Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob, and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is, independent of the outcome of any other toss.

This is how far I got:

A - B - C

Alice has a \(\frac{5}{6}\) chance of NOT rolling a \(6\) , and so does Bob. So if you go in the order of the game, there is a \(\frac{25}{36}\)chance that Alice and Bob won't roll a \(6\) during their turns. Since the next turn is Carol, the probability of her rolling a \(6\) is \(\frac{1}{6}\) . Therefore, the probability of Carol being the first one to toss a \(6\) is \(\frac{1}{6} * \frac{25}{36} = \boxed{\frac{25}{216}}\).

This is incorrect because the correct answer on the Art of Problem Solving website is \(\boxed{\frac{25}{91}}\)

**Can someone please explain why I am incorrect?**

CalculatorUser Feb 11, 2019

#1**+1 **

Here is what might help:

https://math.stackexchange.com/questions/1327201/probability-that-on-three-rolls-of-dice-there-will-be-at-least-one-6-showing-up

ElectricPavlov Feb 11, 2019

#2**+1 **

I understand now, and plus I discovered that its easier to just turn the question around and solve.

CalculatorUser Feb 11, 2019