2014
Problem - 103
Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, find the probability that the third roll will also be a six.
This is a conditional probability. First it is easy to compute that there is $\frac{1}{36}$ chance to have two consecutive rolls of six using the fair dice, and $\frac{4}{9}=\frac{16}{36}$ chance to achieve the same result using the biased dice. Given Charles has two consecutive six, there are $\frac{1}{17}$ chance he chooses the fair dice and $\frac{16}{17}$ chance he chooses the biased dice. Therefore, the expectation for him to get a third six is $$\frac{1}{17}\times \frac{1}{6}+\frac{16}{17}\times\frac{2}{3}=\boxed{\frac{65}{102}}$$