2006 AMC 10A Problem 13

P11-15~2 minPrint

Expected Value Basic Probability Set up the equation/formula and compute; no special trick needed

A player pays $\$5$ 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.)

Show solution without answering

Key insight

Winning needs an even first roll (chance 1/2), then a match (chance 1/6). So the chance is 1/12, and the fair prize is 12 × 5 = 60 dollars.

Solution

First find the chance of winning, then use the rule for a fair game.

1. First roll. It must be even: $2$ , $4$ or $6$ . That is $3$ faces out of $6$ , a chance of $\frac12$ .
2. Second roll. It must match the first number. Whatever that number was, exactly $1$ face out of $6$ matches, a chance of $\frac16$ .
3. Both happen. Half of the games get past the first roll, and one sixth of those win:
$\frac12 \times \frac16 = \frac{1}{12}.$
4. Fair prize. The problem says (chance of winning) $\times$ (prize) must equal the $5$ dollars paid. So
$\frac{1}{12} \times \text{prize} = 5, \qquad \text{prize} = 60.$

A way to picture it: in $12$ games the player pays $12 \times 5 = 60$ dollars and wins about once, so that one win should pay $60$ .

The answer is $\boxed{\textbf{(D)}\ \$60}$ .

Why this works

When winning needs several things to happen in a row, multiply their chances, each one worked out assuming the earlier steps already happened. A fair game is one where, on average, you get back what you pay, so the prize is the cost divided by the chance of winning.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

To win, two things must happen in order. First, the initial roll must be even: probability $\tfrac{3}{6} = \tfrac12$ . Second, the next roll must show that same number: whatever the first number was, there is $1$ matching face out of $6$ , probability $\tfrac16$ .

Because the second roll is independent of the first,
$P(\text{win}) = \frac12 \cdot \frac16 = \frac{1}{12}.$

The game is fair when (probability of winning) $\times$ (prize) equals the $\$5$ entry fee:
$\frac{1}{12} \cdot W = 5 \quad\Longrightarrow\quad W = 60.$

The answer is $\boxed{\textbf{(D)}\ \$60}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 11: Expected Value
Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Intermediate Counting & Probability, Chapter 6: Constructive Expectation
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques