen number is on the bottom with probability $\frac{1}{4}$. Let $X$ be the number of dice showing your number; then $X$ has expected value $2\cdot\frac{1}{4}=\frac{1}{2}$. The payoff is $+\$1$ for each matching die except that going from one match to two adds an extra dollar bonus, and $-\$1$ for zero matches — so instead track payoff directly: $E=\sum (\text{payoff})\cdot P$. Computing $P(X=0)=\frac{9}{16}, P(X=1)=\frac{6}{16}, P(X=2)=\frac{1}{16}$ from the binomial with $n=2, p=\frac14$ gives the same $-\frac{1}{16}$.

CCSS standards used (min grade 7)

7.SP.C.7 Develop a probability model and use it to find probabilities of events (Modeling the two independent dice as 16 equally likely bottom-pairs.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and counting (Counting the outcomes for exactly one match, both matches, and no match.)
7.NS.A.3 Solve real-world problems involving the four operations with rational numbers (Combining the payoffs (including the negative loss) weighted by their probabilities into the expected return.)

⭐ List the $16$ equally likely bottom-pairs, count how many give each prize, then multiply each dollar amount by its chance and add — the small negative total means the game slightly favors the house.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Set up 16 equal outcomes

Count 'exactly one' outcomes

Count 'both' and 'neither'

Average the dollar payoff

Review

CCSS standards used (min grade 7)

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