AMC 10 · 2006 · #13
Grade 7 probabilityA player pays \textdollar5 to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)
Pick an answer.
AMC 10 2006 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.
Toolkit + CCSS Solution
Understand
Restated: You pay $\textdollar 5$ 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: 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.
Givens: The price to play is $\textdollar 5$.; Roll 1: if the number is odd ($1,3,5$) the game is lost right away.; Roll 1 even ($2,4,6$): the die is rolled a second time.; You win only if roll 2 equals roll 1; any other second number loses.; Fair game rule: $P(\text{win}) \times (\text{prize}) = \text{price paid}$.; Answer choices: (A) $\textdollar 12$, (B) $\textdollar 30$, (C) $\textdollar 50$, (D) $\textdollar 60$, (E) $\textdollar 100$.
Unknowns: The prize amount (in dollars) that makes the game fair.
Understand
Restated: You pay $\textdollar 5$ 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: 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.
Givens: The price to play is $\textdollar 5$.; Roll 1: if the number is odd ($1,3,5$) the game is lost right away.; Roll 1 even ($2,4,6$): the die is rolled a second time.; You win only if roll 2 equals roll 1; any other second number loses.; Fair game rule: $P(\text{win}) \times (\text{prize}) = \text{price paid}$.; Answer choices: (A) $\textdollar 12$, (B) $\textdollar 30$, (C) $\textdollar 50$, (D) $\textdollar 60$, (E) $\textdollar 100$.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #11 Work Backwards, #4 Introduce a Variable, #8 Analyze the Units
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 $P(\text{win})$. 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 $P(\text{win}) \times \text{prize} = \text{price}$ as an equation to solve for the prize. Keeping the dollars-and-probability units straight (Tool #8) confirms which quantity is the unknown.
Execute — Answer: D
7.SP.C.7 Step 1 Chance of passing the first roll
- You only get to keep playing if the first roll is even.
- The even faces are $2, 4, 6$ — three of the six equally likely faces.
- So the probability of surviving the first roll is $\tfrac{3}{6} = \tfrac{1}{2}$.
💡 Half the faces of a die are even, so an even opening roll is a coin-flip's worth of likely.
7.SP.C.7 Step 2 Chance the second roll matches
- Suppose the first roll was even — say it landed on some particular number.
- Now roll again.
- To win, the second roll must show that exact same number.
- Only one of the six faces does that, so the chance of matching is $\tfrac{1}{6}$.
- It does not matter whether the first number was $2$, $4$, or $6$: matching a fixed target is always a $1$-in-$6$ chance.
💡 Once the target number is fixed, the die has exactly one winning face out of six.
7.SP.C.8 Step 3 Multiply the two stages
- Winning needs BOTH things to happen: an even first roll AND a matching second roll.
- The rolls are independent, so multiply the two stage-probabilities.
- That gives $\tfrac{1}{2} \times \tfrac{1}{6} = \tfrac{1}{12}$.
- So a player wins about one game in every twelve.
💡 For an 'and' of independent events, the combined chance is the product of the separate chances.
6.EE.B.7 Step 4 Set up the fair-game equation
- Let the prize be $W$ dollars.
- The fair-game rule says the chance of winning times the prize equals the price paid: $\tfrac{1}{12} \cdot W = 5$.
- In words, over many plays the average payout ($\tfrac{1}{12}$ of the prize each game) should exactly cover the $\textdollar 5$ entry fee.
💡 A fair game just means the expected payout per play equals what you pay to play.
5.NF.B.7 Step 5 Solve for the prize
- To undo the $\tfrac{1}{12}$, divide both sides by $\tfrac{1}{12}$, which is the same as multiplying by $12$: $W = 5 \div \tfrac{1}{12} = 5 \times 12 = 60$.
- So the fair prize is $\textdollar 60$, which is choice (D).
💡 If a win happens only $1$ time in $12$, the prize must be $12$ times the fee to break even.
7.SP.C.7 You only get to keep playing if the first roll is even. The even faces are $2, 4 7.SP.C.7 Suppose the first roll was even — say it landed on some particular number. Now r 7.SP.C.8 Winning needs BOTH things to happen: an even first roll AND a matching second ro 6.EE.B.7 Let the prize be $W$ dollars. The fair-game rule says the chance of winning time 5.NF.B.7 To undo the $\tfrac{1}{12}$, divide both sides by $\tfrac{1}{12}$, which is the Review
Reasonableness: Check the break-even directly. If the prize is $\textdollar 60$ and a player pays $\textdollar 5$ each game, then over $12$ typical games they spend $12 \times 5 = \textdollar 60$ and expect to win exactly once, collecting $\textdollar 60$ — money out equals money in, which is exactly what 'fair' means. The winning chance $\tfrac{1}{12}$ is also believable: it sits between $\tfrac{1}{6}$ (matching alone) and $\tfrac{1}{36}$ (rolling a specific pair from scratch), because the odd-first-roll losses cut the pure-match chance in half. A smaller prize like $\textdollar 12$ or $\textdollar 30$ would leave the game favoring the house, so $\textdollar 60$ is the sensible fair value.
Alternative: Count outcomes instead of multiplying fractions. A full play that reaches the second roll has $6 \times 6 = 36$ equally likely ordered pairs $(\text{first}, \text{second})$; but the player only advances on an even first roll, and among all $36$ first-and-second combinations the winning ones are the three matching even pairs $(2,2), (4,4), (6,6)$. That is $3$ winners out of $36$ total ordered pairs, i.e. $\tfrac{3}{36} = \tfrac{1}{12}$, the same winning probability. Then the fair prize is again $5 \times 12 = \textdollar 60$.
CCSS standards used (min grade 7)
7.SP.C.7Develop a uniform probability model and use it to find probabilities of events (Finding the single-roll probabilities $P(\text{even}) = \tfrac{3}{6} = \tfrac{1}{2}$ and $P(\text{match}) = \tfrac{1}{6}$ from equally likely die faces.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Combining the two independent stages by multiplication: $P(\text{win}) = \tfrac{1}{2} \times \tfrac{1}{6} = \tfrac{1}{12}$.)6.EE.B.7Solve real-world and mathematical problems by writing and solving equations of the form px = q (Writing the fair-game condition as the equation $\tfrac{1}{12}\,W = 5$ with the prize $W$ as the unknown.)5.NF.B.7Divide whole numbers by unit fractions and unit fractions by whole numbers (Solving for the prize by computing $5 \div \tfrac{1}{12} = 5 \times 12 = 60$.)
⭐ To win you need an even first roll (chance $\tfrac{1}{2}$) and then a matching second roll (chance $\tfrac{1}{6}$), so winning happens $\tfrac{1}{12}$ of the time — meaning a fair prize must be $12$ times the $\textdollar 5$ fee, or $\textdollar 60$.
⭐ To win you need an even first roll (chance $\tfrac{1}{2}$) and then a matching second roll (chance $\tfrac{1}{6}$), so winning happens $\tfrac{1}{12}$ of the time — meaning a fair prize must be $12$ times the $\textdollar 5$ fee, or $\textdollar 60$.
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