Using a standard pair of dice. What is the probability of roling a $12$ four tosses in a row? This is throwing me off a bit I believe mainly because the way the question is worded? Would this simply be $4$ out of $36$?
 A: 
This is throwing me off a bit i believe mainly because the way the question is worded? 

One toss of a pair of die results in a sum of twelve with probability of $1/36$.
What is the probability for obtaining this result for each of four tosses of these die pairs?
A: You have two dice. 


*

*Let A : the sum of two rolls equals 12 (so both dice have to be 6) 


P(A) = 1/36


*Let B : Event A occurs 4 times in a row.


P(B) = (1/36)^4
A: You have one event that you want it to happen four times in a row. What that event is? Rolling a $12$. Let's solve the problem like this:
Let's denote the result of a single tossing by a set whose members are denoted by $(x,y)$ where $x$ shows the number that faces up on the first die and $y$ denotes the number that faces up on the second die. Then all possible outcomes are
$$\text{ Possible outcomes =}\{(1,1), (1,2),\cdots,(2,1),(2,2),\cdots,(6,5),(6,6)\}$$
So, each time we role a pair of dice, there are $6\times 6 = 36$ possibilities. We are interested in only one particular outcome out of all these possibilities, namely $\{(6,6)\}$. So, the probability of our desired outcome in one roll is $\frac{1}{36}$.
Now we want this event to happen four times in a row. Since we can assume that each time that we roll the pair of dice, the new outcome is not affected by the previous time, this is simply the product of the probability for each single event which is
$$\text{Probability of the same event four times in a row}=\frac{1}{36}\times\frac{1}{36}\times\frac{1}{36}\times\frac{1}{36}=\big(\frac{1}{36}\big)^4$$
