Some simple probability 
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*Brandon has 3 chances to roll a single die. He rolls 2 '6's, and a ‘5’ on his turn. Assuming the dice are fair, what is the probability of (i) ‘6’s on the first two rolls, and a non-six in the third roll, and (ii) getting exactly two ‘6’s?


On first glance, I calculated my answer to be i) 1/6 ^3 and ii) 1/6 ^2. Is this right? I think that each event (roll) is independent.


*In a game of Monopoly, 2 (fair) six-sided dice are rolled. Every face of each die is labelled 1 to 6.
A double occurs if both dice land up on the same number.
Instead of rolling both dice together, Brandon rolls one die at a time. The first die lands on a ‘6’. What is the probability that he rolls a double?
If the first die lands on either ‘1’, ‘2’, ‘3’, ‘4’, ‘5’ or ‘6’, what is the probability that he rolls a double?


My answers to both the questions are 1/6, because I think that the first event where he rolls a specific number, would make up a P of 1. Thus we only need to calculate the second roll. Am I thinking in the right direction?
 A: You are correct that die rolls are assumed to describe events that are independent of each other.  Recall that for independent events, the probability of event $E_1$ and event $E_2$ happening is the product of their probabilities.  Also recall that for disjoint events, the probability of event $E_1$ or event $E_2$ happening is the sum of their probabilities.
For number 1, if we assume that the fair die is 6-sided and if we assume that die rolls are unknown (that is, if we ignore the statement that he has rolled a 6, 6, and a 5), then getting a 6 on the first roll (a $\frac16$ probability), and a 6 on the second roll (a $\frac16$ probability), and a non-6 on the third roll (a $\frac56$ probability) has a probability of $\frac16\cdot\frac16\cdot\frac56=\frac{5}{216}$.
Continuing number 1, getting exactly two 6's means either getting a 6, a 6, then a non-6, or getting a 6, a non-6, then a 6, or getting a non-6, a 6, then a 6.  The probability of this is $\frac16\cdot\frac16\cdot\frac56+\frac16\cdot\frac56\cdot\frac16+\frac56\cdot\frac16\cdot\frac16=\frac{5}{216}+\frac{5}{216}+\frac{5}{216}=\frac{15}{216}=\frac{5}{72}$.
For number 2, your answers are correct.  In the first instance, given that the first die roll is a 6, there is a $\frac16$ probability of getting a 6 in the second roll, that is, rolling a double.  In the second instance, there are $36$ possible outcomes of rolling two dice.  Of these, $6$ are doubles.  The probability is thus $\frac{6}{36}=\frac16$.
