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
You pick one number from 1 to 4. Then two fair four-sided dice with faces 1 to 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, in dollars, for one roll of the two dice.

Pick an answer.

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

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

How to solve
Strategy Make a Systematic List

Tool #2 (Make a Systematic List) — each die's bottom is one of 4 numbers, so the two dice make 4×4=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.

1STEP 1

Set up 16 equal outcomes

Each bottom is equally likely 1,2,3,4 and the dice are independent, so ordered bottom-pairs give 4×4=16 equally likely outcomes.

4×4=16 equally likely outcomes
2STEP 2

Count 'exactly one' outcomes

Exactly one bottom matches in 1×3=3 ways plus 3×1=3 ways, so 6 of the 16 pairs pay +$1.

1·3+3·1=6 → P=6/16
3STEP 3

Count 'both' and 'neither'

Both bottoms match in only 1×1=1 way (+$2); neither matches in 3×3=9 ways (-$1); and 6+1+9=16 covers every case.

1·1=1, 3·3=9, 6+1+9=16
4STEP 4

Average the dollar payoff

Weight each payoff by its count: (6(1)+1(2)+9(-1))/16=(6+2-9)/16=-1/16 dollars, choice (B).

(6(1)+1(2)+9(-1))/16=-1/16 → (B)
Answer
-1/16
The result -1/16 is a small negative number, which fits: the big 9/16 chance of losing $1 nearly cancels the combined 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 -2/16=-1/8 (A), so (B) is the only fit. As a sanity check, the total probability (6+1+9)/16=1, so no outcome was missed or double-counted.
💡Key takeaway

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.

  • Set up 16 equal outcomes
  • Count 'exactly one' outcomes
  • Count 'both' and 'neither'
  • Average the dollar payoff

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