We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

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

Here is what might help:


 Feb 11, 2019

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

Yah, it helped me too.....     cheeky

ElectricPavlov  Feb 11, 2019

3 Online Users