Amy and Brad take turns rolling a fair six-sided die; first to get a "6" win. What are Amy's odds of winning if she goes first?

To solve this problem, we need to understand that the event of Amy winning the game is a geometric distribution problem.

Step 1: Identify the probability of Amy rolling a 6 on her first turn. This is 1/6 because a die has 6 faces.

Step 2: If Amy doesn't roll a 6 on her first turn, then Brad gets a turn. The probability of Brad not rolling a 6 is also 5/6.

Step 3: If neither Amy nor Brad rolls a 6 on their first turn, the game continues with Amy's second turn. The probability of Amy rolling a 6 on her second turn is the product of three probabilities: the probability of Amy not getting a 6 on her first turn (5/6), the probability of Brad not getting a 6 on his first turn (5/6), and the probability of Amy getting a 6 on her second turn (1/6). So, the probability of Amy winning on her second turn is (5/6) * (5/6) * (1/6).

Step 4: We can continue this process for Amy's third turn, fourth turn, and so on. The probability of Amy winning on her third turn is (5/6)^4 * (1/6), and the probability of Amy winning on her fourth turn is (5/6)^6 * (1/6), and so on.

Step 5: The total probability of Amy winning the game is the sum of the probabilities of her winning on each turn. This forms a geometric series with first term (1/6) and common ratio (5/6)^2.

Step 6: The sum of an infinite geometric series is a / (1 - r), where a is the first term and r is the common ratio. So, the total probability of Amy winning the game is (1/6) / (1 - (5/6)^2) = 6/11.

So, Amy's odds of winning if she goes first are 6 to 5.