AMC 10 · 2006 · #13

Grade 7 probability
Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps
probability-basicexpected-valueconditional-probability identify-subproblems ↑ Prerequisites: probability-basic
📏 Medium solution 💡 3 insights
Problem
You pay $5 to play. First you roll a fair die once. If it shows an odd number you lose immediately. If it shows an even number, you roll the die a second time, and you win only if the second roll shows the same number as the first; otherwise you lose. A game is fair when the chance of winning times the prize equals the price paid. Find the prize that makes this game fair.

Pick an answer.

(A)
$\textdollar12$
(B)
$\textdollar30$
(C)
$\textdollar50$
(D)
$\textdollar60$
(E)
$\textdollar 100$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

Winning is a two-stage event, so the cleanest route is to break it into subproblems (Tool #7): first the chance the opening roll is even, then the chance the second roll matches. Each stage is a plain equally-likely die probability; multiplying the two independent stage-chances gives P(win). Once we have that number, the 'fair game' definition is really a target we work backwards from (Tool #11): name the prize with a variable (Tool #4) and read the fair-game equation P(win) × prize = price as an equation to solve for the prize. Keeping the dollars-and-probability units straight (Tool #8) confirms which quantity is the unknown.

1STEP 1

Chance of passing the first roll

You keep playing only if the first roll is even, and 2, 4, 6 are three of the six equally likely faces — chance 1/2.

P(first roll even) = 3/6 = 1/2
2STEP 2

Chance the second roll matches

With the first number now fixed, only one of the six faces repeats it, so the chance the second roll matches is 1/6.

P(second roll matches the first) = 1/6
3STEP 3

Multiply the two stages

Winning needs both stages and the two rolls are independent, so multiply them: 1/2 × 1/6 = 1/12, about one win in twelve games.

P(win) = 1/2 × 1/6 = 1/12
4STEP 4

Set up the fair-game equation

Let the prize be W dollars; a fair game means the average payout equals the fee, so 1/12 · W = 5.

1/12 W = 5
5STEP 5

Solve for the prize

Dividing by 1/12 is the same as multiplying by 12, so W = 5 × 12 = 60 dollars, choice (D).

W = 5 ÷ 1/12 = 5 × 12 = 60 → (D) $ 60
Answer
$60
Check the break-even directly. If the prize is $60 and a player pays $5 each game, then over 12 typical games they spend 12 × 5 = $60 and expect to win exactly once, collecting $60 — money out equals money in, which is exactly what 'fair' means. The winning chance 1/12 is also believable: it sits between 1/6 (matching alone) and 1/36 (rolling a specific pair from scratch), because the odd-first-roll losses cut the pure-match chance in half. A smaller prize like $12 or $30 would leave the game favoring the house, so $60 is the sensible fair value.
💡Key takeaway

To win you need an even first roll (chance 1/2) and then a matching second roll (chance 1/6), so winning happens 1/12 of the time — meaning a fair prize must be 12 times the $5 fee, or $60.

  • Chance of passing the first roll
  • Chance the second roll matches
  • Multiply the two stages
  • Set up the fair-game equation
  • Solve for the prize

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