2006 AMC 10A Problem 13

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 $\boxed{\textbf{(D)}\ \$60}$.

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.

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 $\boxed{\textbf{(D)}\ \$60}$.