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# Help me find my mistake

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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?

Feb 11, 2019

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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

Feb 11, 2019
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I understand now, and plus I discovered that its easier to just turn the question around and solve.

Feb 11, 2019
edited by CalculatorUser  Feb 11, 2019
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Yah, it helped me too.....

ElectricPavlov  Feb 11, 2019